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katrin [286]
1 year 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]1 year 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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Jacobs sand bucket has a top diameter of 10 inches and a bottom diameter of 7 inches. the bucket is 12 inches tall. What’s the v
Elanso [62]
We know that
The volume of a circular truncated cone=(1/3)*pi*[r1²+r1*r2+r2²]*h
where
r1=7/2-----> 3.5 in
r2=10/2----> 5 in
h=12 in
volume of a circular truncated cone=(1/3)*pi*[3.5²+3.5*5+5²]*12
volume=688 in³

the answer is
688 in³

4 0
2 years ago
The box office took in a total of $2905 in paid admissions for the high-school musical. Adult tickets cost $8 each, and student
Svet_ta [14]
There are no equations to choose from but it should look like the following.

a= # of adult tickets
s= # of student tickets

QUANTITY EQUATION
a + s= 560

COST EQUATION
$8a + $3s= $2905

***If you have to also solve for the number of adults and students, here are the steps.

STEP 1:
multiply quantity equation by -8

-8(a + s)= -8(560)
-8a - 8s= -4480

STEP 2:
add cost equation and step 1 equation

$8a + $3s= $2905
-8a - 8s= -4480
a term cancels out to zero
-5s= -1575
divide both sides by -5
s= 315 students

STEP 3:
substitute s=315 in quantity equation
a + s= 560
a + 315= 560
subtract 315 from both sides
a= 245 adults

ANSWER:
Quantity Equation
a + s= 560

Cost Equation
$8a + $3s= $2905

Hope this helps! :)
7 0
2 years ago
A score that is 6 points below the mean corresponds to a z-score of z = 22.00. What is the population standard deviation?
pochemuha

Answer:

<h2>Standard deviation is a measure of how spread out of numbers are:</h2>

Step-by-step explanation:

a score that is 6 points below the mean corresponds to a z score of z= 22 than the population standard deviation will be

----

z = (distance from mean)/(standard deviation)

-----

22 = 6/s

s = 6/22

s=0.272

The population standard deviation deviation is 0.272

4 0
2 years ago
Javier asks his mother how old a tree in their yard is. His mother says, “The sum of 10 and two-thirds of that tree’s age, in ye
sdas [7]

Answer:

10 + (\frac{2}{3}) a = 50 is the CORRECT equation.

Step-by-step explanation:

The given question is INCOMPLETE.

Javier asks his mother how old a tree in their yard is. His mother says, “The sum of 10 and two-thirds of that tree’s age, in years, is equal to 50.” Javier writes the equation { 10 + 2/3} where a is the tree’s age in years. His equation is not correct. What error did he make?

Now here:

a:  The tree’s age in years.

Also,  “The sum of 10 and two-thirds of that tree’s age, in years, is equal to 50.”

⇒ 10 +  two-thirds of that tree’s age  = 50

\implies 10 + (\frac{2}{3}) a = 50

But in the equation written by Javier, the the third fraction is NOT MULTIPLIED by the age of the tree a in Years.

So, the written equation by Javier is Incorrect.

Now, solving the written correct equation for the value of a, we get:

\implies 10 + (\frac{2}{3}) a = 50\\\implies  (\frac{2}{3}) a = 40\\\implies a = 40 \times  (\frac{3}{2})  = 60\\\implies a  = 60

Hence the correct  age of the tree = 60 years

5 0
2 years ago
Mitchel owns a livestock trailer that can hold a maximum of 5,000 pounds. The average weight of each goat is 90 pounds, and the
castortr0y [4]

Answer:

The number of cows and calves Mitchel can transport is determined by the inequity 90g+360c \leq 5000

Step-by-step explanation:

Let g be the number of goats, and c be the number of cows. If Mitchel's livestock trailer can only hold maximum of 5000 pounds. then we have the inequality

90g+360c \leq 5000 <em>(this says that the weight of the goats and the calves cannot exceed 500 pounds.) </em>

Therefore, this inequality determines the number of goats and calves Mitchel can take in a single trip.

3 0
2 years ago
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