If a polynomial "contains", in a multiplicative sense, a factor [tex](x-x_0)[/tex], then the polynomial has a zero at [tex]x=x_0[/tex].
So, you polynomial must contain at least the following:
[tex](x-(-2)),\quad (x-0),\quad (x-3),\quad (x-5)[/tex]
If you multiply them all, you get
[tex]x(x+2)(x-3)(x-5)=x^4 - 6 x^3 - x^2 + 30 x[/tex]
Now, if you want the polynomial to be zero only and exactly at the four points you've given, you can choose every polynomial that is a multiple (numerically speaking) of this one. For example, you can multiply it by 2, 3, or -14.
If you want the polynomial to be zero at least at the four points you've given, you can multiply the given polynomial by every other function.
To find a polynomial with given zeros, each zero can be plugged into the format (x - a). Doing so with the provided zeros in this question, i.e., -2, 0, 3, and 5, represents the zeroes of the polynomial equation x(x + 2)(x - 3)(x - 5) which simplifies to x4 - 6x3 + 3x2 + 60x.
Explanation:The subject of this question is polynomial equations in mathematics. In order to find a polynomial equation that has zeroes at x = -2, x = 0, x = 3, and x = 5, you can use the fact that the equation (x - a)(x - b)(x - c)(x - d) will have zeros at a, b, c, and d.
By substituting the values for a, b, c, and d into the equation with the zeroes provided (-2, 0, 3, and 5), the polynomial equation that satisfies these zeroes can be expressed as: (x - (-2))(x - 0)(x - 3)(x - 5). Simplifying, this gives: x(x + 2)(x - 3)(x - 5).
To further simplify, we need to multiply these expressions: x*(x+2)*(-x+15)-(2*x-10),which simplifies to: x4 - 6x3 + 3x2 + 60x. This is the polynomial equation that has zeros at x = -2, x = 0, x = 3, and x = 5.
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Write and evaluate the definite integral that represents the area of the surface generated by revolving the curve on the indicated interval about the x-axis. y = 81 − x2 , −5 ≤ x ≤ 5
To find the area of the surface generated by revolving the curve about the x-axis, you need to evaluate the definite integral of the square of the given function within the specified interval.
Explanation:Definite integral representing the area:
Given curve: y = 81 - x^2, -5 ≤ x ≤ 5
Revolving this curve about the x-axis generates a surface. The definite integral representing the area is obtained by integrating the square of the function and applying the limits of integration.
Evaluate the integral: ∫[(-5 to 5) of (81 - x^2)^2] dxThe total area of the regions between the curves is 726 square units
Calculating the total area of the regions between the curves
From the question, we have the following parameters that can be used in our computation:
y = 81 - x²
Also, we have the interval to be
−5 ≤ x ≤ 5
So, the area of the regions between the curves is
Area = ∫y dx
This gives
Area = ∫(81 - x²) dx
Integrate
Area = 81x - x³/3
Recall that −5 ≤ x ≤ 5
So, we have
Area = |[81(-5) - (-5)³/3] - [81(5) - (5)³/3]|
Evaluate
Area = 726
Hence, the total area of the regions between the curves is 726 square units
(a) Use Euler's method with step size 0.2 to estimate y(1.4), where y(x) is the solution of the initial-value problem y' = 3x − 3xy, y(1) = 0. (Round your answer to four decimal places.) y(1.4) =
Answer:
Step-by-step explanation:
In monitoring lead in the air after the explosion at the battery factory, it is found that the amounts of lead over a 7 day period had a standard error of 1.91. Find the margin of error that corresponds to a 95% confidence interval. (Round to 2 decimal places)
Final answer:
The margin of error for a 95% confidence interval, given a standard error of 1.91, is approximately 3.74 when rounded to two decimal places.
Explanation:
To find the margin of error for a 95% confidence interval when the standard error is given, we will use the concept of the z-score that corresponds to our desired confidence level. For a 95% confidence interval, the z-score is typically 1.96, which is the critical value for a normal distribution that leaves 2.5% in each tail. The margin of error (MoE) is calculated by multiplying the standard error (SE) by the z-score.
Margin of Error = z * SE
Marginal Error for our scenario = 1.96 * 1.91
Therefore, the Margin of Error = 3.7436. Rounding this to two decimal places gives us a Margin of Error of approximately 3.74.
(8 points) The National Assessment of Educational Progress (NAEP) tested a sample of students who had used a computer in their mathematics classes, and another sample of students who had not used a computer. The mean score for students using a computer was 309, with a standard deviation of 29. For students not using a computer the mean was 303, with a standard deviation of 32. Assume there were 60 students in the computer sample, and 40 students in the sample that hadn’t used a computer. Can you conclude that the population mean scores differ?
Answer:
Step-by-step explanation:
This is a test of 2 independent groups. The population standard deviations are not known. it is a two-tailed test. Let c b the subscript for students using computer and n be the subscript for students not using computer.
Therefore, the population means would be μc and μn.
The random variable is xc - xn = difference in the sample mean scores of students who used computers and those who didn't use computers
We would set up the hypothesis.
The null hypothesis is
H0 : μc = μn H0 : μc - μn = 0
The alternative hypothesis is
Ha : μc ≠ μn Ha : μc - μn ≠ 0
Since sample standard deviation is known, we would determine the test statistic by using the t test. The formula is
(xc - xn)/√(sc²/nc + sn²/nn)
From the information given,
xc = 309
xn = 303
sc = 29
sn = 32
nc = 60
nn = 40
t = (309 - 303)/√(29²/60 + 32²/40)
t = 0.953
The formula for determining the degree of freedom is
df = [sc²/nc + sn²/nn]²/(1/nc - 1)(sc²/nc)² + (1/nn - 1)(sn²/nn)²
df = [29²/60 + 32²/40]²/(1/60 - 1)(29²/60)² + (1/40 - 1)(32²/40)² = 1569.48/20.13
df = 78
We would determine the probability value from the t test calculator. It becomes
p value = 0.344
Assuming a level of significance of 0.05, we would not reject the null hypothesis because the p value, 0.344 is > 0.05
Therefore, we cannot conclude that the population mean scores differ.
