Answer: Our required probability is [tex]\dfrac{1}{7}[/tex]
Step-by-step explanation:
Since we have given that
P(Junior ) = [tex]\dfrac{1}{2}[/tex]
P(Senior) = [tex]\dfrac{1}{2}[/tex]
Let the given event be 'C' taking calculus.
P(C|J) = 10% = 0.10
P(C|S) = 60% = 0.60
We need to find the probability that the student is a junior.
So, our required probability is given by
[tex]P(J|C)=\dfrac{P(J).P(C|J)}{P(S).P(C|S)+P(J).P(C|J)}\\\\P(J|C)=\dfrac{0.5\times 0.1}{0.5\times 0.1+0.5\times 0.6}\\\\P(J|C)=\dfrac{0.05}{0.05+0.3}\\\\P(J|C)=\dfrac{0.05}{0.35}\\\\P(J|C)=\dfrac{5}{35}\\\\P(J|C)=\dfrac{1}{7}[/tex]
Hence, our required probability is [tex]\dfrac{1}{7}[/tex]
Which of the following are solutions to the equation sinx cosx = 1/4? Check all that apply.
A. π/3+nπ/2
B. π/12+nπ
C. π/6+nπ/2
D. 5π/12+nπ ...?
Answer:
B and D
Step-by-step explanation:
sin x cos x = 1/4
Multiply both sides by 2:
2 sin x cos x = 1/2
Use double angle formula:
sin(2x) = 1/2
Solve:
2x = π/6 + 2nπ or 2x = 5π/6 + 2nπ
x = π/12 + nπ or x = 5π/12 + nπ
Membership in an elite organization requires a test score in the upper 30% range. If the mean is equal to 115 and the standard deviation is equal to 12, find the lowest acceptable score that would enable a candidate to apply for membership. Assume the variable is normally distributed. (Show Work)
Answer:
Lowest acceptable score = 121.3
Step-by-step explanation:
Mean test score (μ) = 115
Standard deviation (σ) = 12
The z-score for any given test score 'X' is defined as:
[tex]z=\frac{X-\mu}{\sigma}[/tex]
In this situation, the organization is looking for people who scored in the upper 30% range, that is, people at or above the 70-th percentile of the normally distributed scores. At the 70-th percentile, the corresponding z-score is 0.525 (obtained from a z-score table). The minimum score, X, that would enable a candidate to apply for membership is:
[tex]0.525=\frac{X-115}{12}\\X=121.3[/tex]
Two particles move in the xy-plane. At time t, the position of particle A is given by x(t)=5t−5 and y(t)=2t−k, and the position of particle B is given by x(t)=4t and y(t)=t2−2t−1.
(a) If k=−6, do the particles ever collide?
(b) Find k so that the two particles are certain to collide.
k=
(c) At the time the particle collide in (b), which is moving faster?
A. particle A
B. particle B
C. neither particle (they are moving at the same speed)
Answer:
Part A) Not collide
Part B) k = 4
Part C) Particle B is moving fast.
Step-by-step explanation:
Two particles move in the xy-plane. At time, t
Position of particle A:-
[tex]x(t)=5t-5[/tex]
[tex]y(t)=2t-k[/tex]
Position of particles B:-
[tex]x(t)=4t[/tex]
[tex]y(t)=t^2-2t+1[/tex]
Part A) For k = -6
Position particle A, (5t-5,2t+6)
Position of particle B, [tex](4t,t^2-2t-1)[/tex]
If both collides then x and y coordinate must be same
Therefore,
For x-coordinate:5t - 5 = 4t
t = 5
For y-coordinate:[tex]2t+6=t^2-2t-1[/tex]
[tex]t^2-4t-7=0[/tex]
[tex]t=-1.3,5.3[/tex]
The value of t is not same. So, k = -6 A and B will not collide.
Part B) If both collides then x and y coordinate must be same
For x-coordinate:5t - 5 = 4t
t = 5
For y-coordinate:[tex]2t-k=t^2-2t-1[/tex]
Put t = 5
[tex]10-k=25-10-1[/tex]
[tex]k=4[/tex]
Hence, if k = 4 then A and B collide.
Part C)
Speed of particle A, [tex]\dfrac{dA}{dt}[/tex]
[tex]\dfrac{dA}{dt}=\dfrac{dy}{dt}\cdot \dfrac{dt}{dx}[/tex]
[tex]\dfrac{dA}{dt}=2\cdot \dfrac{1}{5}\approx 0.4[/tex]
Speed of particle B, [tex]\dfrac{dB}{dt}[/tex]
[tex]\dfrac{dB}{dt}=\dfrac{dy}{dt}\cdot \dfrac{dt}{dx}[/tex]
[tex]\dfrac{dB}{dt}=2t-2\cdot \dfrac{1}{4}[/tex]
At t = 5
[tex]\dfrac{dB}{dt}=10-2\cdot \dfrac{1}{4}=2[/tex]
Hence, Particle B moves faster than particle A
To determine if the particles collide, we set up equations with their x-coordinates and y-coordinates. Part (a) asks if they collide when k = -6. Part (b) asks for the value of k that guarantees a collision, and part (c) compares the speeds at the time of collision.
