Answer:
a) 0
b) 0.579 is the probability that chloride concentration is less than 105.
c) 0.32 is the probability that chloride concentration differs from the mean by more than 1 standard deviation.
Step-by-step explanation:
We are given the following information in the question:
Mean, μ = 104
Standard Deviation, σ = 5
We are given that the distribution of blood chloride concentration is a bell shaped distribution that is a normal distribution.
Formula:
[tex]z_{score} = \displaystyle\frac{x-\mu}{\sigma}[/tex]
a) Since normal distribution s a continuous distribution, the probability for a particular value is zero. Therefore,
[tex]P(x =105) = 0[/tex]
b) P(chloride concentration is less than 105)
[tex]P(x < 105) = P(z < \displaystyle\frac{105-104}{5}) = P(z < 0.2)[/tex]
Calculating the value from the standard normal table we have,
[tex]P(z<0.2) = 0.579 = 57.9\%\\P( x < 105) = 57.9\%[/tex]
c) P(chloride concentration differs from the mean by more than 1 standard deviation)
Since it is a normal distribution, the Empirical rule shows that 68% falls within the first standard deviation, 95% within the first two standard deviations, and 99.7% within the first three standard deviations.
= 1 - P(chloride concentration within the mean by 1 standard deviation)
= 1 - 0.68 = 0.32
Final answer:
The probability of a single value is 0 in a normal distribution. The probability that the chloride concentration is less than 105 mmol/L can be found using the z-score and CDF of the normal distribution. The probability of the concentration being more than one standard deviation from the mean involves calculating the z-scores for one standard deviation away from the mean and using standard normal distribution probabilities.
Explanation:
The question involves the application of probability theory and statistics, specifically the properties of the normal distribution. Here are the answers to the student's questions:
(a) In a normal distribution, the probability of a continuous random variable taking on any single value is technically zero because there are an infinite number of potential values it could take. Therefore, the probability that the chloride concentration equals exactly 105 mmol/L is 0.
(b) To find the probability that the chloride concentration is less than 105 mmol/L, we use the cumulative distribution function (CDF) for a normal distribution. The z-score is calculated by subtracting the mean from the observed value and dividing by the standard deviation: z = (105 - 104) / 5 = 0.20. Using standard normal distribution tables or a calculator, we find the probability corresponding to a z-score of 0.20.
(b) The probability that the chloride concentration differs from the mean by more than 1 standard deviation involves finding the probability of being outside the interval [99, 109], since one standard deviation from the mean is 104 ± 5. We calculate the z-scores for 99 and 109, which are -1 and +1, respectively. Using the normal distribution, we find the probabilities associated with these z-scores and then subtract them from 1 to get the probability of being more than one standard deviation away from the mean.
Solve the following problems : Given: S, T, and U are the midpoints of RP , PQ , and QR respectively. Prove: △SPT≅△UTQ.
Answer:
Hence Proved △ SPT ≅ △ UTQ
Step-by-step explanation:
Given: S, T, and U are the midpoints of Segment RP , segment PQ , and segment QR respectively of Δ PQR.
To prove: △ SPT ≅ △ UTQ
Proof:
∵ T is is the midpoint of PQ.
Hence PT = PQ ⇒equation 1
Now,Midpoint theorem is given below;
The Midpoint Theorem states that the segment joining two sides of a triangle at the midpoints of those sides is parallel to the third side and is half the length of the third side.
By, Midpoint theorem;
TS║QR
Also, [tex]TS = \frac{1}{2} QR[/tex]
Hence, TS = QU (U is the midpoint QR) ⇒ equation 2
Also by Midpoint theorem;
TU║PR
Also, [tex]TU = \frac{1}{2} PR[/tex]
Hence, TU = PS (S is the midpoint QR) ⇒ equation 3
Now in △SPT and △UTQ.
PT = PQ (from equation 1)
TS = QU (from equation 2)
PS = TU (from equation 3)
By S.S.S Congruence Property,
△ SPT ≅ △ UTQ ...... Hence Proved
To prove the congruency of the triangles, we use the Midpoint Theorem and the Side-Side-Side (SSS) criterion. According to the Midpoint Theorem, the line segments are parallel and half the length of the third side of the triangle. Since the triangles share a side and the other sides are proportional, the SSS criterion allows us to conclude that the triangles are congruent.
Explanation:To prove that the triangles △SPT and △UTQ are congruent, we need to first understand that S, T, and U are midpoints. According to the Midpoint Theorem, a line segment connecting the midpoints of two sides of a triangle is parallel to the third side and half its length. So, consider the segments ST and TU, which are connecting the midpoints of the sides of △PQR.
By the Midpoint Theorem, we know ST is parallel to UQ and also ST = 0.5*UQ. Similarly, TU is parallel to SP and TU = 0.5*SP.
Moreover, since T is the midpoint of PQ, we know PT = QT. Therefore, △SPT and △UTQ share a side (PT), and their other corresponding sides are proportional due to the Midpoint Theorem. Consequently, by Side-Side-Side (SSS) criterion, we can conclude that △SPT is congruent to △UTQ.
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Here are summary statistics for randomly selected weights of newborn girls: nequals=174174, x overbarxequals=30.930.9 hg, sequals=7.57.5 hg. Construct a confidence interval estimate of the mean. Use a 9595% confidence level. Are these results very different from the confidence interval 29.629.6 hgless than
Answer:
The 95% confidence interval would be given by (29.780;32.020)
Step-by-step explanation:
A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".
The margin of error is the range of values below and above the sample statistic in a confidence interval.
Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".
[tex]\bar X=30.9[/tex] represent the sample mean for the sample
[tex]\mu[/tex] population mean (variable of interest)
s=7.5 represent the sample standard deviation
n=174 represent the sample size
The confidence interval for the mean is given by the following formula:
[tex]\bar X \pm t_{\alpha/2}\frac{s}{\sqrt{n}}[/tex] (1)
In order to calculate the critical value [tex]t_{\alpha/2}[/tex] we need to find first the degrees of freedom, given by:
[tex]df=n-1=174-1=173[/tex]
Since the Confidence is 0.95 or 95%, the value of [tex]\alpha=0.05[/tex] and [tex]\alpha/2 =0.025[/tex], and we can use excel, a calculator or a tabel to find the critical value. The excel command would be: "=-T.INV(0.025,173)".And we see that [tex]t_{\alpha/2}=1.97[/tex], this value is similar to the obtained with the normal standard distribution since the sample size is large to approximate the t distribution with the normal distribution.
Now we have everything in order to replace into formula (1):
[tex]30.9-1.97\frac{7.5}{\sqrt{174}}=29.780[/tex]
[tex]30.9+1.97\frac{7.5}{\sqrt{174}}=32.020[/tex]
So on this case the 95% confidence interval would be given by (29.780;32.020)
The value 29.6 is not contained on the interval calculated.
Assume each person makes one cola purchase per week in Turlock. Suppose 60% of all people now drink Coke, and 40% drink Pepsi. How many people (roughly) will be drinking Coke five weeks from now in Turlock if the current population of Turlock is 70,000?
Answer:
In think the answer is 42,000
Step-by-step explanation:
Final answer:
To determine how many people will be drinking Coke in Turlock five weeks from now, multiply the percentage of Coke drinkers (60%) by the town's population (70,000). This gives us 0.60 * 70,000 = 42,000 people drinking Coke.