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A school allots £2000 to spend on a trip to the theatre.
Theatre tickets have a regular cost of £36 each and are on offer for off.
A train ticket for the day will cost £12 each.
if 2 teachers and the maximum number of students attend, how many students go on the trip?
The maximum number of students that attend the trip is 160.
To determine the number of students that go on the trip, to consider the cost of theatre tickets, train tickets, and the allotted budget.
Let's denote the number of students as S. We know that the cost of each theatre ticket is £36 and that there are 2 teachers attending. Therefore, the total cost of theatre tickets for the teachers is 2 × £36 = £72.
The remaining budget after accounting for the cost of theatre tickets for the teachers is £2000 - £72 = £1928.
Each student will require a train ticket costing £12. So, the cost of train tickets for S students will be 12S.
to find the maximum number of students (S) that can be accommodated within the remaining budget.
From the remaining budget, the cost of train tickets for the students should be less than or equal to the available funds:
12S ≤ £1928
To solve for S, we divide both sides of the inequality by 12:
S ≤ £1928 / 12
S ≤ 160.67
Since the number of students must be a whole number, the maximum number of students that can go on the trip is 160.
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4,3,2,5, 6, 6, 10, 5, 6, 2, 3, 4, 6, 7, 14,5
5. What is the 3rd (upper) quartile of the data set?
Answer:
14
Step-by-step explanation:
A 3 meter chain with linear mass density rho(x) = 2x(4 − x) kg/m lies on the ground, where x = 0 is the top of the chain. Calculate the work required to lift the chain from its top end so that its bottom is 2 meters above the ground.
Answer:
the total workdone required to to lift the chain from its top end so that its bottom is 2 meters above the ground = 661 J
Step-by-step explanation:
Given that:
The mass density of the chain is [tex]\rho (x) = 2x( 4-x) \ kg/m[/tex]
It is pertinent and crucial to consider the determination of the work-done in lifting the chain from its front in that its bottom is 2 meters away from the ground.
Consider a cross section portion of the chain of length Δx that has to be lifted to a height [tex]x_k[/tex]
The required wok to be done for this work is [tex]W_k = \rho(x_k)g \delta x(x_k)[/tex]
combining the segments of the chain and taking the [tex]\delta x[/tex]→0
Integrally, the work-done can be illustrated as :
[tex]\int\limits^3_0 \rho {(x_k)} \, gxdx \ \ = \ \ \int\limits^3_0 (9.8) 2x^2 (4-x)dx \\\\= 19.6 \int\limits^3_0 (4x^2-x^3)dx\\\\\\= 19.6 [ \frac{4}{3}x^2- \frac {x^4}{4}]^3__0}}dx\\\\\\= 19.6 [ \frac{4}{3}(3)^2- \frac {3^4}{4}]}dx\\\\= 19.6 (36- \frac{81}{4})\\\\\\= 308.7 \ \ J[/tex]
Furthermore, there is need to lift the chain up to 2 meters . So, calculating the weight of the chain ; we have:
Weight = [tex]\int\limits^3_0 \rho {x} \, gdx[/tex]
[tex]= \int\limits^3_0 (9.8) {2x}(4-x) \, dx\\ \\\\= 19.6\int\limits^3_0 {(4x-x^2)} \, dx \\\\= 19.6 [ 2x^2 - \frac{x^3}{3}]^3_0\\\\= 19.6 [2(3)^2 - \frac{3^3}{3}]\\\\=19.6 [18-9]\\\\= 176.4 \ \ J[/tex]
Finally .the work-done is said to be equal to the potential energy
∴ W = mgh
W = (176.4)×2
W = 352.8 J
Total workdone = (308.7 + 352.8 ) J
Total workdone = 661 J
Thus, the total workdone required to to lift the chain from its top end so that its bottom is 2 meters above the ground = 661 J
Answer:
Step-by-step explanation:
Answer:
Step-by-step explanation:
Given that,
Length of chain
L = 3m
Linear mass density is
ρ(x) = 2x(4 — x) kg/m lies on the ground
When x = 0, is top of the chain
Work done to lift the chain from top end so that the bottom is 2m above the ground.
Considered the segment of the chain of length ∆x that will be lifted in the positive y direction (+j)m from the foot.
The work needed to lift this segment is given as
Work = mass density × ∆x × gravity
W = ρ × ∆x × g
g is acting downward = 9.8j
Summing over all segment of the chain and passing to the limit as ∆x→0.
Therefore, the total work done needed to full extend the chain is
W = ∫ ρ × ∆x × g x = 0 to 3
Since g is constant
∆x = xdx
Then,
W = g ∫ 2x(4—x)x dx. x= 0 to x = 3
W = 9.81 ∫ (8x² — 2x³)x dx
W = 9.81 ( 8x³/3 — 2x⁴/4)
W=9.81(8x³/3— ½x⁴) from x=0 to x=3
W = 9.81[8(3)³/3 — ½(3⁴)] — 0
W = 9.81 × (72 —40.5)
W = 9.81 × 31.5
Work done= 309.015 J
Lifting the entire chain requires to light the weight
Weight = ∫ρgdx
Weight = g ∫ρ dx. From x=0 to x=3
Weight = g ∫ 2x(4-x) dx
Weight = 9.81 ∫(8x-2x²)dx
Weight = 9.81 [ 8x²/2 - 2x³/3]
Weight = 9.81[4x²-⅔x³] x=0 to x=3
Weight = 9.81[4(3²) — ⅔(3³)]
Weight = 9.81(36—18)
Weight = 9.81 × 18
Weight = 176.58N
Now, this weight is lifted to a height of 2m, then using potential energy formula, we have
P.E = Work = mgh = Weight ×height
Work = W×h = 176.58 × 2
Work = 353.16 J
Then, total workdone is
W = 353.16 + 309.015
W = 662.18 J
The required Work done required to lift the chain from top so that it's bottom is 2m from the ground is 662.18J
what is 20000 x 50000
Answer:
1000000000
Step-by-step explanation:
20000 times 50000 equals 1000000000
when you multiply 20,000 by 50,000, you get the product of 1,000,000,000 or 1 billion.