Explanation:To determine if the particles collide, we need to find out if their x-coordinates and y-coordinates are equal at any given time. Given the positions of particles A and B, we can set up two equations by equating their x-coordinates and y-coordinates. For part (a), when k = -6 we can solve the equations to find if they intersect. For part (b), we need to find the value of k that makes the two particles collide. Finally, for part (c), we compare the speeds of particles A and B when they collide.
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In professional basketball games during 2009-2010, when Kobe Bryant of the Los Angeles Lakers shot a pair of free throws, 8 times he missed both, 152 times he made both, 33 times he only made the first, and 37 times he made the second. Is it plausible that the successive free throws are independent?
Answer:
Is plausible that the successive throws are independent
Step-by-step explanation:
1) Table with info given
The observed values are given by the following table
__________________________________________________
First shot Made Second shot missed Total
__________________________________________________
Made 152 33 185
Missed 37 8 45
__________________________________________________
Total 189 41 230
2) Calculations and test
We are interested on check independence and for this we need to conduct a chi square test, the next step would be find the expected value:
Null hypothesis: Independence between two successive free throws
Alternative hypothesis: No Independence between two successive free throws
_____________________________________________________
First shot Made Second shot missed
_____________________________________________________
Made 189(185)/230=152.0217 41(185)/230=32.9783
Missed 189(45)/230=36.9783 41(45)/230=8.0217
_____________________________________________________
On this case all the expected values are higher than 5 and the sample size 230 is enough to apply the chi squared test.
3) Calculate the chi square statistic
The statistic for this case is given by:
[tex]\chi_{cal}^2 =\sum \frac{(O_i -E_i)}{E_i}[/tex]
Where O represent the observed values and E the expected values. Replacing the values that we got we have this
[tex]\chi_{cal}^2 =\frac{(152-152.0217)^2}{152.0217}+\frac{(33-32.9783)^2}{32.9783}+\frac{(37-36.9783)^2}{36.9783}+\frac{(8-8.0217)^2}{8.0217}=0.000003098+0.00001428+0.00001273+0.0.00005870=0.00008881[/tex]
Now with the calculated value we can find the degrees of freedom
[tex]df=(r-1)(c-1)=(2-1)(2-1)=1[/tex] on this case r means the number of rows and c the number of columns.
Now we can calculate the p value
[tex]p_v =P(\chi^2 >0.00008881)=0.9925[/tex]
On this case the pvalue is a very large value and that indicates that we can fail to reject the null hypothesis of independence. So is plausible that the successive throws are independent.
medical school claims that more than 28% of its students plan to go into general practice. It is found that among a random sample of 130 of the school's students, 39% of them plan to go into general practice. Find the P-Value for a test of the school's claim.
0.9974
0.1635
0.3078
0.0026
Answer: 0.0026
Step-by-step explanation:
Let p denotes the proportion of students plan to go into general practice.
As per given , we have
Alternative hypothesis : [tex] H_a: p>0.28[/tex]
Since the alternative hypothesis [tex](H_a)[/tex] is right-tailed so the test is a right-tailed test.
Also , it is given that ,
i.e. sample size : = 130
x= 490
[tex]\hat{p}=0.39[/tex]
Test statistic(z) for population proportion :
[tex]z=\dfrac{\hat{p}-p}{\sqrt{\dfrac{p(1-p)}{n}}}[/tex]
, where p=population proportion.
[tex]\hat{p}[/tex]= sample proportion
n= sample size.
[tex]z=\dfrac{0.39-0.28}{\sqrt{\dfrac{0.28(1-0.28)}{130}}}\\\\=\dfrac{0.11}{0.0393798073988}=2.79330975101\approx2.79[/tex]
P-value for right-tailed test = P(z>2.79)=1-P(z≤ 2.79) [∵P(Z>z)=1-P(Z≤z)]
=1- 0.9974=0.0026 [using z-value table]
Hence, the P-Value for a test of the school's claim = 0.0026
To find the p-value for a test of the school's claim, we can use a hypothesis test. The null hypothesis (H0) is that the proportion of students planning to go into general practice is equal to or less than 28%. The alternative hypothesis (Ha) is that the proportion of students planning to go into general practice is greater than 28%. Given that 39% of the sample of 130 students plan to go into general practice, we calculated the test statistic and the p-value to determine that the p-value is 0.3078.
Explanation:To find the p-value for a test of the school's claim, we can use a hypothesis test. The null hypothesis (H0) is that the proportion of students planning to go into general practice is equal to or less than 28%. The alternative hypothesis (Ha) is that the proportion of students planning to go into general practice is greater than 28%.
Given that 39% of the sample of 130 students plan to go into general practice, we can calculate the test statistic and the p-value. Using a chi-square test, we calculate the test statistic to be approximately 1.307. The degrees of freedom for this test is 1.
Looking up the critical value for a one-tailed test with an alpha level of 0.05 and 1 degree of freedom, the critical value is approximately 3.841. Since the test statistic is less than the critical value, we fail to reject the null hypothesis. Therefore, the p-value is greater than 0.05.
Therefore, the correct answer for the P-value is 0.3078.
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which graph represents this function?
f(x) = 1/2 x – 5
O
Answer: Slope 1/2
y-int -5
graph (0,-5),(1,-9/2)
Step-by-step explanation:
Can someone please help me with that one problem!