Explanation:
The question involves a numerical problem where we have to determine how many people will be drinking Coke in Turlock five weeks from now if the current population is 70,000 and 60% of them drink Coke. To find the answer, we simply need to calculate 60% of 70,000.
Calculation steps:
Convert 60% to a decimal by dividing 60 by 100, which gives us 0.60
Multiply 0.60 by the total population of Turlock, which is 70,000
The calculation will be 0.60 * 70,000
Now, let's do the calculation:
0.60 * 70,000 = 42,000
Therefore, if the population of Turlock stays constant at 70,000 and the cola drinking habits remain the same, approximately 42,000 people will be drinking Coke five weeks from now.
Use pigeonhole principle to prove the following (need to identify pigeons/objects and pigeonholes/boxes): a. How many cards must be drawn from a standard 52-card deck to guarantee 2 cards of the same suit?Note that there are 4 suits b. Prove that if four numbers are chosen from the set {1, 2, 3, 4,5,6), at least one pair must add up to 7.
Answer:
Step-by-step explanation:
a. The pidgeons/objects here are the cards and the pigeonholes/boxes here are the suits. We are trying to draw cards/pidgeon that have the same suits/pidgeonholes. Since there are 4 different suits, we need to draw at least 5 cards to guarantee 2 cards of the same suit.
b. Let there be 3 holes, which represent 3 pairs of number with sum equals to 7. They are:
- {1,6} as 1 + 6 = 7
- {2,5} as 2 + 5 = 7
- {3,4} as 3 + 4 = 7
You can see that if we pick randomly 4 numbers, we will always end up with 2 numbers in the same hole, which means there will be at least a pair and add up to 7.
If f(x)=3x^2-5, then what is f(2)
Answer:
7
Step-by-step explanation:
substitute 2 for x
3(2^2) - 5
3 * 4 - 5
12 - 5 = 7
Answer:
Step-by-step explanation:
7....
Sample spaces For each of the following, list the sample space and tell whether you think the events are equally likely:
(a) Toss 2 coins; record the order of heads and tails.
(b) A family has 3 children; record the number of boys.
(c) Flip a coin until you get a head or 3 consecutive tails; record each flip.
(d) Roll two dice; record the larger numbe
Answer and explanation:
To find : List the sample space and tell whether you think the events are equally likely ?
Solution :
a) Toss 2 coins; record the order of heads and tails.
Let H is getting head and t is getting tail.
When two coins are tossed the sample space is {HH,HT,TH,TT}.
Total number of outcome = 4
As the outcome HT is different from TH. Each outcome is unique.
Events are equally likely since their probabilities [tex]\frac{1}{4}[/tex] are same.
b) A family has 3 children; record the number of boys.
Let B denote boy and G denote girl.
If there are 3 children then the sample space is
{GGG,GGB,GBG,BGG,BBG,GBB,BGB,BBB}
The possible number of boys are 0,1,2 and 3.
Number of boys Favorable outcome Probability
0 GGG [tex]\frac{1}{8}[/tex]
1 GGB,GBG,BGG [tex]\frac{3}{8}[/tex]
2 GBB,BGB,BBG [tex]\frac{3}{8}[/tex]
3 BBB [tex]\frac{1}{8}[/tex]
Since the probabilities are not equal the events are not equally likely.
c) Flip a coin until you get a head or 3 consecutive tails; record each flip.
Getting a head in a trial is dependent on the previous toss.
Similarly getting 3 consecutive tails also dependent on previous toss.
Hence, the probabilities cannot be equal and events cannot be equally likely.
d) Roll two dice; record the larger number
The sample space of rolling two dice is
(1,1) (1,2) (1,3) (1,4) (1,5) (1,6)
(2,1) (2,2) (2,3) (2,4) (2,5) (2,6)
(3,1) (3,2) (3,3) (3,4) (3,5) (3,6)
(4,1) (4,2) (4,3) (4,4) (4,5) (4,6)
(5,1) (5,2) (5,3) (5,4) (5,5) (5,6)
(6,1) (6,2) (6,3) (6,4) (6,5) (6,6)
Now we form a table that the number of time each number occurs as maximum number then we find probability,
Highest number Number of times Probability
1 1 [tex]\frac{1}{36}[/tex]
2 3 [tex]\frac{3}{36}[/tex]
3 5 [tex]\frac{5}{36}[/tex]
4 7 [tex]\frac{7}{36}[/tex]
5 9 [tex]\frac{9}{36}[/tex]
6 11 [tex]\frac{11}{36}[/tex]
Since the probabilities are not the same the events are not equally likely.
The sample spaces for different scenarios include (a) {HH, HT, TH, TT}, (b){0, 1, 2, 3}, (c){H, TH, TTH, TTTH}, and (d){1, 2, 3, 4, 5, 6}.
Let's explore the sample spaces and determine if the events are equally likely for each scenario:
(a) Toss 2 Coins; Record the Order of Heads and Tails
The sample space is: {HH, HT, TH, TT}. Each of these outcomes has an equal probability of occurring, since each coin flip is independent and there are two coins, making each combination equally likely.
(b) A Family Has 3 Children; Record the Number of Boys
The sample space can be represented by the number of boys: {0, 1, 2, 3}. Assuming that each child is equally likely to be a boy or a girl, the probabilities of each outcome differ. For example, having 1 boy (and thus 2 girls) has more combinations than having 3 boys or no boys at all.
(c) Flip a Coin Until You Get a Head or 3 Consecutive Tails; Record Each Flip
The sample space includes sequences that stop as soon as we get a head or reach 3 tails: {H, TH, TTH, TTTH}. These outcomes are not equally likely because the sequence lengths vary, affecting their probabilities.
(d) Roll Two Dice; Record the Larger Number
The sample space is: {1, 2, 3, 4, 5, 6}. However, the events are not equally likely. For example, getting a 6 as the larger number is more probable since it can occur in more pairs (e.g., (6,5), (6,4), etc.) compared to lower numbers.
Plot the point (3, 2pi/3), given in polar coordinates, and find other polar coordinates (r, θ )of the point for which the following are true.
(a) r>0, -2 pi < 0
(b) r < 0, 0 < 2 pi
(c) r > 0, 2 pi < 4 pi
Answer:
(a) [tex](3, -\frac{4\pi}{3})[/tex]
(b) [tex](-3, \frac{5\pi}{3})[/tex]
(c)[tex](3, \frac{8\pi}{3})[/tex]
Step-by-step explanation:
All polar coordinates of point (r, θ ) are
[tex](r,\theta+2n\pi)[/tex] and [tex](-r,\theta+(2n+1)\pi)[/tex]
where, θ is in radian and n is an integer.
The given point is [tex](3, \frac{2\pi}{3})[/tex]. So, all polar coordinates of point are
[tex](3, \frac{2\pi}{3}+2n\pi)[/tex] and [tex](-3, \frac{2\pi}{3}+(2n+1)\pi)[/tex]
(a) [tex]r>0,-2\pi\leq \theta <0[/tex]
Substitute n=-1 in [tex](3, \frac{2\pi}{3}+2n\pi)[/tex], to find the point for which [tex]r>0,-2\pi\leq \theta <0[/tex].
[tex](3, \frac{2\pi}{3}+2(-1)\pi)[/tex]
[tex](3, -\frac{4\pi}{3})[/tex]
Therefore, the required point is [tex](3, -\frac{4\pi}{3})[/tex].