How to determine the product of 20000 x 50000The The product of 20,000 multiplied by 50,000 is:
20,000 x 50,000 = 1,000,000,000
Therefore, 20,000 multiplied by 50,000 equals 1 billion. of 20,000 multiplied by 50,000 is:
20,000 x 50,000 = 1,000,000,000
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The time taken to prepare the envelopes to mail a weekly report to all executives in a company has a normal distribution, with a mean of 35 minutes and a standard deviation of 2 minutes. On 95% of occasions the mailing preparation takes less than a) 38.29 minutes b) 31.71 minutes c) 35.25 minutes d) 34.75 minutes
Answer:
Mailing preparation takes 38.29 min max time to prepare the mails.
Step-by-step explanation:
Given:
Mean:35 min
standard deviation:2 min
and 95% confidence interval.
To Find:
In normal distribution mailing preparation time taken less than.
i.eP(t<x)=?
Solution:
Here t -time and x -required time
mean time 35 min
5 % will not have true mean value . with 95 % confidence.
Question is asked as ,preparation takes less than time means what is max time that preparation will take to prepare mails.
No mail take more time than that time .
by Z-score or by confidence interval is
Z=(X-mean)/standard deviation.
Z=1.96 at 95 % confidence interval.
1.96=(X-35)/2
3.92=(x-35)
X=38.29 min
or
Confidence interval =35±Z*standard deviation
=35±1.96*2
=35±3.92
=38.29 or 31.71 min
But we require the max time i.e 38.29 min
And by observation we can also conclude the max time from options as 38.29 min.
In a normal distribution, the 95th percentile is calculated by adding twice the standard deviation to the mean. Based on the given mean (35 minutes) and standard deviation (2 minutes), 95% of the time the envelope preparation should take less than 39 minutes.
The problem asks what time, in percentage terms, the preparation for mailing the weekly report to the executives is less than.
Given that the time taken to prepare the envelopes follows a normal distribution with a mean of 35 minutes and a standard deviation of 2 minutes, we need to determine the time that occurs at the 95th percentile of this distribution.
The 95th percentile is found by adding 2 standard deviations to the mean in a normal distribution (this is due to the empirical rule, which states that about 95% of values lie within 2 standard deviations of the mean in a normal distribution).
Here's how you calculate it:
Determine the mean: The mean is given as 35 minutes.
Determine the standard deviation: The standard deviation is given as 2 minutes.
Calculate the 95th percentile: Mean + 2(Standard Deviation) = 35 + 2(2) = 39 minutes.
So, on 95% of occasions the mailing preparation takes less than 39 minutes. None of the options (a, b, c, and d) provided are correct based on the calculations.
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Samuel is at the top of the tower and will ride down a zip line to a lower tower. The total vertical drop of the zip line is 40ft. The zip line's angle of elevation from the lower tower is 11.5° . To the nearest tenth, what is the horizontal distance between the towers?
Answer:
196.6 feet
Step-by-step explanation:
Draw the right triangle formed by the zip-line. Use SOH-CAH-TOA to write and solve an equation:
tan 11.5° = 40 / x
x = 40 / tan 11.5°
x = 196.6
The horizontal distance between the towers is 196.60 ft.
What are heights and distances in trigonometry?Distance is the measurement of an object from a specific point in the horizontal direction, and height is the measurement of an object in the vertical direction.
Here, ∅₁ is called the angle of elevation and ∅₂ is called the angle of depression. For one specific type of problem in height and distances, we have a generalized formula. Height = Distance moved / [cot(original angle) – cot(final angle)] => h = d / (cot ∅₁ – cot ∅₂)
Given:
The vertical height of the zip line = 40 ft
The angle of elevation = 11.5°
We know, according to trigonometric values tan n = height / base
∴ tan 11.5 = 40 / base
base = 40/ tan 11.5
base = 40 / 0.203
base = 196. 60 ft
Therefore, the horizontal distance between the towers is 196.60 ft.
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Work out the percentage change to 2 decimal places when a price of £60 is increased to £89.99
Answer:
that is the solution to the question
The percentage change to 2 decimal places when a price of £60 is increased to £89.99 is 49.98%.
Given that, a price of £60 is increased to £89.99.
What is percentage?Percentage is defined as a given part or amount in every hundred. It is a fraction with 100 as the denominator and is represented by the symbol "%".
Here, old price =£60 and new price =£89.99
Change in price = 89.99-60
= 29.99
Percentage = Change in price/Old price ×100
= 29.99/60 ×100
= 0.4998 ×100
= 49.98%
Therefore, the percentage change to 2 decimal places when a price of £60 is increased to £89.99 is 49.98%.
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Write a rule for the nth term of the geometric sequence if r = 1/4 and a3 = 2.
Answer:
[tex]a_{n} = 32(\frac{1}{4})^{n-1}[/tex]
Step-by-step explanation:
The nth term of a geometric sequence is given by the following equation.
