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katrin [286]
2 years ago
14

High school students from track teams in the state participated in a training program to improve running times. Before the train

ing, the mean running time for the students to run a mile was 402 seconds with standard deviation 40 seconds. After completing the program, the mean running time for the students to run a mile was 368 seconds with standard deviation 30 seconds. Let X represent the running time of a randomly selected student before training, and let Y represent the running time of the same student after training. Which of the following is true about the distribution of X-Y?a. The variables X and Y are independent, therefore, the meanis 34 seconds and the standard deviation is 10 seconds.b. The v ales X and Y are independent therefore, the meanis 34 seconds and the standard deviation is 50 secondsc. The variables X and Y are not independent, therefore, the standard deviation is 50 seconds and the mean cannot be determined with the information given.d. The variables and are not independent, therefore, the meanis 3 seconds and the standard deviation cannot be determined with the information givene. The variables X and Y We not independent, therefore, neither the mean nor the standard deviation can be determined with the informantion given.
Mathematics
1 answer:
Nataly [62]2 years ago
3 0

Answer:

b. The values X and Y are independent therefore, the mean is 34 seconds and the standard deviation is 50 seconds

Step-by-step explanation:

Subtraction between normal variables:

When two normal variables are subtracted, the mean is the difference of the means, while the standard deviation is the square root of the sum of the variances.

Before the training, the mean running time for the students to run a mile was 402 seconds with standard deviation 40 seconds.

This means that \mu_X = 402, \sigma_X = 40

After completing the program, the mean running time for the students to run a mile was 368 seconds with standard deviation 30 seconds.

This means that \mu_Y = 368, \sigma_Y = 30

Which of the following is true about the distribution of X-Y?

They are independent, so:

\mu = \mu_X - \mu_Y = 402 - 368 = 34

\sigma = \sqrt{\sigma_X^2+\sigma_Y^2} = \sqrt{40^2+30^2} = 50

This means that the correct answer is given by option b.

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Is 8456 mL greater than 9L
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Read 2 more answers
23% of U.S adults say they are more likely to make purchases during a sales tax holiday. You randomly select 10 adults. Find the
algol13

Answer:

a) 0.294

b) 0.414

c) 0.694

Step-by-step explanation:

Use binomial probability.

P = nCr pʳ qⁿ⁻ʳ

where n is the number of trials,

r is the number of successes,

p is the probability of success,

and q is the probability of failure (1−p).

In this case, n = 10, p = 0.23, and q = 0.77.

a) If r = 2, then the probability is:

P = ₁₀C₂ (0.23)² (0.77)¹⁰⁻²

P = 45 (0.23)² (0.77)⁸

P = 0.294

b) If r > 2, then the probability is 1 − P(r ≤ 2).

P = 1 − ₁₀C₀ (0.23)⁰ (0.77)¹⁰⁻⁰ − ₁₀C₁ (0.23)¹ (0.77)¹⁰⁻¹ − ₁₀C₂ (0.23)² (0.77)¹⁰⁻²

P = 1 − 1 (0.23)⁰ (0.77)¹⁰ − 10 (0.23)¹ (0.77)⁹ − 45 (0.23)² (0.77)⁸

P = 1 − 0.073 − 0.219 − 0.294

P = 0.414

c) If 2 ≤ r ≤ 5, then the probability is:

P = ₁₀C₂ (0.23)² (0.77)¹⁰⁻² + ₁₀C₃ (0.23)³ (0.77)¹⁰⁻³ + ₁₀C₄ (0.23)⁴ (0.77)¹⁰⁻⁴ + ₁₀C₅ (0.23)⁵ (0.77)¹⁰⁻⁵

P = 45 (0.23)² (0.77)⁸ + 120 (0.23)³ (0.77)⁷ + 210 (0.23)⁴ (0.77)⁶ + 252 (0.23)⁵ (0.77)⁵

P = 0.294 + 0.234 + 0.122 + 0.044

P = 0.694

8 0
2 years ago
The closing price (in dollars) per share of stock of tempco electronics on the tth day it was traded is approximated by p(t) = 2
Vaselesa [24]

solution:

The closing price (in dollars) per share of stock of Tempco Electronics on the tth day it was  

traded is approximated by  

P(t) = 20 + 12 sin πt/30 − 6 sin πt/15 + 4 sin πt/10 − 3 sin 2πt/15 (0 ≤ t ≤ 24)  

where t = 0 corresponds to the time the stock was first listed on a major stock exchange.  

What was the rate of change of the stock's price at the close of the 15th day of trading?  

P'(t) = 12(cos πt/30) - 6 (cos πt/15) + 4(cos πt/10) - 3(cos 2πt/15)  

t = 15  

P'(t) = 12(cos 15π/30) - 6 (cos 15π/15) + 4(cos 15π/10) - 3(cos 30π/15)  

P'(t) = 12(cos π/2) - 6 (cos π) + 4(cos 3π/2) - 3(cos 2π)  

P'(t) = 12(0) - 6(-1) + 4(0) - 3(1) = 6 - 3  

P'(t) = $3 per day RATE OF CHANGE  

the closing price on that day

P(t) = 20 + 12 sin πt/30 − 6 sin πt/15 + 4 sin πt/10 − 3 sin 2πt/15  

t = 15  

P(t) = 20 + 12 sin 15π/30 − 6 sin 15π/15 + 4 sin 15π/10 − 3 sin 30πt/15  

P(t) = 20 + 12 sin π/2 − 6 sin π + 4 sin 3π/2 − 3 sin 2π  

P(t) = 20 + 12(1) − 6(0) + 4(-1) − 3(0) = 20 + 12 - 0 - 4 - 0 = 20 + 12 - 4  

P(t) = $28 per share close price



4 0
2 years ago
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