(b) [tex]r<0,0\leq \theta <2\pi[/tex]
Substitute n=0 in [tex](-3, \frac{2\pi}{3}+(2n+1)\pi)[/tex], to find the point for which [tex]r>0,-2\pi\leq \theta <0[/tex].
[tex](-3, \frac{2\pi}{3}+(2(0)+1)\pi)[/tex]
[tex](-3, \frac{2\pi}{3}+\pi)[/tex]
[tex](-3, \frac{5\pi}{3})[/tex]
Therefore, the required point is [tex](-3, \frac{5\pi}{3})[/tex].
(c) [tex]r>0,2\pi \leq \theta <4\pi[/tex]
Substitute n=1 in [tex](3, \frac{2\pi}{3}+2n\pi)[/tex], to find the point for which [tex]r>0,2\pi \leq \theta <4\pi[/tex].
[tex](3, \frac{2\pi}{3}+2(1)\pi)[/tex]
[tex](3, \frac{2\pi}{3}+2\pi)[/tex]
[tex](3, \frac{8\pi}{3})[/tex]
Therefore, the required point is [tex](3, \frac{8\pi}{3})[/tex].
A study was conducted in order to estimate ?, the mean number of weekly hours that U.S. adults use computers at home. Suppose a random sample of 81 U.S. adults gives a mean weekly computer usage time of 8.5 hours and that from prior studies, the population standard deviation is assumed to be ? = 3.6 hours.
Based on this information, what would be the point estimate for ??
(a) 81
(b) 8.5
(c) 3.6
(d) None of the above.
We are 95% confident that the mean number of weekly hours that U.S. adults use computers at home is:
(a) between 8.1 and 8.9.
(b) between 7.8 and 9.2.
(c) between 7.7 and 9.3.
(d) between 7.5 and 9.5.
(e) between 7.3 and 9.7.
Which of the following will provide a more informative (i.e., narrower) confidence interval than the one in problem 3?
(a) Using a sample of size 400 (instead of 81).
(b) Using a sample of size 36 (instead of 81).
(c) Using a different sample of size 81.
(d) Using a 90% level of confidence (instead of 95%).
(e) Using a 99% level of confidence (instead of 95%).
(f) Both (a) and (d) are correct.
(g) Both (a) and (e) are correct.
How large a sample of U.S. adults is needed in order to estimate ? with a 95% confidence interval of length 1.2 hours?
(a) 6
(b) 12
(c) 20
(d) 36
(e) 144
Answer:
a) 8.5
b) between 7.7 and 9.3.
c)Both using a sample of size 400 (instead of 81) and using a 90% level of confidence (instead of 95%) are correct.
(e) 144
Step-by-step explanation:
Explanation for a)
The point estimate for the population mean μ is the sample mean, x ¯ . In this case, to estimate the mean number of weekly hours of home-computer use among the population of U.S. adults, we used the sample mean obtained from the sample, therefore x ¯ = 8.5.
Explanation for b)
The 95% confidence interval for the mean, μ, is x ¯ ± 2 ⋅ σ n = 8.5 ± 2 ⋅ 3.6 81 = 8.5 ± . 8 = ( 7.7 , 9.3 ) .
Explanation for c)
In general, a more concise (narrower) confidence interval can be achieved in one of two ways: sacrificing on the level of confidence (i.e. selecting a lower level of confidence) or increasing the sample size
Explanation for d)
We would like our confidence interval to be a 95% confidence interval (implying that z* = 2) and the confidence interval length should be 1.2, therefore the margin of error (m) = 1.2 / 2 = .6. The sample size we need in order to obtain this is: 144.
You and your best friend Janine decide to play a game. You are in a land of make believe where you are a function, f(t), and she is a function, g(t). The two of you move together throughout this land with you (that is, f(t) ) controlling your East/West movement and Janine (that is, g(t) ) controlling your North/South movement. If your identity, f(t), is given by f(t)=(t2+10)323 and Janine's identity, g(t), is given by g(t)=5t then how many units of distance do the two of you cover between the Most Holy Point o' Beginnings, t=0, and The Buck Stops Here, t=20?
Answer:
12920
Step-by-step explanation :
distance covered = d = ∫[tex]\int\limits^a_b ({sqrt{f'(t)^{2} +g'(t)^2\\}}) \, dt[/tex]
a = t = 20
b = t = 0
f'(t) = 646 : f'(t)² = 417316
g'(t) = 5 : g'(t)² = 25
d = [tex]\int\limits^a_b {646} \, dt[/tex]
t = 20 & t = 0
d = 646t
d = 646*20 = 12920
The problem is about calculating distance based on movement rates represented by mathematical functions in an imaginary game. This is done by integrating the given functions over the stipulated time period and then summing the distances covered in each direction.
Explanation:In this mathematical scenario where you and your best friend Janine are functions, you are given an identity f(t)=(t2+10)323 which represents your East/West movement, and Janine's identity is given by g(t)=5t which represents North/South movement. To calculate the units of distance you two cover between t=0 to t=20, we need to integrate the rate of movement of each function over that time period.
Step 1: Integrate your function f(t) from t=0 to t=20. This will give the total distance you moved in the East/West direction.
f(t) = ∫(0 to 20) (t2+10)323 dt
Step 2: Integrate Janine's function g(t) from t=0 to t=20. This will give the total distance Janine moved in the North/South direction.
g(t) = ∫(0 to 20) 5t dt
Step 3: The total distance covered will then be the summation of the distances covered in each direction, as calculated above.
Remember, these are integrations because the functions f(t) and g(t) represents the rate of movement (velocity), and integrating velocity over time gives distance.
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I need help with 8 and 9 please!
Answer:
Step-by-step explanation:
8) looking at the figure at number 8,
It is a quadrilateral. In a quadrilateral, the opposite angles are supplementary. This means that the sum of all the angles is 360 degrees. We have assigned y degrees to the remaining unknown angle.
y = 180 - 71 = 109 degrees( this is because the sum of angles on a straight line is 180 degrees.
Therefore
10x + 6 + 8x - 1 + 13x - 2 + 109 = 360
31x + 112 = 360
31x = 360 - 112 = 248
x = 248/31 = 8
9) looking at the figure, we have assigned alphabets a, b and c to represent the inner angles that form the triangle.
a + 9x + 1 = 180
a = 180 - 1 -9x = 179 - 9x(this is because the sum of the angles on a straight line is 180 degrees)
b + 5x + 12 = 180
b = 180 - 12 - 5x
b = 168 - 5x
c + 10x -37 = 180
c = 180 - 10x + 37
c = 217 - 10x
a + b + c = 180( sum of angles in a triangle is 180 degrees)
179 - 9x + 168 - 5x + 217 - 10x = 180
-9x - 5x - 10x = 180 - 179 - 168 - 217
-24x = -384
x = -384/-24
x = 16
Consider an unreliable communication channel that can successfully send a message with probability 1/2, or otherwise, the message is lost with probability 1/2. How many times do we need to transmit the message over this unreliable channel so that with probability 63/64 the message is received at least once? Explain your answer.
Hint: treat this as a Bernoulli process with a probability of success 1/2. The question is equivalent to: how many times do you have to try until you get at least one success?
Answer:
6 times we need to transmit the message over this unreliable channel so that with probability 63/64.