[tex]a_{n+1} = ra_{n}[/tex]
In which r is the common ratio.
This can be expanded for the nth term in the following way:
[tex]a_{n} = a_{1}r^{n-1}[/tex]
In which [tex]a_{1}[/tex] is the first term.
This means that for example:
[tex]a_{3} = a_{1}r^{3-1}[/tex]
So
[tex]a_{3} = a_{1}r^{2}[/tex]
[tex]2 = a_{1}(\frac{1}{4})^{2}[/tex]
[tex]2 = \frac{a_{1}}{16}[/tex]
[tex]a_{1} = 32[/tex]
Then
[tex]a_{n} = 32(\frac{1}{4})^{n-1}[/tex]
Determine which is the appropriate approach to conduct a hypothesis test. Claim: The AI, or Adequate Intake of water, for pregnant women is a mean of 3L/d, liters per day. Sample data: nequals200, x overbar equals2.5, sequals1. The sample data appear to come from a normally distributed population with sigma equals1.2.
Answer:
One sample z-test for population mean would be the best approach to conduct a hypothesis test.
Step-by-step explanation:
Following is the data available to us:
Mean amount of water = u = 3
Sample mean = x = 2.5
Sample Size = n = 200
Sample Standard Deviation = s = 1
Population Standard Deviation = [tex]\sigma[/tex] = 1.2
Population is normally distributed.
We need to find the best approach to conduct a hypothesis test. Since only one sample is involved, it is a One-Sample test about population mean. for conducting hypothesis test for One-Sample about population mean we have following two options:
One Sample z-test for population meanOne sample t-test for population meanSelecting the best approach:
The first thing to check is if our data from a population which is normally distributed. Which in this case is. Next we check if the value of population standard deviation is known or unknown. the rule is:
If value of population standard deviation is known, then we use One sample z-testIf value of population standard deviation is unknown and only value of sample standard deviation is known, then we use one-sample t-test.Since, in this case we know the value of Population Standard Deviation which is 1.2, One sample z-test for population mean would be the best approach to conduct a hypothesis test.
The appropriate approach to conduct a hypothesis test for the claim that the AI, or Adequate Intake of water, for pregnant women is a mean of 3L/d, liters per day, is to use a one-sample t-test.
The one-sample t-test is a statistical test that is used to compare the sample mean to a known population mean. It is appropriate to use this test when the sample size is small (n < 30) and the population standard deviation is unknown.
In this case, the sample size is 200, which is greater than 30, but the population standard deviation is known (1.2). However, it is still appropriate to use the one-sample t-test because the sample size is not large enough to reliably estimate the population standard deviation.
To conduct the one-sample t-test, we will follow these steps:
State the null and alternative hypotheses.
Null hypothesis (H0): The mean water intake for pregnant women is 3L/d.
Alternative hypothesis (Ha): The mean water intake for pregnant women is not 3L/d.
Calculate the test statistic.
The test statistic for the one-sample t-test is calculated as follows:
t = (x - μ) / (s / √n)
where:
x is the sample mean
μ is the known population mean
s is the sample standard deviation
n is the sample size
In this case, the test statistic is calculated as follows:
t = (2.5 - 3) / (1 / √200) = -4.17
Find the p-value.
The p-value is the probability of obtaining a test statistic as extreme or more extreme than the one observed, assuming that the null hypothesis is true.
We can use a t-table to find the p-value. The p-value for a t-statistic of -4.17 with 199 degrees of freedom is less than 0.001.
Make a decision.
Since the p-value is less than the significance level (α = 0.05), we reject the null hypothesis and conclude that there is sufficient evidence to support the claim that the mean water intake for pregnant women is not 3L/d.
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Dayren is driving ti visit family. She drives at an average of 65 miles per hour. She drives 227.5 miles before lunch and then 97.5 miles after lunch. How many hours did she spend driving?
Answer:
5 hours
Step-by-step explanation:
227.5+97.5=325
325/65=5
factor 3x^2 +30x + 75
I WILL MARK BRAINLIEST IF ANSWER IS RIGHT
Answer:
3(x+5)^2
Step-by-step explanation:
Step 1:
Extract the obvious common constant factor
3(x^2+10x+25)
Step 2:
Recognize the second term as a square
3(x+5)^2
The selling price for a classic car is $12,000, which is $500 less than two times its original price. What was the original price of the car?
Answer:$6250
Step-by-step explanation:
selling price(sp)=12000
Original price(op)=?
sp+500=2 x op
12000+500=2 x op
12500=2 x op
Divide both sides by 2
12500/2=(2 x op)/2
6250=op
Therefore original price is $6250
To find the original price of the car, set up an equation with the given information and solve for X. The original price of the car was $6,250.
Explanation:
To find the original price of the car, we need to set up an equation based on the given information. Let's assume the original price of the car is X. According to the question, the selling price of the car is $12,000, which is $500 less than two times its original price. So, we can write the equation as:
2X - $500 = $12,000
To solve the equation, we add $500 to both sides:
2X = $12,500
Finally, we divide both sides by 2 to find the original price of the car:
X = $6,250
Therefore, the original price of the car was $6,250.
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What is the answer to 2(4+9w)-4
Answer:
=2(4+9w)
=2(4) 2(+9w)
=8 +18w
=18w+8
Step-by-step explanation:
Many realistic application involve sampling without replacement. For example, in manufacturing, quality control inspectors sample items from a finite production run without replacement. For such a finite population. we have to adjust the value of sigma(X bar). Take (without replacement) samples of size 2 from the above population of N= 5 objects {1, 2, 3, 4, 5) a. How many such samples are possibleb. List all the Xs, i.e. the means of all these samples of size 2c. Find the mean of these means, i.e. E(X)d. Is E(X)=Mu?e. Find the standard error sigma(X), i.e. the std deviation of all these meansf. Show that the standard error equals Note: is called the finite population correction factor. Typically it is used when the sample size n is greater than 5% of the finite population size.
