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Finger [1]
2 years ago
14

Student researchers Haley, Jeff, and Nathan saw an article on the Internet claiming that the average vertical jump for teens was

15 inches. They wondered if the average vertical jump of students at their school differed from 15 inches, so they obtained a list of student names and selected a random sample of 20 students. After contacting these students several times, they finally convinced them to allow their vertical jumps to be measured. Here are the data (in inches): 11.0 11.5 12.5 26.5 15.0 12.5 22.0 15.0 13.5 12.0 23.0 19.0 15.5 21.0 12.5 23.0 20.0 8.5 25.5 20.5 A visual of the data distribution is below: Do these data provide convincing evidence at the α = 0.10 level that the average vertical jump of students at this school differs from 15 inches?
Mathematics
1 answer:
abruzzese [7]2 years ago
7 0

Answer:

There is no convincing evidence at α = 0.10 level that the average vertical jump of students at this school differs from 15 inches.

Step-by-step explanation:

We have to make a hypothesis test to prove the claim that the average vertical jump of students differs from 15 inches.

The null and alternative hypothesis are:

H_0: \mu=15\\\\H_a: \mu\neq15

The significance level is 0.10.

The sample mean is 17 and the sample standard deviation is 5.37.

The degrees of freedom are df=(20-1)=19.

The t-statistic is:

t_{19}=\frac{M-\mu}{s_M/\sqrt{n}} =\frac{17-15}{5.37/\sqrt{20}}=\frac{2}{5.37/4.47} =2/1.20=1.67

The two-sided P-value for t=1.67 is P=0.11132.

This P-value is bigger than the significance level, so the effect is not significant. The null hypothesis can not be rejected.

There is no convincing evidence at α = 0.10 level that the average vertical jump of students at this school differs from 15 inches.

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Ester's choir wants to learn a new song for the school concert in 7 weeks. The song has 3016 lines. The choir learns an equal nu
andriy [413]

Hello!

We are asked to find the number of lines the choir has to learn in time for a concert.

We're given that the concert is in 7 weeks, but they learn new lines each day, so we have to convert it to days. We know that there is 7 days in a week, we would multiply 7 weeks by 7 days, which is 49 days.

Now, we would take the total number of lines, which is 3016, and divide that by 49 days.

3016 lines / 49 days ≈ 61.55 lines/day

Since this is a real-world situation, we should round our answer is to nearest whole number, which would be 62 lines/day.

Therefore, Ester's choir needs to learn about 62 lines per day in time for the concert.

3 0
2 years ago
What is 62.19 x 32.5
navik [9.2K]
You have to multilpy them . answer would be 2021.175 
8 0
2 years ago
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A library buys 36 English books, 48 Science books and 72 Mathematics books. The thickness of each book is the same. Now, the lib
fomenos

Answer:

13

Step-by-step explanation:

The GCF of 36, 48, and 72 is 12 so there will be 36 / 12 = 3 stacks of English books, 48 / 12 = 4 stacks of science books and 72 / 12 = 6 stacks of math books for a total of 3 + 4 + 6 = 13 stacks.

5 0
2 years ago
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A hole is drilled in a sheet-metal component, and then a shaft is inserted through the hole. The shaft clearance is equal to dif
VashaNatasha [74]

Answer:

(A)

P(X \geq 0.8) = \int\limits_{0.8}^{\infty} f(x) \, dx  = \int\limits_{0.8}^{1} 1.25(1-x^4) \, dx = 0.08192

(B)

Then the cumulative function would be

CF(x) = 1.25x - 0.25x^5       if   0<x<1

0 otherwise.

Step-by-step explanation:

(A)

We are looking for the probability that the random variable X is greater than 0.8.

P(X \geq 0.8) = \int\limits_{0.8}^{\infty} f(x) \, dx  = \int\limits_{0.8}^{1} 1.25(1-x^4) \, dx = 0.08192

(B)

For any  x you are looking for the probability P(X \geq x)  which is

P(X \geq x)  = \int\limits_{-\infty}^{x}  1.25(1-t^4) dt = \int\limits_{0}^{x}  1.25(1-t^4) dt = 1.25x - 0.25x^2

Then the cumulative function would be

CF(x) = 1.25x - 0.25x^5       if   0<x<1

0 otherwise.

5 0
2 years ago
Jar A contains four liters of a solution that is 45% acid. Jar B contains five liters of a solution that is 48% acid. Jar C cont
castortr0y [4]

Answer:

k= 80%

Step-by-step explanation:

Jar A contains 4*0.45 L acid, and 4 L of a solution  of acid.

Jar B contains 5*0.48 L acid., and 5 L of a solution of acid.

Jar C contains 1*k/100 = k/100 acid, and 1 L of a solution.

50% = 0.5

For jar A.

(2/3)*k/100 L acid  is added to jar A.

Now jar A contains   4*0.45 L + (2/3)*k/100 L acid, and it has (4+2/3)L of a solution.

L solute/L solution = 0.5

[4*0.45 L + (2/3)*k/100 L]/(4+2/3)L = 0.5

[1.8 + (2k/300)]/[(12+2)/3] = 0.5

[1.8 + (2k/300)]/[14/3] = 0.5

[1.8 + (2k/300)]= 0.5*(14/3)

(2k/300) = 0.5*(14/3) - 1.8

2k = (0.5*(14/3) - 1.8)*300

k = (0.5*(14/3) - 1.8)*300/2 =80

k= 80%

We also can find k using jar B.

(1/3)k/100 L acid is added  to jar B.

Now jar B contains 5*0.48 L+ (1/3)k/100 L acid, and it has (5+1/3) L of a solution.

L solute/L solution = 0.5

[5*0.48 L+ (1/3)k/100 L ]/(5+1/3)L= 0.5

[5*0.48 + (1/3)k/100 ]/(5+1/3)= 0.5

This equation also gives k=80%

Check.

We can check at least for jar A.

Jar A has 4L solution and 4*0.45=1.8 L acid.

2/3 L of the solution from jar C was added, and now we have 4 2/3 L of solution.

(2/3)* 80%= (2/3)*0.8 acid was added from jar C.

Now we have [1.8 +(2/3)*0.8] L acid in jar A.

L solute/L solution =  [1.8 +(2/3)*0.8] L /(4 2/3) L = 0.5 or 50%  as it is given that jar A has 50% at the end.

7 0
2 years ago
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