Step-by-step explanation:
Consider the provided information.
Let x is the number of times massage received.
It is given that the probability of successfully is 1/2.
Thus p = 1/2 and q = 1/2
We want the number of times do we need to transmit the message over this unreliable channel so that with probability 63/64 the message is received at least once.
According to the binomial distribution:
[tex]P(X=x)=\frac{n!}{r!(n-r)!}p^rq^{n-r}[/tex]
We want message is received at least once. This can be written as:
[tex]P(X\geq 1)=1-P(x=0)[/tex]
The probability of at least once is given as 63/64 we need to find the number of times we need to send the massage.
[tex]\frac{63}{64}=1-\frac{n!}{0!(n-0)!}\frac{1}{2}^0\frac{1}{2}^{n-0}[/tex]
[tex]\frac{63}{64}=1-\frac{n!}{n!}\frac{1}{2}^{n}[/tex]
[tex]\frac{63}{64}=1-\frac{1}{2}^{n}[/tex]
[tex]\frac{1}{2}^{n}=1-\frac{63}{64}[/tex]
[tex]\frac{1}{2}^{n}=\frac{1}{64}[/tex]
By comparing the value number we find that the value of n should be 6.
Hence, 6 times we need to transmit the message over this unreliable channel so that with probability 63/64.
Suppose that a one-way ANOVA is being performed to compare the means of five populations and that the sample sizes are 17 comma 15 comma 16 comma 18 comma and 11. Determine the degrees of freedom for the F-statistic.
(a) the degree of freedom of the numerator _______________________
(b) the degree of freedom of the denominator______________________
Answer:
a) [tex]df_{num}=k-1=5-1=4[/tex]
b) [tex]df_{den}=N-k=77-5=72[/tex]
Step-by-step explanation:
Analysis of variance (ANOVA) "is used to analyze the differences among group means in a sample".
The sum of squares "is the sum of the square of variation, where variation is defined as the spread between each individual value and the grand mean"
If we assume that we have [tex]p[/tex] groups and on each group from [tex]j=1,\dots,p[/tex] we have [tex]n_j[/tex] individuals on each group we can define the following formulas of variation:
[tex]SS_{total}=\sum_{j=1}^p \sum_{i=1}^{n_j} (x_{ij}-\bar x)^2 [/tex]
[tex]SS_{between}=SS_{model}=\sum_{j=1}^p n_j (\bar x_{j}-\bar x)^2 [/tex]
[tex]SS_{within}=SS_{error}=\sum_{j=1}^p \sum_{i=1}^{n_j} (x_{ij}-\bar x_j)^2 [/tex]
And we have this property
[tex]SST=SS_{between}+SS_{within}[/tex]
The degrees for the numerator are [tex]df_{num}=k-1=5-1=4[/tex], where k represent the number of groups on this case k =5.
The degrees for the denominator are [tex]df_{den}=N-k=77-5=72[/tex], where N represent the total number of people N=17+15+16+18+11=77.
And the total degrees of freedom are given by: [tex]df_{tot}=df_{num}+df_{den}=N-1=4+72=76[/tex]
Final answer:
The degrees of freedom for the F-statistic in a one-way ANOVA with five populations and sample sizes of 17, 15, 16, 18, and 11 are 4 and 72.
Explanation:
The degrees of freedom for the F-statistic in a one-way ANOVA are determined by the sample sizes of the populations being compared. In this case, the sample sizes are 17, 15, 16, 18, and 11.
The degrees of freedom for the numerator (df(num)) is equal to the number of groups being compared minus 1, which is 5-1=4.
The degrees of freedom for the denominator (df(denom)) is equal to the total number of observations minus the number of groups, which is 77-5=72. Therefore, the degrees of freedom for the F-statistic are 4 and 72.
A research report from an independent-measures study states that there are significant differences between treatments, F(2, 36)3.45, p <.05. How many treatment conditions were compared in the study? Q 40 treatment conditions O 36 treatment conditions O 2 treatment conditions 3 treatment conditions What was the total number of participants in the study? O The study used a total of N 36 participants. O The study used a total of N 13 participants. O The study used a total of N 39 participants. O The study used a total of N 38 participants.
Answer: The study used a total of N 36 participants.
Step-by-step explanation:
There are 3 treatment conditions were compared in the study and the total number of participants in the study is 39.
What is degree of freedom?Degree of freedom, in mathematics, any of the number of independent quantities necessary to express the values of all the variable properties of a system.
Given the ratio F(2, 36)
F takes in degree of freedom values; degree of freedom of groups ; degree of freedom of error
Hence,
df group or treatment = 2
df error = 36
df treatment = k - 1
k = Number of groups
2 = k - 1
k = 3
Number of treatment condition = 3
df error = N-k
N = total number of observations
36 = N - 3
N = 39
Hence, there are 3 treatment conditions were compared in the study and the total number of participants in the study is 39.
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3 quarts of maple syrup cost $19.47. How much do 3 gallons of maple syrup cost?
Answer:
$ 77.88
Step-by-step explanation:
1 Gallon = 4 Quarts
3 Gallons = 3*4=12 Quarts
thus (19.47/3)*12 =$ 77.88 (we know 3 quarts cost $19.47, so divide by 3 and multiply by 12 gives the result.
14 students have volunteered for a committee. Eight of them are seniors and six of them are juniors.
(a) How many ways are there to select a committee of 5 students?
(b) How many ways are there to select a committee with 3 seniors and 2 juniors?
(c) Suppose the committee must have five students (either juniors or seniors) and that one of the five must be selected as chair. How many ways are there to make the selection?
Answer: a) 2002, b) 840, c) 10010.
Step-by-step explanation:
Since we have given that
Number of students = 14
Number of students senior = 8
Number of students junior = 6
(a) How many ways are there to select a committee of 5 students?
Here, n = 14
r = 5
We will use "Combination" for choosing 5 students from 14 students.
[tex]^{14}C_5=2002[/tex]
(b) How many ways are there to select a committee with 3 seniors and 2 juniors?
Again we will use "combination":
[tex]^8C_3\times ^6C_2\\\\=56\times 15\\\\=840[/tex]
(c) Suppose the committee must have five students (either juniors or seniors) and that one of the five must be selected as chair. How many ways are there to make the selection?
So, number of ways would be
[tex]^{14}C_5\times ^5C_1\\\\=2002\times 5\\\\=10010[/tex]
Hence, a) 2002, b) 840, c) 10010.
There are 2002 ways to select a committee of any 5 students, 840 ways to select a committee specifically with 3 seniors and 2 juniors, and 10010 ways to select a committee and designate one as chair.
Explanation:The subject of this question is combinatorics, which is part of mathematics. It relates to counting, specifically concerning combinations and permutations. Combinations represent the number of ways a subset of a larger set can be selected, while permutations are the number of ways to arrange a subset.
To select a committee of 5 students out of 14, we find the combination, represented mathematically as C(14,5) = 2002 ways. To select a committee with 3 seniors and 2 juniors, we find the combination separately for seniors and juniors and multiply the results, which is C(8,3) * C(6,2) = 56 * 15 = 840 ways. If one of the five students must be the chair, first, the committee of 5 is selected from 14 students as 2002 ways, then one of the five selected students is chosen as chair in 5 ways. These are permutations in action. The total number of ways is then 2002 * 5 = 10010 ways. Learn more about Combinatorics here:
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A survey of 400 college seniors resulted in the following crosstabulation regarding their undergraduate major and whether or not they plan to go to graduate school. What percentage of the undergraduates surveyed are majoring in Engineering?