Answer:
a. possible sample size is 10
b. mean is 3
c standard deviation is 0.9
Step-by-step explanation:
Mean of any given set of numbers is the "average" you're used to, where you add up all the numbers and then divide by the number of numbers.
Standard deviation is a measure of the amount of variation or dispersion of a set of values.
Go to attachment for detailed analysis.
The number of possible samples of size 2 without replacement from a population of 5 objects is 10. The means of these samples can be calculated by taking the combinations of 2 objects and finding their mean. The mean of these means is equal to the population mean. The standard error of these means can be calculated using the variance of the means and the finite population correction factor.
Explanation:a. To calculate the number of possible samples of size 2 without replacement from a population of 5 objects, we use the combination formula: C(n, r) = n! / (r! * (n-r)!), where n is the population size and r is the sample size. In this case, n = 5 and r = 2, so the number of possible samples is C(5, 2) = 5! / (2! * (5-2)!) = 10.
b. To list all the means of these samples, we take each combination of 2 objects from the population and calculate their mean. For example, one possible sample is {1, 2}, and its mean is (1 + 2) / 2 = 1.5. Similarly, we calculate the means for the other 9 possible samples.
c. To find the mean of these means, we calculate the average of all the means calculated in part b. d. The mean of these means, denoted as E(X), is equal to the population mean, denoted as Mu, when the sampling is done without replacement.
e. To find the standard error sigma(X), we need to calculate the standard deviation of all these means. We can do this by calculating the variance of the means and taking its square root. f. The standard error sigma(X) equals the square root of the variance of the means divided by the square root of the sample size, multiplied by the finite population correction factor.
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A 29-m tall building casts a shadow. The distance from the top of the building to the tip of the shadow is 34 m. Find the length of the shadow. If necessary, round your answer to the nearest tenth.
The length of the shadow cast by the 29-m tall building is approximately 13.9 m. This is derived by using the Pythagorean theorem, considering the details of the problem as elements of a right triangle.
Explanation:In this problem, we are given a 29-m tall building and the distance from the top of the building to the tip of the shadow which is 34 m. We are asked to find the length of the shadow. To solve for this, we can use the knowledge of right triangles from geometry.
In this case, the height of the building is one side of a right triangle, the distance from the top of the building to the end of the shadow is the hypotenuse, and the shadow length would be the other side. Since we know the length of the hypotenuse (34 m) and one side (29 m), we can solve for the other side (the shadow length) by using the Pythagorean theorem which states that the square of the hypotenuse (side opposite the right angle) is equal to the sum of the squares of the other two sides.
Expressed in the formula: c^2 = a^2 + b^2. Where 'c' is the hypotenuse and 'a' and 'b' are the other two sides.
Rearranging for 'a', which is the length of the shadow, we get: a^2 = c^2 - b^2. Substituting with the given values results in: a = sqrt(34^2 - 29^2). After calculating, the approximate length of the shadow is 13.9 m (rounded to the nearest tenth).
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As predictors become more highly correlated, I. the p-values of the beta estimates become smaller II. it becomes more difficult to determine which predictor is actually producing the effect on the response III. the values of the beta estimates all approach zero
Answer:
Hence P values of beta becomes smaller(< 0.0001). and doest affect the mean response
Step-by-step explanation:
Given:
AS Predictor become more highly correlated .
To find:
Descriptive Nature of high correlated Predictor .
Solution:
A predictor is high correlated means:
1)It means that the two variables are strongly related to each other.
2)This is also called as problem of multicollinearity when two variables are
in Regression.
Effects when predictor are highly correlated ;
The estimated coefficient of one any one variable depends on the other predictor variable in model.Estimated coefficient of regression decrease as predictor variables are added.Hypothesis test Beta = zero gives different conclusion depending upon variable.High correlated of predictor variable does not provide good precision of predication of response in within model.In short ,Mulitcollinearity does not affect the mean response and new response of the model.
Hence P values of beta becomes smaller(< 0.0001). and doest affect the mean response
(-4x⁴+6x²+3)-(5x²-3) answer should be a standard polynomial
Answer: -4x⁴ + x² + 6
Step-by-step explanation:
Remove parentheses: (-a) = -a
-4x⁴ + 6x²+3 - (5x²-3)
-4x⁴ + 6x² + 3 - 5x² + 3
Simplify -4x⁴+ 6x² + 3 - 5x² + 3: -4x⁴ + x² + 6
-4x⁴ + x² + 6
A small appliance manufacturer is planning to open a repair facility that would receive broken appliances from customers and repair them. Customers would send their broken appliances to the facility, and when repaired, they would be returned directly to the customer. There are five different appliance models that would be repaired, identified here as A, B, C, D, and E. The industrial engineering department has provided estimates of the average time to repair each model. These mean times are, respectively, 23 min, 42 min, 19 min, 27 min, and 35 min. The management wants to staff the facility in anticipation of a weekly return rate of 100 appliances per week for models A, B, and C, and a weekly return rate of 150 appliances for models D and E. The facility will operate one shift five days per week, but it is anticipated that each repairperson will spend an average of only 6.0 hours per day repairing appliances. How many repairpersons will be required for the facility
Answer:
There will be required 590 repair persons.
Step-by-step explanation:
Lets count the amount of minutes that are going to be used per week repairing. The amount of minutes of repair work required per week is
100*23 + 100*42+ 100*19 + 150*27 + 150*35 = 17700.