Answer:
AAS method can be used to prove that the two triangles are congruent.
Step-by-step explanation:
According to the question for the two triangles one pair of opposite angles are equal. One another pair of angles are equal for the two and one pair of sides are also equal of the two.
Hence, the two given triangles are congruent by AAS rule.
Hence, AAS method can be used to prove that the two triangles are congruent.
Black perch are ocean fish found on reefs near shore. A graduate student at CSULB is interested in testing whether black perch from two different populations behave differently, on average. In her experiment, she measured routine swimming velocity (to the nearest 0.001 cm/s) for 18 fish from one population, and 21 fish from another. a) Which statistical test would you use to analyze these data?
One sample t-test
Step-by-step explanation:
In this statistical test, you will be able to test if a sample mean, significantly differs from a hypothesized value.Here you can test if the average swimming velocity differs significantly from an identified value in the hypothesis.Then you can conclude whether the group of 18 fish or that of 21 fish has a significantly higher or lower mean velocity than the one in the hypothesis.
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7) Lindsay has some yellow and red flowers in a vase. The ratio of yellow to red flowers is 4:7. The number of yellow flowers in Lindsay’s vase is shown. How many red flowers are in the vase?
Answer:
[tex]\frac{7}{4}[/tex]×x
Step-by-step explanation:
given ratio of yellow to red flowers is 4:7.
let number of yellow flowers in Lindsay’s vase is x
let number of red flowers in Lindsay’s vase is y
[tex]\frac{x}{y}[/tex]=[tex]\frac{4}{7}[/tex]
number of red flowers such that x is known is y= [tex]\frac{7}{4}[/tex]×x
Peter and Amanda collect seashells on a beach. They both had 0 seashells when they started. Peter collects 5 seashells every minute, and Amanda collects 8 seashells every minute. Which graph and explanation correctly represent the number of shells Peter and Amanda have collected?
Just wondering I know the equation but I don't get the graphs??? which graph would it be
Answer:
;-; I belive it's the 1st or 2nd one
Step-by-step explanation:
hope this helps you.
ask me for explanation if you want and I'll try to explane it the best I Can.
(P.S;please mark me as brainlyest,ty in advaince)
The average GRE score at the University of Pennsylvania for the incoming class of 2016-2017 was 311. Assume that the standard deviation was 13.
If you select a random sample of 40 students, what is the probability that the sample mean will be greater than 308? Round your answer to three decimal places.
Answer:
[tex]P(\bar X>308)=1-0.0721=0.928[/tex]
Step-by-step explanation:
1) Previous concepts
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".
The Z-score is "a numerical measurement used in statistics of a value's relationship to the mean (average) of a group of values, measured in terms of standard deviations from the mean".
Let X the random variable that represent the GRE score at the University of Pennsylvania for the incoming class of 2016-2017, and for this case we know the distribution for X is given by:
[tex]X \sim N(\mu=311,\sigma=13)[/tex]
And let [tex]\bar X[/tex] represent the sample mean, the distribution for the sample mean is given by:
[tex]\bar X \sim N(\mu,\frac{\sigma}{\sqrt{n}})[/tex]
On this case [tex]\bar X \sim N(311,\frac{13}{\sqrt{40}})[/tex]
2) Calculate the probability
We want this probability:
[tex]P(\bar X>308)=1-P(\bar X<308)[/tex]
The best way to solve this problem is using the normal standard distribution and the z score given by:
[tex]z=\frac{x-\mu}{\frac{\sigma}{\sqrt{n}}}[/tex]
If we apply this formula to our probability we got this:
[tex]P(\bar X >308)=1-P(Z<\frac{308-311}{\frac{13}{\sqrt{40}}})=1-P(Z<-1.46)[/tex]
[tex]P(\bar X>308)=1-0.0721=0.9279[/tex] and rounded would be 0.928
Over the past several years, the proportion of one-person households has been increasing. The Census Bureau would like to test the hypothesis that the proportion of one-person households exceeds 0.27. A random sample of 125 households found that 43 consisted of one person. To conduct the hypothesis test, what distribution would you use to calculate the critical value and the p-value?
Answer:
To conduct the hypothesis test I would use the standard normal distribution to calculate the critical value and the p-value.
Step-by-step explanation:
To conduct the hypothesis test I would use the standard normal distribution because there is a large sample size of n = 125 households. This because a point estimator for the true proportion p of one-person households is [tex]\hat{p} = Y/n[/tex] which is normally distributed with mean p and standard error [tex]\sqrt{p(1-p)/n}[/tex] when the sample size n is large. Here Y is the random variable that represents the number of one-person households observed. Then the test statistic is [tex]Z = \frac{\hat{p}-0.27}{\sqrt{p(1-p)/n}}[/tex] which has a standard normal distribution under the null hypothesis.
A roulette wheel has the numbers 1 through 36, 0, and 00. A bet on five numbers pays 6 to 1 (that is, if one of the five numbers you bet comes up, you get back your $1 plus another $6). How much do you expect to win with a $1 bet on five numbers? HINT [See Example 4.] (Round your answer to the nearest cent.)
On average, when placing a $1 bet on five numbers in roulette, you would expect to lose about 53 cents due to the game's probability structure.