Answer:
38 %
Step-by-step explanation:
Survey 400 college seniors resulted in the following cross tabulation . regarding their undergraduate major and graduate.
Missing details
Undergraduate Major
Graduate School || Business || Engineering || Others
Yes ----- 35 ---------42 ---------- 63
No ----- ----91 ------- 104 --------- 65
Answer:
The percentage of the undergraduates surveyed who major in Engineering is 36.5%
Step-by-step explanation:
Given: The above table
Required: Percentage of undergraduates that majored in Engineering
We'll solve this question irrespective of whether the undergraduate student plans to go to a graduate school or not.
Follow the steps below
Step 1: Calculate the total engineering students
Tot Engineering students = Undergraduate student that plan to go to graduate school + those that have no plans for graduate school
From the engineering column
Undergraduate student that plan to go to graduate school = 42
those that have no plans for graduate school = 104
Total = 42 + 104
Total = 146
Step 2: Calculate percentage of this category of undergraduates
Percentage is calculated as a ratio of required population to total population.
In other words,
Percentage = Required Population ÷ Total Population
Here,
Required population = total engineering students
Required population = 146
Total Population = 400
Percentage = 146/400
Percentage = 0.365
Percentage. = 36.5%
Hence, the percentage of the undergraduates surveyed who major in Engineering is 36.5%
A New York Times article titled For Runners, Soft Ground Can Be Hard on the Body considered two perspectives on whether runners should stick to hard surfaces or soft surfaces following an injury. One position supported running on soft surfaces to relieve joints that were in recovery from injury. The second position supported running on hard surfaces since soft surfaces can be uneven, which may make worse those injuries a soft surface was intended to help. Suppose we are given sufficient funds to run an experiment to study this topic. With no studies to support either position, which of the following hypotheses would be appropriate? A.The second position makes the more sense, so this should be a one-sided test. In this case, we should form the alternative hypothesis around the first position.
B. The first position is more sensible, so this should postpone defining the hypotheses until after we collect data to guide the rest.
C. Because there is uncertainty, we should postpone defining the hypotheses until after we collect data to guide the test.
D. Because we would be interested in any difference between running on hard and soft surfaces, we should use a two-sided hypothesis test
Answer:
D. Because we would be interested in any difference between running on hard and soft surfaces, we should use a two-sided hypothesis test
Step-by-step explanation:
Hello!
When planning what kind of hypothesis to use, you have to take into account any other studies that were made about that topic so that you can decide the orientation you will give them.
Normally, when there is no other information available to give an orientation to your experiment, the first step to take is to make a two-tailed test, for example, μ₁=μ₂ vs. μ₁≠μ₂, this way you can test whether there is any difference between the two stands. Only after having experimental evidence that there is any difference between the treatments is there any sense into testing which one is better than the other.
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A clinical trial tests a method designed to increase the probability of conceiving a girl. In the study 300 babies were born, and 270 of them were girls. Use the sample data to construct a 99% confidence interval estimate of the percentage of girls born. Based on the result, does the method appear to be effective?
The 99% confidence interval for the percentage of girls born using this method is 86.7% - 93.3%.
Step-by-step calculation:
1. Calculate the sample proportion of girls:
Divide the number of girls by the total number of babies:
Sample proportion ([tex]\hat p[/tex]) = 270 girls / 300 babies = 0.9
2. Determine the z-score for a 99% confidence level:
Using a z-score table or calculator, find the z-score that corresponds to a 99% confidence level. This value is approximately 2.576.
3. Calculate the margin of error:
Margin of error (E) = z-score × [tex]\sqrt {(\hat p(1 - \hat p) / n)[/tex]
E = 2.576×√(0.9 × 0.1 / 300) ≈ 0.033
4. Construct the confidence interval:
Lower limit = [tex]\hat p[/tex] - E = 0.9 - 0.033 = 0.867
Upper limit = [tex]\hat p[/tex] + E = 0.9 + 0.033 = 0.933
5. Interpretation:
We are 99% confident that the true proportion of girls born using this method lies between 86.7% and 93.3%.
6. While the upper limit suggests a potential increase in girls, the confidence interval remains wide, and further studies are needed for a conclusive effectiveness claim.
Use technology and a t-test to test claim about the population mean mu at the given level of significance alpha using the given sample statistics. Assume the population is normally distributed. Claim mu > 76 alpha = 0.01 Sample statistics x = 77.5, s= 3.3, n=29 What are the null and alternative hypotheses? Choose the correct answer below What is te value of the standardized test statistic? The standardized test statistic is (Round to two decimal places as needed) What is the P-value of the test statistic? P-value = (round to three decimal places as needed.) What is the value of the standardized test statistic? The standardized test statistic is . (Round to two decimal places as needed.) What is the P-value of the test statistic? P-value = (Round to three decimal places as needed.) Decide whether to reject or fail to reject the null hypothesis. Choose the correct answer below. Reject H0. There is enough evidence to support the claim. Fail to reject H0. There is not enough evidence to support the claim. Fail to reject H0 There is enough evidence to support the claim. Fail to reject H0 there is not enough evidence to support the claim.
The null hypothesis is that the population mean is equal to or less than 76, while the alternative hypothesis is that it is greater than 76. The standardized test statistic is 2.089, and the P-value is approximately 0.022. Therefore, we reject the null hypothesis and conclude that there is enough evidence to support the claim.
Explanation:The null hypothesis, denoted as H0, states that the population mean (mu) is equal to or less than 76. The alternative hypothesis, denoted as H1, states that the population mean (mu) is greater than 76. To test the claim, we can perform a one-sample t-test.
The standardized test statistic, also known as the t-value, can be calculated using the formula t = (x - mu) / (s / sqrt(n)), where x is the sample mean, mu is the population mean under the null hypothesis, s is the sample standard deviation, and n is the sample size. Plugging in the values x = 77.5, mu = 76, s = 3.3, and n = 29 into the formula gives us a t-value of 2.089 (rounded to two decimal places).
The P-value of the test statistic can be determined by comparing the t-value to the critical value of the t-distribution with (n - 1) degrees of freedom at the given level of significance (alpha). Since the alternative hypothesis is one-sided (mu > 76), we need to find the right-tail area of the t-distribution. Consulting a t-table or using statistical software, we find that the P-value is approximately 0.022 (rounded to three decimal places).
To make a decision, we compare the P-value to the significance level (alpha). If the P-value is less than alpha, we reject the null hypothesis. In this case, the P-value is 0.022, which is less than alpha = 0.01. Therefore, we reject the null hypothesis and conclude that there is enough evidence to support the claim that the population mean (mu) is greater than 76.
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The null and alternative hypotheses can be set, and the t-value can be calculated using the provided sample statistics. The P-value and the decision to reject or fail to reject the null hypothesis cannot be determined without further calculations.