Each worker works 6 hours per day in average and 5 times per week. This means that each worker will do 30 hours of work each week. Therefore, we will need 17700/30 = 590 workers (In practice you should contract a few more, like 600, just in case).
Suppose any one round of a gambling game pays 3 to 1 and the odds are 4 to 1 against you. Suppose you will play for 100 rounds and bet one dollar on each round. We want the probability you will come out ahead.
a. Determine what the possible net gains are. (Not all integers from −100 to 300 are possible.)
b. What is the smallest possible positive net gain?
c. What is the chance you will come out ahead?
Answer:
Check the explanation
Step-by-step explanation:
The odds are 4 to 1 against, so we can estimate the probability of success (p) as
[tex]\frac{p}{q}=\frac{p}{1-p}=\frac{1}{4}\\\\4p=1-p\\\\5p=1\\\\p=0.2[/tex]
The expected pay for every success is 3 to 1, so we lose $1 for every lose and we gain $3 for every win.
The number of winnings in the 100 rounds to be even can be calculated as:
[tex]W+L=100\\\\L=100-W\\\\\\Payoff=0=3*W-1*L=3W-1*(100-W)=3W+W-100\\\\0=4W-100\\\\W=25[/tex]
We have to win at least 25 rounds to have a positive payoff.
As the number of rounds is big, we will approximate the binomial distribution to a normal distribution with parameters:
[tex]\mu=np=100*0.2=20\\\\\ \sigma=\sqrt{npq}=\sqrt{100*0.2*0.8}=4[/tex]
The z-value for x=25 is
[tex]z=\frac{X-\mu}{\sigma}=\frac{25-20}{4}=1.25[/tex]
The probability of z>1.25 is
P(X>25)=P(z>1.25)=0.10565
There is a 10.5% chance of having a positive payoff.
NOTE: if we do all the calculations for the binomial distribution, the chances of having a net payoff are 13.1%.
Use the t-distribution to find a confidence interval for a mean μ given the relevant sample results. Give the best point estimate for μ, the margin of error, and the confidence interval. Assume the results come from a random sample from a population that is approximately normally distributed.1. A 90% confidence interval for μ using the sample results x^- =143.0, s=56.7, and n=50Round your answer for the point estimate to one decimal place, and your answers for the margin of error and the confidence interval to two decimal places.i. point estimate = ii. margin of error = iii. The 90% confidence interval is_______ to _________.
Answer:
i. Point of estimate:
[tex] \hat \mu = \bar X =143.0[/tex]
ii. Margin of error:
[tex] ME = 2.01 *\frac{56.7}{\sqrt{50}}= 16.12[/tex]
iii. The 90% confidence interval
Replacing in the confidence interval formula we got
[tex]143.0-16.12=126.88[/tex]
[tex]143.0+16.12=159.12[/tex]
The 90% confidence interval is 126.88 to 159.12
Step-by-step explanation:
Information given
[tex]\bar X=143.0[/tex] represent the sample mean for the variable of interest
[tex]\mu[/tex] population mean
s=56.7 represent the sample standard deviation
n=50 represent the sample size
Confidence interval
The confidence interval for the true mean when we don't know the deviation 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 for the confidence interval [tex]t_{\alpha/2}[/tex] we need to find the degrees of freedom, with this formula:
[tex]df=n-1=50-1=49[/tex]
The Confidence level provided is 0.90 or 90%, the value for the significance is [tex]\alpha=1-0.9=0.1[/tex] and [tex]\alpha/2 =0.05[/tex], and we can use excel, a calculator or a table to find the critical value. The excel command would be: "=-T.INV(0.05,49)".And we see that [tex]t_{\alpha/2}=2.01[/tex]
i. Point of estimate:
[tex] \hat \mu = \bar X =143.0[/tex]
ii. Margin of error:
[tex] ME = 2.01 *\frac{56.7}{\sqrt{50}}= 16.12[/tex]
iii. The 90% confidence interval
Replacing in the confidence interval formula we got
[tex]143.0-16.12=126.88[/tex]
[tex]143.0+16.12=159.12[/tex]
The 90% confidence interval is 126.88 to 159.12
10 points) National Institutes of Health recommends that the healthy adults take at least 50 nmol/L of vitamin D per day. Suppose that you are interested in finding out if the BSU students are suffering from vitamin D deficiency or not. (a) (5 points) Formulate the appropriate null and alternative hypotheses. Clearly define the parameter of interest in the context of the problem.
Answer:
Check the explanation
Step-by-step explanation:
Given that:
The National Institutes of Health has recommended that the healthy adults are to take at least 50 nmol/L of vitamin D per day. Let’s assume that you are interested in discovering if the BSU students are suffering from vitamin D deficiency or not.
a) Formulate the appropriate null and alternative hypotheses
Null hypothesis
H o : u >= 50
Alternative hypothesis
Ha : u < 50
Consider the function Upper F (x comma y )equalse Superscript negative x squared divided by 6 minus y squared divided by 6 and the point Upper P (negative 3 comma 3 ). a. Find the unit vectors that give the direction of steepest ascent and steepest descent at P. b. Find a vector that points in a direction of no change in the function at P.
Answer:
a.
[tex]\vec{u}=\frac{\bigtriangledown F(-3,3)}{|\bigtriangledown F(-3,3)|}=\frac{1}{\sqrt{2}}[\hat{i}-\hat{j}][/tex] (ascent)
[tex]\vec{u}=-\frac{\bigtriangledown F(-3,3)}{|\bigtriangledown F(-3,3)|}=-\frac{1}{\sqrt{2}}[\hat{i}-\hat{j}][/tex] (descent)
b.