Explanation:In order to calculate the expected winnings in a roulette game, consider a $1 bet on five numbers. The roulette wheel has 38 possibilities (1 to 36, 0, and 00). So, the chance that your bet will win is 5 out of 38. If it wins, the game pays off 6 to 1, which means you get your $1 bet back and win $6 additional, for a total of $7. The expected value of this bet can be calculated as follows: (probability of winning * amount won if bet is successful) - (probability of losing * amount lost if bet is not successful). In other words, E(X) = [5/38 * $7] - [33/38 * $1]. This equates to an expected value of -$0.53 (rounded to the nearest cent). So, on average, with each $1 bet on five numbers, you would expect to lose about 53 cents.
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In the game of roulette with 38 outcomes, if you place a $1 bet on five numbers, your expected winnings are -$0.08, which means you lose 8 cents per game on average.
Explanation:To calculate the expected winnings of a bet in this roulette scenario, you need to know the probability of winning and the associated payout, and also the probability of losing. In a roulette wheel with numbers 1-36 plus 0 and 00, there are a total of 38 possible outcomes. When you place a 5-number bet, the probability of winning is 5 out of 38 or approximately 0.1316, and the probability of losing is 33 out of 38 or approximately 0.8684.
Now we add the expected winnings and losses. The expected gain from a win is the probability of winning times the payout, which is $6 in this case. So, 0.1316 * $6 = $0.79, rounded to the nearest cent. The expected loss from a bet is the probability of loss times the amount of the bet which is $1. So, 0.8684 * -$1 = -$0.87, rounded to the nearest cent. Total expected value or earnings from a single $1 bet on five numbers would be the sum of expected gains and losses, or $0.79 - $0.87 = -$0.08.
So, with a $1 bet on five numbers on this roulette wheel, you can expect, on average, to lose 8 cents per game. This negative figure indicates that this is not a good bet if you're expecting to make money on average.
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At 6:00 PM, a flagpole that is 35 feet tall casts a shadow that is 50 feet long. At the same time, how long will a person's shadow be if they are 4 feet tall?
Please help ASAP!!! :(
Answer:
The length of the person’s shadow is 5.7ft
Explanation:
Length of the flagpole =a= 35ft
Length of the shadow of the flagpole= b=50ft
Length of the person=c= 4ft
Suppose the length of the person’s shadow is=d
According to the rules of trigonometry
[tex]\frac{\text { Length of the flagpole }}{\text { Length of the shadow of the flagpole }}=\frac{\text { Length of the person }}{\text { Length of the person's shadow }}[/tex]
[tex]\frac{a}{b}=\frac{c}{d}[/tex]
[tex]\frac{35}{50}=\frac{4}{d}[/tex]
35d=200
d=[tex]\frac{200}{35}[/tex]
d=5.7ft
Hence, The length of the person’s shadow is 5.7ft.
A group of 56 computer science students were taught introductory computer programming class with an innovative teaching method that used a graphical interface and drag-and-drop methods of creating computer programs. At the end of the class, 43 of these students said that they felt confident in their ability to write computer programs. Another group of 41 students were taught the same material using a standard method. At the end of class, 25 of these students said they felt confident. Assume that each class contained a simple random sample of students. Let pX represent the population proportion of students taught by the innovative method who felt confident and let pY represent the population proportion of students taught by the standard method who felt confident. Find a 99% confidence interval for the difference pX−pY . Round the answers to four decimal places.
Answer:
(-0.0861, 0.4023)
Step-by-step explanation:
We have large sample sizes [tex]n_{x} = 56[/tex] and [tex]n_{y} = 41[/tex]. A [tex]100(1-\alpha)[/tex]% confidence interval for the difference [tex]p_{x}-p_{y}[/tex] is given by [tex](\hat{p}_{x}-\hat{p}_{y})\pm z_{\alpha/2}\sqrt{\frac{\hat{p}_{x}(1-\hat{p}_{x})}{n_{x}}+\frac{\hat{p}_{y}(1-\hat{p}_{y})}{n_{y}}}[/tex]. [tex]\hat{p}_{x}=43/56 = 0.7679[/tex] and [tex]\hat{p}_{y}=25/41=0.6098[/tex]. Because we want a 99% confidence interval for the difference [tex]p_{x}-p_{y}[/tex], we have that [tex]z_{\alpha/2}=z_{0.01/2}=z_{0.005}=2.5758[/tex] (The area above 2.5758 and below the curve of the stardard normal density is 0.005) and the confidence interval is [tex](0.7679-0.6098)\pm (2.5758)\sqrt{\frac{0.7679(1-0.7679)}{56}+\frac{0.6098(1-0.6098)}{41}}[/tex] = (-0.0861, 0.4023).
A laboratory tested twelve chicken eggs and found that the mean amount of cholesterol was 185 milligrams with sequals17.6 milligrams. Construct a 95% confidence interval for the true mean cholesterol content of all such eggs.
Answer: 95% confidence interval would be (175.04,194.96).
Step-by-step explanation:
Since we have given that
Mean = 185 mg
Standard deviation = 17.6 mg
At 95% confidence level, z = 1.96
So, Interval would be
[tex]\bar{x}\pm z\dfrac{\sigma}{\sqrt{n}}\\\\=185\pm 1.96\times \dfrac{17.6}{\sqrt{12}}\\\\=185\pm 9.958\\\\=(185-9.958,185+9.958)\\\\=(175.042,194.958)\\\\=(175.04,194.96)[/tex]
Hence, 95% confidence interval would be (175.04,194.96).
Using the provided values and steps for constructing a 95% confidence interval, we estimate, with 95% confidence, that the true mean cholesterol content of all such eggs lies between 146.5 and 223.5 milligrams.
Explanation:To construct a 95% confidence interval for the mean cholesterol content of all such eggs, we will use the provided sample mean (185 milligrams), standard error (17.6 milligrams), and the fact that the sample size (12 eggs) is relatively small, so we use a t-distribution.
Firstly, we need to find the t-score for a 95% confidence level. Since degrees of freedom (df) is n-1 -> 12-1=11, and at a 95% confidence level, the t-score (from t-distribution table) is approximately 2.201 for a two-tailed test.
Next, we use the formula for confidence interval:
Lower bound = Sample mean - (t-score * standard error) Upper bound = Sample mean + (t-score * standard error)
Calculating these gives:
Lower bound = 185 - (2.201 * 17.6) ≈ 146.5 milligrams Upper bound = 185 + (2.201 * 17.6) ≈ 223.5 milligrams
So, we estimate with 95 percent confidence that the true mean cholesterol content of all such eggs is between 146.5 and 223.5 milligrams.
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What is the common difference in the sequence 7,12,17,22,27,...
Answer:
5
Step-by-step explanation:
27-22= 22-17 = 17-12 =12-7=d
d= 5
The common difference of the arithmetic sequence 7, 12, 17, 22, 27,... is 5, which is the constant difference between consecutive terms of the sequence.
The sequence given is an arithmetic sequence, which is a sequence of numbers where the difference between consecutive terms is constant. To find the common difference in the arithmetic sequence 7, 12, 17, 22, 27,..., we subtract any term from the preceding term in the sequence. Let's perform this calculation using the first two terms:
Thus, the common difference is 5. We can confirm this by checking the difference between other consecutive terms:
All differences are the same, confirming that the common difference of the sequence is indeed 5.
A right triangle whose hypotenuse is StartRoot 7 EndRoot7 m long is revolved about one of its legs to generate a right circular cone. Find the radius, height, and volume of the cone of greatest volume that can be made this way.
Rotate the right triangle with hypotenuse √7 about one leg to create a cone. Use the Pythagorean theorem to express the radius as a function of the other leg. To find the max volume, differentiate the volume formula and set equal to zero to solve for the radius and height.
Explanation:The problem presents a right triangle with hypotenuse of √7 meters. This right triangle is rotated around one of its legs to form a right circular cone. To find the radius, height, and volume of the cone, we can use the properties of the triangle.
Imagine that the triangle is rotated about the shorter leg. The height of the cone (h) is the length of the shorter leg, and the radius (r) is the longer leg. By the Pythagorean theorem, a² + b² = c², where c is the hypotenuse, a and b represent the legs of the triangle. Assume that a is the longer leg and b is the shorter leg.
From the Pythagorean theorem we get a² = (√7)² - b² = 7 - b² and so a = √(7 - b²).
The volume of the cone V = (1/3)πr²h = (1/3)π(√(7 - b²))²b = (1/3)π(7 - b²)b. Differentiating and setting the derivative equal to zero provides the desired maximum volume, giving us the values for the radius and height of the cone.
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A lakefront resort is planning for its summer busy season. It wishes to estimate with 95% confidence the average number of nights each guest will stay for a consecutive visit. Using a sample of guests who stayed last year, the average number of nights per guest is calculated at 5 nights. The standard deviation of the sample is 1.5 nights. The size of the sample used is 120 guests and the resort desires a precision of plus or minus .5 nights. What is the standard error of the mean in the lakefront resort example? Within what range below can the resort expect with 95% confidence for the true population means to fall? Show the calculation; otherwise, the answer will not be accepted.
Answer:
[tex]SE=\frac{1.5}{\sqrt{120}}=0.137[/tex]
The 95% confidence interval would be given by (4.729;5.271)
Step-by-step explanation:
1) Previous concepts
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".
The standard error of a statistic is "the standard deviation of its sampling distribution or an estimate of that standard deviation"
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[/tex] represent the sample mean for the sample
[tex]\mu[/tex] population mean (variable of interest)
s represent the sample standard deviation
n 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)
We use the t distirbution for this case since we don't know the population standard deviation [tex]\sigma[/tex].
Where the standard error is given by: [tex]SE=\frac{s}{\sqrt{n}}[/tex]
And the margin of error would be given by: [tex]ME=t_{\alpha/2}\frac{s}{\sqrt{n}}[/tex]
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=120-1=119[/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 table to find the critical value. And we see that [tex]t_{\alpha/2}=1.98[/tex]
The standard error would be given by:
[tex]SE=\frac{1.5}{\sqrt{120}}=0.137[/tex]
Now we have everything in order to replace into formula (1) and calculate the interval:
[tex]5-1.98\frac{1.5}{\sqrt{120}}=4.729[/tex]
[tex]5+1.98\frac{1.5}{\sqrt{120}}=5.271[/tex]
So on this case the 95% confidence interval would be given by (4.729;5.271)
Final answer:
The standard error of the mean (SE) is calculated to be 0.137, allowing the resort to expect the true average number of nights per guest to fall within a range of approximately 4.73 to 5.27 nights with 95% confidence.
Explanation:
The standard error of the mean (SE) is calculated by dividing the sample standard deviation by the square root of the sample size. In this case, the standard error is 1.5 / sqrt(120) = 0.137. With a confidence level of 95%, the resort can expect the true average number of nights to fall within approximately 5 - 1.96 * 0.137 nights to 5 + 1.96 * 0.137 nights, which is roughly between 4.73 and 5.27 nights.
help please
1 thought 3
A function is a relation that has one output for a given input.
For the first one, there is no one x value with two or more y values so it is a function.
The second example is also a function because a certain value can only have one cube root.
For problem number 3 input "-3" for every instance of x in h(x).
So, h(-3)=2(3^2)-1= 17
Lynbrook West, an apartment complex, has 100 two-bedroom units. The monthly profit (in dollars) realized from renting out x apartments is given by the following function. P(x) = -11 x^2 + 1804 x - 43,000 To maximize the monthly rental profit, how many units should be rented out?
units What is the maximum monthly profit realizable? $
Answer:
82 apartments should be rented.
Maximum profit realized will be $30964.
Step-by-step explanation:
Monthly profit realized from renting out x apartments is modeled by
P(x) = -11x² + 1804x - 43000
To maximize the profit we will take the derivative of the function P(x) with respect to x and equate it to zero.
P'(x) = [tex]\frac{d}{dx}(-11x^{2}+1804x-43000)[/tex]
= -22x + 1804
For P'(x) = 0,
-22x + 1804 = 0
22x = 1804
x = 82
Now we will take second derivative,
P"(x) = -22
(-) negative value of second derivative confirms that profit will be maximum if 82 apartments are rented.
For maximum profit,
P(82) = -11(82)² + 1804(82) - 43000
= -73964 + 147928 - 43000
= $30964
Therefore, maximum monthly profit will be $30964.
The number of units that should be rented out to maximize the monthly rental profit is 82 units.
The maximum monthly profit realizable is [tex]\( \$30,964 \).[/tex]
To find the number of units that should be rented out to maximize the monthly rental profit and the maximum monthly profit, we need to analyze the given profit function:
[tex]\[ P(x) = -11x^2 + 1804x - 43,000 \][/tex]
This is a quadratic function of the form [tex]\( P(x) = ax^2 + bx + c \)[/tex], where [tex]\( a = -11 \), \( b = 1804 \)[/tex], and c = -43,000 .
For a quadratic function [tex]\( ax^2 + bx + c \)[/tex] with a < 0 (which opens downwards), the maximum value occurs at the vertex. The x-coordinate of the vertex for the function P(x) is given by:
[tex]\[ x = -\frac{b}{2a} \][/tex]
Substituting a = -11 and b = 1804 :
[tex]\[ x = -\frac{1804}{2(-11)} = \frac{1804}{22} = 82 \][/tex]
Therefore, the number of units that should be rented out to maximize the profit is x = 82 .
Next, we calculate the maximum monthly profit by substituting x = 82 back into the profit function:
[tex]\[ P(82) = -11(82)^2 + 1804(82) - 43,000 \][/tex]
Calculating step by step:
[tex]\[ 82^2 = 6724 \]\[ -11(6724) = -73,964 \]\[ 1804(82) = 147,928 \]\[ P(82) = -73,964 + 147,928 - 43,000 \]\[ P(82) = 30,964 \][/tex]
So, the maximum monthly profit realizable is [tex]\( \$30,964 \).[/tex]
In summary:
- The number of units that should be rented out to maximize the monthly rental profit is 82 units.
- The maximum monthly profit realizable is [tex]\( \$30,964 \).[/tex]
The package of a particular brand of rubber band says that the bands can hold a weight of 7 lbs. Suppose that we suspect this might be an overstatement of the breaking weight So we decide to take a random sample of 36 of these rubber bands and record the weight required to break each of them. The mean breaking weight of our sample of 36 rubber bands is 6.6 lbs. Assume that the standard deviation of the breaking weight for the entire population of these rubber bands is 2 lbs.Finding a random sample with a mean this low in a population with mean 7 and standard deviation 2 is very unlikely. a. True b. False
Answer:
False this value is very likely to find in this distribution
Step-by-step explanation:
With mean 7 ( μ ) and standad deviation (σ ) 2 we can observe, value 6.6 is close to the lower limit of the interval
μ ± 0,5 σ 7 ± 1 in which we should find 68,3 % of all values
(just 6 tenth to the left)
And of course 6.6 is inside the interval
μ ± 1 σ where we find 95.7 % of th value
We conclude this value is not unlikely at all
Final answer:
To determine whether it is unlikely to find a random sample with a mean breaking weight of 6.6 lbs in a population with a mean of 7 lbs and a standard deviation of 2 lbs, we can use hypothesis testing. The calculated z-score is less than the critical z-value, indicating that it is unlikely to find a random sample with a mean breaking weight of 6.6 lbs in the population. Therefore, the statement is true.
Explanation:
To determine whether it is unlikely to find a random sample with a mean breaking weight of 6.6 lbs in a population with a mean of 7 lbs and a standard deviation of 2 lbs, we need to use hypothesis testing. We can set up the null and alternative hypotheses as follows:
Null hypothesis (H0): The population mean breaking weight is equal to 7 lbs.
Alternative hypothesis (Ha): The population mean breaking weight is less than 7 lbs.
Next, we can calculate the z-score using the formula (sample mean - population mean)/(standard deviation/sqrt(sample size)). Substituting the values, we get (6.6 - 7)/(2/sqrt(36)) = -0.6/ (2/6) = -0.6/ (1/3) = -1.8.
We can now look up the critical z-value for a one-tailed test at a significance level of 0.05. The critical z-value is -1.645. Since the calculated z-score (-1.8) is less than the critical z-value (-1.645), we can reject the null hypothesis. Therefore, it is true that finding a random sample with a mean breaking weight of 6.6 lbs in a population with a mean of 7 lbs and a standard deviation of 2 lbs is very unlikely.
Quadrilateral EFGH is on a coordinate plane. Which statement is true?
Answer:
see below
Step-by-step explanation:
Opposite sides are parallel in a parallelogram, so if they have different slope, the figure will not be a parallelogram.
_____
Comments on other answer choices
If the slope of diagonal EG is perpendicular to that of diagonal FH, it only proves the figure is some sort of kite. The figure may or may not be a parallelogram.
Adjacent sides being different lengths does not prove anything (except that the figure is not a rhombus).
Proving angle F is a right angle does not prove anything else about the shape of the figure. The figure may or may not be a parallelogram. (If it is a parallelogram, it is also a rectangle.)
Answer:
Have no fear, the answer is top left GIVE BRANLIEST PLZ
Step-by-step explanation:
A boutique handmade umbrella factory currently sells 37500 umbrellas per year at a cost of 7 dollars each. In the previous year when they raised the price to 15 dollars, they only sold 17500 umbrellas that year. Assuming the amount of umbrellas sold is in a linear relationship with the cost, what is the maximum revenue?
Answer:
$302,500
Step-by-step explanation:
If cost (C) = $7, then Sales (S) = 37,500 units
If cost (C) = $15, then Sales (S) = 17,500 units
The slope of the linear relationship between units sold and cost is:
[tex]m=\frac{37,500-17,500}{7-15}\\m= -2,500[/tex]
The linear equation that describes this relationship is:
[tex]s-s_0 =m(c-c_0)\\s-17500 =-2500(c-15)\\s(c)=-2500c + 55,000[/tex]
The revenue function is given by:
[tex]R(c) = c*s(c)\\R(c)=-2500c^2 + 55,000c[/tex]
The cost at which the derivative of the revenue equals zero is the cost that yields the maximum revenue.
[tex]\frac{d(R(c))}{dc}=0 =-5000c + 55,000\\c=\$11[/tex]
The optimal cost is $11, therefore, the maximum revenue is:
[tex]R(11)=-2500*11^2 + 55,000*11\\R(11)=\$ 302,500[/tex]
Express the sum of the polynomial 5x^2+6x−17 and the square of the binomial (x+6) as a polynomial in standard form.
Answer:
6x^2 + 18x + 19.
Step-by-step explanation:
5x^2 + 6x - 17 + (x + 6)^2
= 5x^2 + 6x - 17 + x^2 + 12x + 36
= 6x^2 + 18x + 19.
Answer:
6x^2 + 18x + 19
Step-by-step explanation:
The standard form of a polynomial depends on the degree of the polynomial. The polynomial is in the standard form when it is arranged such that the first term contains the highest degree, and it decreases with the consecutive terms.
The square of the binomial is (x+6)^2
(x+6)^2 = (x+6)(x+6) = x^2 + 6x + 6x + 36 = x^2 + 12x + 36
The sum of the two polynomials will be 5x^2+6x−17 + x^2 + 12x + 36
Collecting like terms,
5x^2 + x^2 + 6x + 12x + 36 -17
= 6x^2 + 18x + 19
The median of a continuous random variable having distribution function F is that value m such that F(m) = 1/2 . That is, a random variable is just as likely to be larger than its median as it is to be smaller. The mode of a continuous random variable having pdf f(x) is the value of x for which f(x) attains its maximum. Find the median and the mode of X if X is(a) uniformly distributed over (a, b)(b) normal with parameters μ, σ2(c) exponential with parameter λ
Answer:
Step-by-step explanation:
To find median and mode for
a) In a uniform distribution median would be
(a+b)/2 and mode = any value
b) X is N
we know that in a normal bell shaped curve, mean = median = mode
Hence mode = median = [tex]\mu[/tex]
c) Exponential with parameter lambda
Median = [tex]\frac{ln2}{\lambda }[/tex]
Mode =0
The median of a distribution is the middle value while the mode is the highest occuring value
(a) uniformly distributed over (a, b)The median (M) of a uniform distribution is:
[tex]M = \frac{a +b}2[/tex]
A uniform distribution has no mode
(b) normal with parameters μ, σ2For a normal distribution with the given parameters, we have:
Median = Mean = Mode = μ
Hence, the median and the mode are μ
(c) exponential with parameter λFor an exponential distribution with the given parameter, we have:
[tex]Median = \frac{\ln 2}{\lambda}[/tex]
The mode of an exponential distribution is 0
Read more about median and mode at:
https://brainly.com/question/14532771
Find z such that 22% of the area under the standard normal curve lies to the right of z. (Round your answer to two decimal places.)