Explanation:The null and alternative hypotheses can be defined as follows:
Null Hypothesis (H0): Mu is less than or equal to 76 (mu <= 76)Alternative Hypothesis (Ha): Mu is greater than 76 (mu > 76)The standardized test statistic (t-value) can be calculated using the formula:
t = (x - mu) / (s / sqrt(n))
Using the given sample statistics:
t = (77.5 - 76) / (3.3 / sqrt(29)) = 1.69 (rounded to two decimal places)
The P-value for the test statistic can be found using a t-distribution table or a statistical software. The P-value represents the probability of obtaining a t-value as extreme as the calculated one, assuming the null hypothesis is true. Since the P-value is not provided in the question, it cannot be determined without further calculations.
To decide whether to reject or fail to reject the null hypothesis, compare the P-value to the significance level (alpha). If the P-value is less than alpha, reject the null hypothesis. If the P-value is greater than or equal to alpha, fail to reject the null hypothesis.
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Apply the Gram-Schmidt orthonormalization process to transform the given basis for Rn into an orthonormal basis. Use the Euclidean inner product on Rn and use the vectors in the order in which they are given. B = {(8, −6, 0), (5, 1, 0), (0, 0, 2)}U1=??U2=??U3=??
Answer:
Remember, if [tex]\{x_1,x_2,...,x_k\}[/tex] is a basis for a subspace W of [tex]\mathbb{R}^n[/tex] then [tex]\{q_1,q_2,...,q_k\}[/tex] is an orthonormal basis of W, where
[tex]q_i=\frac{1}{||v_i||}[/tex] and [tex]v_i[/tex] is defined as:
[tex]v_1=x_1\\v_2=x_2-\frac{v_1\cdot x_2}{v_1\cdot v_1}v_1\\v_k=x_k-\frac{v_1\cdot x_k}{v_1\cdot v_1}v_1 - \cdots - \frac{v_{k-1}\cdot x_k}{v_{k-1}\cdot v_{k-1}}v_{k-1}[/tex]
Then, to find a orthonormal basis of [tex]\{x_1=(8, -6, 0), x_2=(5, 1, 0), x_3=(0, 0, 2)\}[/tex] we will find first the [tex]v_i[/tex]'s.
[tex]v_1=x_1[/tex]
[tex]v_2=x_2-\frac{v_1\cdot x_2}{v_1\cdot v_1}v_1=(5,1,0)-\frac{(8,-6,0)\cdot (5,1,0)}{(8,-6,0)\cdot (8,-6,0)}(8,-6,0)=\\=(5,1,0)-\frac{17}{50}(8,-6,0)=(\frac{57}{25},\frac{76}{25},0)[/tex]
[tex]v_3=x_3-\frac{v_1\cdot x_3}{v_1\cdot v_1}v_1-\frac{v_2\cdot x_3}{v_2\cdot v_2}v_2=\\=(0,0,2)-\frac{(8-6,0)\cdot (0,0,2)}{(8,-6,0)\cdot (8,-6,0)}(8,-6,0) -\frac{(\frac{57}{25},\frac{76}{25},0)\cdot (0,0,2)}{(\frac{57}{25},\frac{76}{25},0)\cdot (\frac{57}{25},\frac{76}{25},0)}(\frac{57}{25},\frac{76}{25},0)=\\=(0,0,2)-0-0=\\=(0,0,2)[/tex]
Therefore, [tex]\{v_1=(8,-6,0),v_2=(\frac{57}{25},\frac{76}{25},0),v_3=(0,0,2)\}[/tex] is a ortogonal basis for [tex]\mathbb{R}^3[/tex]. But we need a orthonormal basis. Then is enough find the corresponding unit vector of the ortogonal basis found.
[tex]q_1=\frac{1}{||v_1||}v_1=\frac{1}{\sqrt{100}}(8,-6,0)\\q_2=\frac{1}{||v_2||}v_2=\frac{1}{\frac{19}{5}}(\frac{57}{25},\frac{76}{25},0)=\frac{5}{19}(\frac{57}{25},\frac{76}{25},0)\\q_3=\frac{1}{||v_3||}v_3=\frac{1}{2}(0,0,2)[/tex]
Hence
[tex]\{q_1=({\frac{8}{\sqrt{100}},\frac{-6}{\sqrt{100}},0), q_2=(\frac{3}{5},\frac{4}{5},0), q_3=(0,0,1)\}[/tex] is a orthonormal basis for [tex]\mathbb{R}^3[/tex]
Final answer:
To transform the given basis B into an orthonormal basis using the Gram-Schmidt orthonormalization process, we can follow the steps outlined.
Explanation:
To apply the Gram-Schmidt orthonormalization process to the given basis B = {(8, −6, 0), (5, 1, 0), (0, 0, 2)} in Rn, we'll follow these steps:
Take the first vector from the basis, let's call it U1, and normalize it to obtain the first vector of the orthonormal basis, which we'll also call U1.For the second vector, U2, subtract the projection of U2 onto U1 from U2 to obtain a vector orthogonal to U1. Normalize this resulting vector to obtain U2 of the orthonormal basis.For the third vector, U3, subtract the projection of U3 onto U1 and U2 from U3 to obtain a vector orthogonal to U1 and U2. Normalize this resulting vector to obtain U3 of the orthonormal basis.By applying this process to B, we find that U1 is (8/10, -6/10, 0), U2 is (1/2, 7/10, 0), and U3 is (0, 0, 2/√10).
Natalie is skiing along a circular ski trail that has a radius of 2.6 km. She starts at the 3-o'clock position and travels in the CCW direction. Natalie stops skiing when she is 0.942 km to the right and 2.423 km above the center of the ski trail. Imagine an angle with its vertex at the center of the circular ski trail that subtends Natalie's path.
a. How many radians has the angle swept out since Natalie started skiing? ______radians
b. How many km has Natalie skied since she started skiing? _________km
Answer:
1.20 radians3.12 kmStep-by-step explanation:
a. The tangent of the central angle is the ratio ...
tan(θ) = (2.423 km)/(0.942 km) . . . . . definition of tangent
θ = arctan(2.423/.942) ≈ 1.200 radians
__
b. The length of the arc is ...
s = rθ = (2.6 km)(1.2 radians) = 3.12 km
_____
The attached output from a graphics program shows the angle as 68.76°. Multiplying by π/180°, we can convert that to radians:
θ = 68.76π/180 = 216.0/180 = 1.200 radians
To solve this, we use trigonometry and the relationships between parts of a circle. The angle subtended at the center of the trail by Natalie's path is 0.966 radians. The distance Natalie has skied since starting is 2.512 km.
Explanation:The subject of this question is Mathematics, specifically Trigonometry, which deals with angles, lengths, and relationships in triangles. With a radius of 2.6 km, as Natalie moves along the circular ski trail, she traces out an arc and subtends an angle at the center of the circle.
(a) To find the angle she's covered in radians: consider the endpoint of her path which forms a triangle with the center of the circle and the initial position (3-o'clock). The angle she's subtended can be found using trigonometry (inverse tangent). The tangent of the angle equals the vertical component (2.423 km) divided by the horizontal component (2.6 km - 0.942 km). Giving us:
Angle in radians = tan-inverse of (2.423/1.658) = 0.966 radians.
(b) The distance she's covered along the circumference of the trail is the length of the arc she has skied, given by:
Arc length = radius * angle (in radians).
So, Distance covered = 2.6 km * 0.966 = 2.512 km.
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Two triangles are shown to be congruent by identifying a combination of translations, rotations, or reflections that move one figure
onto the other. If AACB ADFE, which line segment must be congruent to FD? Why?
A) CA because corresponding parts of congruent triangles are congruent.
B) BC because corresponding parts of congruent triangles are congruent.
C) BA because the triangles are similar and these sides must be congruent.
D) CB because the triangles are similar and these sides must be congruent.
Answer:
the anwser is c
Step-by-step explanation:
Answer:
The correct answer is A
Step-by-step explanation:
A) CA because corresponding parts of congruent triangles are congruent.
A coin-operated drink machine was designed to discharge a mean of 6 ounces of coffee per cup. In a test of the machine, the discharge amounts in 20 randomly chosen cups of coffee from the machine were recorded. The sample mean and sample standard deviation were 6.13 ounces and 0.23 ounces, respectively. If we assume that the discharge amounts are normally distributed, is there enough evidence, at the 0.05 level of significance, to conclude that the true mean discharge, differs from 6 ounces?
Answer:
Step-by-step explanation:
An instructor was interested in seeing if there was a difference in the average amount of time that men and women anticipate studying for an Introduction to Statistics course in the summer. A group of men and women were randomly selected from the University of Florida. The Minitab results are below. What is the best interpretation of the results below?Difference = mu (F) - mu (M)T-Test of difference = 0 (vs not =): T-Value = 0.23 P-Value = 0.817 DF = 46A. With a p-value of 0.817, that there is no statistically significant evidence of a difference in average anticipated amount of time studying between the men and women.B. With a p-value of 0.23, that there is no statistically significant evidence of a difference in average anticipated amount of time studying between men and women.C. We are 23% confidence that there is no statistically significant difference between the anticipated study time between men and women for this course.D. With a p-value of 0.817, there is statistically significant evidence of a difference in average anticipated amount of time studying between men and women.E. We are 81.7% confidence that there is no statistically significant difference between the anticipated study time between men and women for this course.F. With a p-value of 0.23, there is statistically significant evidence of a difference in average anticipated amount of time studying between men and women.
Answer:
.Option A
Step-by-step explanation:
Given that an instructor was interested in seeing if there was a difference in the average amount of time that men and women anticipate studying for an Introduction to Statistics course in the summer.
Minitab results are
[tex]Difference = mu (F) - mu (M)\\T-Test of difference = 0 (vs not =): T-Value = 0.23 \\P-Value = 0.817 \\DF = 46[/tex]
Since p value >0.05 our alpha significant level we accept null hypothesis that
difference in means =0
With a p-value of 0.817, that there is no statistically significant evidence of a difference in average anticipated amount of time studying between the men and women
.Option A
A researcher took a random sample of 100 students from a large university. She computed a 95% confidence interval to estimate the average weight of the students at this university. The confidence interval was too wide to provide a precise estimate.True or false? The researcher could produce a narrower confidence interval by increasing the sample size to 150.A medical researcher is investigating the effect of drinking coffee on systolic blood pressure. The researcher assumes the average systolic blood pressure is 120 mmHg. For a random sample of 200 patients, the researcher takes two measurements of systolic blood pressure. The first systolic blood pressure measurement is taken during a week when the patients drink no coffee, and the second systolic blood pressure measurement is taken during a week when the patients drink at least two cups of coffee. The medical researcher wonders whether there is a significant difference between the blood pressure measurements. Which of the following is the correct null and alternative hypothesis for the medical researchers study?
Answer:
Part (A): The correct option is true.
Part (B): The null and alternative hypothesis should be:
[tex]H_o: \mu =0 ; H_a:\mu \neq0[/tex]
Step-by-step explanation:
Consider the provided information.
Part (A)
A random sample of 100 students from a large university.
Increasing the sample size decreases the confidence intervals, as it increases the standard error.
If the researcher increase the sample size to 150 which is greater than 100 that will decrease the confidence intervals or the researcher could produce a narrower confidence interval.
Hence, the correct option is true.
Part (B)
The researcher wants to identify that whether there is any significant difference between the measurement of the blood pressure.
Therefore, the null and alternative hypothesis should be:
[tex]H_o: \mu =0 ; H_a:\mu \neq0[/tex]
The statement is true. The researcher could produce a narrower confidence interval by increasing the sample size. The correct null hypothesis for the medical researcher's study is that there is no significant difference between the blood pressure measurements with and without coffee.
Explanation:The statement is true. A wider confidence interval means that there is more uncertainty in the estimate. To make the confidence interval narrower, the sample size needs to be increased. Increasing the sample size to 150 would likely result in a narrower confidence interval.
The correct null hypothesis for the medical researcher's study would be: There is no significant difference between the blood pressure measurements when patients drink coffee and when they don't. The alternative hypothesis would be: There is a significant difference between the blood pressure measurements when patients drink coffee and when they don't.
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Monthly sales of a particular personal computer are expected to decline at the following rate of S'(t) computers per month, where t is time in months and S(t) is the number of computers sold each month.
S'(t)= -30t^(2/3)
The company plans to stop manufacturing this computer when monthly sales reach 1,000 computers. If monthly sales now (t=0) are 2,440 computers, find S(t). How long will the company continue to manufacture this computer?
The function S(t) that represents the number of computers sold each month is S(t) = -18t^(5/3) + 2,440.
The company will continue to manufacture this computer for 5.656 months.
We have,
To find the function S(t), which represents the number of computers sold each month, we need to integrate the rate of change function S'(t).
Given:
S'(t) = -30t^(2/3)
Integrating S'(t) with respect to t will give us S(t):
∫S'(t) dt = ∫-30t^(2/3) dt
Using the power rule for integration:
S(t) = -30 * (3/5) * t^(5/3) + C
Simplifying:
S(t) = -18t^(5/3) + C
Now, we need to find the value of C, which represents the constant of integration.
Given that monthly sales at t = 0 (S(0)) are 2,440 computers, we can substitute this value into the equation:
S(0) = -18(0)^(5/3) + C
2,440 = 0 + C
C = 2,440
Substituting the value of C back into the equation, we get:
S(t) = -18t^(5/3) + 2,440
To find how long the company will continue to manufacture this computer, we need to solve for t when S(t) = 1,000:
-18t^(5/3) + 2,440 = 1,000
Simplifying the equation and rearranging:
-18t^(5/3) = -1,440
Dividing by -18:
t^(5/3) = 80
Taking the 3rd root of both sides to isolate t:
t = (80)^(3/5)
t = 5.656
Therefore,
The function S(t) that represents the number of computers sold each month is S(t) = -18t^(5/3) + 2,440.
The company will continue to manufacture this computer for 5.656 months.
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The function S(t) is obtained via integration of S'(t) with respect to t. We find that S(t) = -45t^(5/3) + 2440. When the computer sales fall to a 1000, it takes approximately 5.71 months, given the decline rate.
Explanation:To find the function S(t) from its derivative, S'(t), we need to perform an integral of S'(t) which is -30t^(2/3). By performing the integral:
∫ S'(t)dt = ∫ -30t^(2/3) dt = -45t^(5/3) + C
Where C is the constant of integration.
To find C, we know from the question that at t=0, S(t) = 2440. So:
2440 = -45*(0)^(5/3) + C
2440 = C
Now we have the function: S(t) = -45t^(5/3) + 2440
We also know from the question that the company will stop manufacturing the computer when the monthly sales reach 1000 computers. We set our function equal to 1000 to find the time t:
1000 = -45t^(5/3) + 2440
We can solve for t and we get: t = ((2440-1000)/45)^(3/5), which is approximately 5.71 months.
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[Second Order Differential Equations] Linear constant-coefficient 2nd order differential are very important in Electrical and Computer Engineering (ECE), yet recitation quiz #5 shows that many of you are still struggling with that concept. Here we start with such a differential equation, first a homogeneous solution and then with a forcing function for which we know the form of the particular solution, but then we move on to a more complicated forcing function and then finally to a problem with time-varying coefficients: Find the entire solution for each of the following: (a) y 00 − 4y 0 + 5y = 0, y(0) = 1, y0 (0) = 0 (b) y 00 − 4y 0 + 5y = 5t 2 , y(0) = 2, y0 (0) = 0 (fractions get a little messy) Find the particular solution to the following using the variation of parameters technique: (a) y 00 − 4y 0 + 5y = e 2t csc t (b) t 2y 00 − 4ty0 + 6y = t −4 .
Answer:
Please see attachment
Step-by-step explanation:
Please see attachment
You wish to test whether the number of fouls called in regular season games is different than during the NCAA tournament. The mean number of fowls called during all regular season gams is u = 40.1 During 16 randomly selected playoff games n = 16 the mean number of fowls is 38.1 and the standard deviation of the sample = 5.8 You calculate a 95% Confidence Interval for the mean number of fouls called during playoff games The lower value of the confidence interval is:______
Answer:
the lower value of the interval = 35.26
Step-by-step explanation:
The mean number is u = 40.1
n = 16
the mean number called x is 38.1
the standard deviation = 5.8
given a 95% Confidence Interval
The lower value of the confidence interval is?
solution
There is a infinite population and the standard deviation of the population is known,
the below formula is used for determining an estimate of the confidence limits of the population mean, i.e.
x ± ( zₐσ)/ √n
For a 95% confidence level, the value of za is taken from the confidence interval table = 1.96.
the confidence limits of the population=
x ± ( zₐσ)/ √n
38.1 ± (1.96*5.8)/ √16
38.1 ± 11.368/4
38.1 ± 2.842
40.942 or 35.258
Thus, the 95% confidence limits are 40.942 or 35.258
this prediction is made with confidence that it will be correct nine five times out of 100.
finally, the lower value of the interval = 35.26
Are most student government leaders extroverts? According to Myers-Briggs estimates, about 82% of college student government leaders are extroverts.† Suppose that a Myers-Briggs personality preference test was given to a random sample of 74 student government leaders attending a large national leadership conference and that 57 were found to be extroverts. Does this indicate that the population proportion of extroverts among college student government leaders is different (either way) from 82%? Use α = 0.01.
Answer:
[tex]z=-1.12[/tex]
[tex]p_v =2*P(z<-1.12)=0.263[/tex]
So with the p value obtained and using the significance level given [tex]\alpha=0.01[/tex] we have [tex]p_v>\alpha[/tex] so we can conclude that we don't have enough evidence to reject the null hypothesis, and we can said that at 1% of significance the population proportion of extroverts among college student government leaders is not different from 0.82.
Step-by-step explanation:
1) Data given and notation
n=74 represent the random sample taken
X=57 represent the number of people found extroverts
[tex]\hat p=\frac{57}{74}=0.770[/tex] estimated proportion of people found extroverts.
[tex]p_o=0.82[/tex] is the value that we want to test
[tex]\alpha=0.01[/tex] represent the significance level
Confidence=99% or 0.959
z would represent the statistic (variable of interest)
[tex]p_v[/tex] represent the p value (variable of interest)
2) Concepts and formulas to use
We need to conduct a hypothesis in order to test if population proportion of extroverts among college student government leaders is different (either way) from 82%, the system of hypothesis would be on this case:
Null hypothesis:[tex]p= 0.82[/tex]
Alternative hypothesis:[tex]p \neq 0.82[/tex]
When we conduct a proportion test we need to use the z statisitc, and the is given by:
[tex]z=\frac{\hat p -p_o}{\sqrt{\frac{p_o (1-p_o)}{n}}}[/tex] (1)
The One-Sample Proportion Test is used to assess whether a population proportion [tex]\hat p[/tex] is significantly different from a hypothesized value [tex]p_o[/tex].
3) Calculate the statistic
Since we have all the info requires we can replace in formula (1) like this:
[tex]z=\frac{0.770 -0.82}{\sqrt{\frac{0.82(1-0.82)}{74}}}=-1.12[/tex]
4) Statistical decision
It's important to refresh the p value method or p value approach . "This method is about determining "likely" or "unlikely" by determining the probability assuming the null hypothesis were true of observing a more extreme test statistic in the direction of the alternative hypothesis than the one observed". Or in other words is just a method to have an statistical decision to fail to reject or reject the null hypothesis.
The significance level provided [tex]\alpha=0.01[/tex]. The next step would be calculate the p value for this test.
Since is a bilateral test the p value would be:
[tex]p_v =2*P(z<-1.12)=0.263[/tex]
So with the p value obtained and using the significance level given [tex]\alpha=0.01[/tex] we have [tex]p_v>\alpha[/tex] so we can conclude that we don't have enough evidence to reject the null hypothesis, and we can said that at 1% of significance the population proportion of extroverts among college student government leaders is not different from 0.82.
The attention span of little kids (ages 3–5) is claimed to be Normally distributed with a mean of 15 minutes and a standard deviation of 4 minutes. A test is to be performed to decide if the average attention span of these kids is really this short or if it is longer. You decide to test the hypotheses H0: μ = 15 versus Ha: μ > 15 at the 5% significance level. A sample of 10 children will watch a TV show they have never seen before, and the time until they walk away from the show will be recorded. If, in fact, the true mean attention span of these kids is 18 minutes, what is the probability of a Type II error?
Answer:
P(Type ll error) = 0.2327
Step-by-step explanation:
We are given the following in the question:
Population mean, μ = 15 minutes
Sample size, n = 10
Alpha, α = 0.05
Population standard deviation, σ = 4 minutes
First, we design the null and the alternate hypothesis
[tex]H_{0}: \mu = 18\\H_A: \mu > 18[/tex]
We use One-tailed z test to perform this hypothesis.
Formula:
[tex]z_{stat} = \displaystyle\frac{\bar{x} - \mu}{\frac{\sigma}{\sqrt{n}} }[/tex]
[tex]z_{critical} \text{ at 0.05 level of significance } = 1.645[/tex]
Putting the values, we get,
[tex]z_{stat} = \displaystyle\frac{\bar{x}- 15}{\frac{4}{\sqrt{10}} } > 1.645\\\\\bar{x} -1 5 > 1.645\times \frac{4}{\sqrt{10}}\\\\\bar{x} -15 > 2.08\\\bar{x} = 17.08[/tex]
Type ll error is the error of accepting the null hypothesis when it is not true.
P(Type ll error)
[tex]P(\bar{x}<17.07 \text{ when mean is 18})\\\\= P(z < \frac{\bar{x}-18}{\frac{4}{\sqrt{10}}})\\\\= P(z < \frac{17.08-18}{\frac{4}{\sqrt{10}}})\\\\= P(z<-0.7273)[/tex]
Calculating value from the z-table we have,
[tex]P(z<-0.7273) = 0.2327[/tex]
Thus,
P(Type ll error) = 0.2327