[tex]\vec{v}=\frac{1}{\sqrt{2}}[\hat{i}+\hat{j}][/tex]
Step-by-step explanation:
a. The function is given by:
[tex]F(x,y)=e^{-(x^2/6+y^2/6)}[/tex]
the point is P(-3,3)
a. The unit vector that gives the direction of the steepest ascent is necessary to compute the gradient of F(x,y):
[tex]\bigtriangledown F(x,y)=e^{-(x^2/6+y^2/6)}(-\frac{x}{3})\hat{i}+e^{-(x^2/6+y^2/6)}(-\frac{y}{3})\hat{j}\\\\\bigtriangledown F(x,y)=-\frac{1}{3}e^{-(x^2/6+y^2/6)}[x\hat{i}+y\hat{j}][/tex]
The, it is necessary to evaluate in the point P, and to compute the norm of the vector in order to get the unit vector:
[tex]\bigtriangledown F(-3,3)=-\frac{1}{3}e^{-(\frac{9}{6}+\frac{9}{6})}[-3\hat{i}+3\hat{j}]\\\\\bigtriangledown F(-3,3)=e^{-3}[\hat{i}-\hat{j}]\\\\|\bigtriangledown F(-3,3)|=\sqrt{(e^{-3})^2+(e^{-3})^2}=\sqrt{2}e^{-3}\\\\\vec{u}=\frac{\bigtriangledown F(-3,3)}{|\bigtriangledown F(-3,3)|}=\frac{1}{\sqrt{2}}[\hat{i}-\hat{j}][/tex] (ascent)
for the steepest descend you have
[tex]\vec{u}=-\frac{\bigtriangledown F(-3,3)}{|\bigtriangledown F(-3,3)|}=-\frac{1}{\sqrt{2}}[\hat{i}-\hat{j}][/tex]
b.
the vector with the direction of no change is a vector perpendicular to grad(F):
[tex]\bigtriangledown F(-3,3)\cdot \vec{v}=0\\\\e^{-3}v_1-e^{-3}v_2=0\\\\v_1=v_2[/tex]
furthermore, v is an unit vector:
[tex]\sqrt{v_1^2+v_2^2}=1\\\\v_1 ^2+v_1^2=1\\\\2v_1^2=1\\\\v_1=\frac{1}{\sqrt{2}}=v_2[/tex]
then, the vector is:
[tex]\vec{v}=\frac{1}{\sqrt{2}}[\hat{i}+\hat{j}][/tex]
The chess club and the ballet club of Nahk University were founded at the same time, and they each accept new members at a constant rate. The chess club accepts 5 new members each week, and it had 37 members after 6 weeks. The number of members in the ballet club as a function of time (in weeks) is given by the following table of values:
Answer:
1:B the ballet club
2.B the ballet club.
Step-by-step explanation:
It is correct on khan :)
The chess club accepts 5 new members each week, and had 37 members after 6 weeks. The club started with 7 members. Without more data, a similar function for the ballet club cannot be determined.
Explanation:This question requires a bait to understand the concept of linear functions. The chess club at Nahk University accepts 5 new members each week and had 37 members after 6 weeks. We can express this as a linear function, with the number of weeks as the independent variable and the number of members as the dependent variable:
Members = 5 * Weeks + Initial Members
Here, we know from the information given that after 6 weeks, the chess club had 37 members. So if we plug those numbers into the equation:
37 = 5 * 6 + Initial Members
Then solving for the number of initial members gives us 7. Hence, the chess club started with 7 members.
To find a similar function for the ballet club, we would need more specific data to map time (in weeks) to the number of members in the same way.
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9. Traditional medical treatment will cure about 88% of all people who suffer from a certain eye
disorder. Suppose a large medical clinic treats 93 people with this disorder using a new state-of-the-
art technique.
a.
Estimate the probability that fewer than 80 patients will recover.
b.
The director of the hospital recommends that all hospitals adopt this new technique
since more than 85 patients in the sample have recovered after being treated with the
new technique. Confirm/disprove the director's claim. Show all your work and interpret
your results.
Answer:
a) P (Y less than or equal to 79 ) = 0.226
b) Z = 1.008 less than or equal to Zc = 1.64
So, it means null hypothesis is not rejected.
Step-by-step explanation:
a) Probability that fewer than 80 patients will recover.
As we have:
p = 88% = 0.88 = population success proportion
n = 93 = Sample size
mean = n x p , mean = 93 x 0.88 , mean = 81.84
So, we need P (Y less than or equal to 79) fewer than 80 patients.
Standard Deviation of the population =
SD of population = [tex]\sqrt{n . p . (1-p)}[/tex] , SD of population = [tex]\sqrt{93 .0.88 .(1-0.88)}[/tex]
SD of population = 3.13
As we know that,
probability in binomial distribution is approximately equal to normal distribution so, we will use normal distribution from here. In addition, due to this shift 79 will be equal to 79.5 in normal distribution.
P(Y less than or equal to 79) ≈ P(Y less than or equal to 79.5)
P([tex]\frac{Y - mean }{SD}[/tex]) , P([tex]\frac{79.5 - 81.84}{3.1336}[/tex]) , P(Z less than or equal to -0.75)
So, when will you check the Z score of -0.75 in the Z table you will get the probability which is :
P (Y less than or equal to 79 ) = 0.226
So, 0.226 is the probability that fewer than 80 patients will recover.
b) Z test is used in this part to check the claim:
Favorable Cases sample (recovered) = 85 = Y
Sample size = N = 93
Significance level = 0.05
Proportion Sample = p = Y/N = 85/93 = 0.914.
1. Null hypothesis and Alternative Hypothesis:
H1: p less than or equal to 88% or p less than or equal to 0.88
H2: p greater than 88% or p greater than 0.88
2. Rejection Region:
Zc value is = 1.64 for a right tail test.