Answer:
z=0.77
Step-by-step explanation:
If 22% of the area under the standard normal curve lies to the right of z, 78% lies to the left.
According to a one-sided z-score table:
If z-score = 0.77 the area under the curve is 77.94%
If z-score = 0.78 the area under the curve is 78.23%
Since only two decimal plates are required, there is no need to actually interpolate the values, just verify which one is closer to 78%
Since 77.94% is closer, a z-score of 0.77 represents a 22% area under the standard normal curve to the right of z.
To find z such that 22% of the area under the standard normal curve lies to the right of z, the value of z is approximately 0.81.
Explanation:To find the value of z such that 22% of the area under the standard normal curve lies to the right of z, we can use a z-table or a calculator. First, we need to find the area under the curve to the left of z, which is 1 - 0.22 = 0.78. We can then look up the z-score that corresponds to this area in the table. The z-score is approximately 0.81. Therefore, the value of z is approximately 0.81.
A tire company measures the tread on newly-produced tires and finds that they are normally distributed with a mean depth of 0.98mm and a standard deviation of 0.35mm. Find the probability that a randomly selected tire will have a depth less than 0.70mm. Would this outcome warrant a refund (meaning that it would be unusual)?
Answer:
0.212 is the probability that a randomly selected tire will have a depth less than 0.70 mm.
Step-by-step explanation:
We are given the following information in the question:
Mean, μ = 0.98 mm
Standard Deviation, σ = 0.35 mm
We are given that the distribution of tire tread is a bell shaped distribution that is a normal distribution.
Formula:
[tex]z_{score} = \displaystyle\frac{x-\mu}{\sigma}[/tex]
a) P(depth less than 0.70 mm)
P(x < 0.70)
[tex]P( x < 0.70) = P( z < \displaystyle\frac{0.70 - 0.98}{0.35}) = P(z < -0.8)[/tex]
Calculating from normal z table, we have:
[tex]P(z<-0.8) = 0.212[/tex]
[tex]P(x < 0.70) = 0.212 = 21.2\%[/tex]
Thus, this event is not unusual and will not warrant a refund.
The parametric equations below describe the line segment that joins the points P1(x1,y1) and P2(x2,y2). Find parametric equations to represent the line segment from (-3, 5) to (1, -2). x = x1 + (x2 – x1)t y = y1 + (y2 – y1)t 0 ≤ t ≤ 1
Answer:
x=-3+4t\\
y =5-7t
Step-by-step explanation:
we are given that when two point (x1,y1) and (x2,y2) are joined the parametric equation representing the line segment would be
[tex]x = x_1 + (x_2 - x_1)t y = y_1 + (y_2 - y_1)t, 0 \leq ≤t \leq ≤ 1[/tex]
Our points given are
[tex](-3, 5) to (1, -2)[/tex]
So substitute the values to get
[tex]x=-3+(1+3)t =-3+4t\\y = 5+(-2-3)t =5-7t[/tex]
Hence parametric equations for the line segment joining the points
(-3,5) and (1,-2) is
[tex]x=-3+4t\\y =5-7t[/tex]
Final answer:
The parametric equations for the line segment from (-3, 5) to (1, -2) are x = -3 + 4t and y = 5 - 7t for 0 ≤ t ≤ 1.
Explanation:
The question asks for parametric equations to describe the line segment joining the points P1(-3, 5) and P2(1, -2). According to the formula provided, we have:
x = x1 + (x2 – x1)t
y = y1 + (y2 – y1)t
Substituting the values of P1 and P2 into the equations:
x = -3 + (1 – (-3))t = -3 + 4t
y = 5 + (-2 – 5)t = 5 - 7t
This means for 0 ≤ t ≤ 1, the parametric equations representing the line segment from (-3, 5) to (1, -2) are:
x = -3 + 4t
y = 5 - 7t
In a survey of 40 Clemson students, it was found that the mean age (in years) when they would like to get married is 27.4 with a standard deviation of 6. How many Clemson students would need to be surveyed to estimate the mean age at which Clemson students would like to get married to within 1.5 years with 90% confidence?
Answer:
Step-by-step explanation:
To estimate the mean age at which Clemson students would like to get married to within 1.5 years with 90% confidence, we need to survey at least 92 Clemson students.
Explanation:To estimate the mean age at which Clemson students would like to get married to within 1.5 years with 90% confidence, we need to calculate the sample size needed for this level of precision. The formula to determine the sample size is:
n = (z * σ / E)^2
where n is the required sample size, z is the z-score corresponding to the desired confidence level (in this case 90% confidence level which corresponds to a z-score of 1.645), σ is the standard deviation, and E is the desired margin of error (1.5 years).
Plugging in the values, we get:
n = (1.645 * 6 / 1.5)^2
n = 91.154
Rounding up to the nearest whole number, we need to survey at least 92 Clemson students in order to estimate the mean age at which Clemson students would like to get married to within 1.5 years with 90% confidence.