Rejection region is such that z value must be greater than Zc value i.e 1.64. If it is not greater, it will not lie in rejection region.
3. Statistics Test:
In this we have to calculate value of Z and compare it with value of Zc.
Z = [tex]\frac{P-p1}{\sqrt{p1(1-p1)/n} }[/tex]
Small p1 = 0.88 , Capital P = 0.914 Z = [tex]\frac{0.914-0.88}{\sqrt{0.88.(1-0.88)/93} }[/tex]
Z = 1.008
4. Final Decision about the null hypothesis:
As we can see, Z value is less than Zc value, hence it does not lie in the rejection region.
Z = 1.008 less than or equal to Zc = 1.64
So, it means null hypothesis is not rejected.
H1 is not rejected, it means there is not sufficient proof to claim that population proportion is larger than population success proportion p1.
a. Estimate the probability that fewer than 80 patients will recover: Approximately 0.2743.
b. Confirm/disprove the director's claim:
The probability of observing more than 85 patients recovering out of 93 treated with the new technique is approximately 0.1515. Therefore, there isn't sufficient evidence to support the director's claim that all hospitals should adopt the new technique based solely on this sample data.
To solve this problem, we'll use the normal approximation to the binomial distribution since the sample size (93) is relatively large.
Let's break it down step by step.
Given:
Probability of traditional medical treatment curing the disorder = 0.88
Number of patients treated with the new technique = 93
a. Estimate the probability that fewer than 80 patients will recover.
To estimate this probability, we'll use the normal approximation to the binomial distribution.
1. Calculate the mean [tex](\(\mu\))[/tex] and standard deviation [tex](\(\sigma\))[/tex] of the binomial distribution:
[tex]\[\mu = np = 93 \times 0.88 = 81.84\][/tex]
[tex]\[\sigma = \sqrt{np(1-p)} = \sqrt{93 \times 0.88 \times 0.12} \approx 3.076\][/tex]
2. Standardize the value of 80 using the z-score formula:
[tex]\[z = \frac{x - \mu}{\sigma} = \frac{80 - 81.84}{3.076} \approx -0.600\][/tex]
3. Use the standard normal distribution table or calculator to find the probability corresponding to z = -0.600.
[tex]\[P(Z < -0.600) \approx 0.2743\][/tex]
So, the estimated probability that fewer than 80 patients will recover is approximately 0.2743.
b. The director of the hospital recommends that all hospitals adopt this new technique since more than 85 patients in the sample have recovered after being treated with the new technique.
Confirm/disprove the director's claim.
To confirm/disprove the director's claim
We'll calculate the probability of observing more than 85 successes (patients recovering) out of 93 trials (patients treated) under the assumption that the success rate is the same as the traditional treatment (0.88).
1. Calculate the mean [tex](\(\mu\))[/tex] and standard deviation [tex](\(\sigma\))[/tex] of the binomial distribution:
[tex]\[\mu = np = 93 \times 0.88 = 81.84\][/tex]
[tex]\[\sigma = \sqrt{np(1-p)} = \sqrt{93 \times 0.88 \times 0.12} \approx 3.076\][/tex]
2. Standardize the value of 85 using the z-score formula:
[tex]\[z = \frac{x - \mu}{\sigma} = \frac{85 - 81.84}{3.076} \approx 1.031\][/tex]
3. Calculate the probability of observing more than 85 successes:
[tex]\[P(Z > 1.031) \approx 1 - P(Z < 1.031) \approx 1 - 0.8485 \approx 0.1515\][/tex]
The probability of observing more than 85 successes out of 93 trials is approximately 0.1515, which means there's about a 15.15% chance of observing this many or more successes if the true success rate is 0.88.
Since [tex]\(15.15\% > 5\%\)[/tex] (common significance level),
We fail to reject the null hypothesis that the success rate of the new technique is the same as the traditional treatment.
Therefore, there isn't sufficient evidence to support the director's claim that all hospitals should adopt the new technique based solely on this sample data.
The meat department of a local supermarket packages ground beef using meat trays of two sizes: 1 designed to hold 1 lb of meet and other hold 3lbs.A random sample of 35 packages in small meat trays produced weight with an average of 1.01 lbs and standard deviation
of 0.18 lbs.
For a 99% confidence interval for average weights of all packages sold in small meat trays, what is the lower limit?
a.
1.088 lbs
b.
0.546 lbs
c.
0.932 lbs
d.
1.01 lbs
Answer:
c) 0.932
99% confidence interval for average weights of all packages sold in small meat trays.
(0.932 ,1.071)
Step-by-step explanation:
Explanation:-
Given random sample of 35 packages in small meat trays produced weight with an average of 1.01 lbs. and standard deviation of 0.18 lbs.
size of the sample 'n' = 35
mean of the sample x⁻= 1.01lbs
standard deviation of the sample 'S' = 0.18lbs
The 99% confidence intervals are given by
[tex](x^{-} - t_{\alpha } \frac{S}{\sqrt{n} } , x^{-} +t_{\alpha } \frac{S}{\sqrt{n} } )[/tex]
The degrees of freedom γ=n-1 =35-1=34
tₐ = 2.0322
99% confidence interval for average weights of all packages sold in small meat trays
[tex](1.01 - 2.0322 \frac{0.18}{\sqrt{35} } , 1.01+2.0322 \frac{0.18}{\sqrt{35} } )[/tex]
( 1.01 - 0.06183 , 1.01+0.06183)
(0.932 ,1.071)
Final answer:-
99% confidence interval for average weights of all packages sold in small meat trays.
(0.932 ,1.071)
What should you always check for when solving rational equations?
Answer-
You should always check the solution is a rational equation. It would be easier to get an answer... Idk if this helps, but thats what I was taught>>>
Step-by-step explanation: