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Lunna [17]
3 years ago
10

PLLLLEEEAAASSSEEE HHHHEEEELLLPPPPPP

Mathematics
1 answer:
Alona [7]3 years ago
3 0

Answer:

looks like 410,526 books in total

Step-by-step explanation:

lmk if you need work

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Randomly selected 110 student cars have ages with a mean of 8 years and a standard deviation of 3.6 years, while randomly select
monitta

Answer:

1. Yes, there is sufficient evidence to support the claim that student cars are older than faculty cars.

2. The 98% confidence interval for the difference between the two population means is [1.432 years, 3.968 years].

Step-by-step explanation:

We are given that randomly selected 110 student cars to have ages with a mean of 8 years and a standard deviation of 3.6 years, while randomly selected 75 faculty cars to have ages with a mean of 5.3 years and a standard deviation of 3.7 years.

Let \mu_1 = <em>mean age of student cars.</em>

\mu_2   = <em>mean age of faculty cars.</em>

So, Null Hypothesis, H_0 : \mu_1 \leq \mu_2      {means that the student cars are younger than or equal to faculty cars}

Alternate Hypothesis, H_A : \mu_1>\mu_2      {means that the student cars are older than faculty cars}

(1) The test statistics that will be used here is <u>Two-sample t-test statistics</u> because we don't know about the population standard deviations;

                             T.S.  =  \frac{(\bar X_1-\bar X_2)-(\mu_1-\mu_2)} {s_p \times \sqrt{\frac{1}{n_1}+\frac{1}{n_2} } }   ~   t_n_1_+_n_2_-_2

where, \bar X_1 = sample mean age of student cars = 8 years

\bar X_2 = sample mean age of faculty cars = 5.3 years

s_1 = sample standard deviation of student cars = 3.6 years

s_2 = sample standard deviation of student cars = 3.7 years

n_1 = sample of student cars = 110

n_2 = sample of faculty cars = 75

Also, s_p=\sqrt{\frac{(n_1-1)\times s_1^{2}+(n_2-1)\times s_2^{2} }{n_1+n_2-2} }  = \sqrt{\frac{(110-1)\times 3.6^{2}+(75-1)\times 3.7^{2} }{110+75-2} }  = 3.641

So, <u><em>the test statistics</em></u> =  \frac{(8-5.3)-(0)} {3.641 \times \sqrt{\frac{1}{110}+\frac{1}{75} } }  ~ t_1_8_3

                                     =  4.952    

The value of t-test statistics is 4.952.

Since the value of our test statistics is more than the critical value of t, so <u><em>we have sufficient evidence to reject our null hypothesis</em></u> as it will fall in the rejection region.

Therefore, we support the claim that student cars are older than faculty cars.

(2) The 98% confidence interval for the difference between the two population means (\mu_1-\mu_2) is given by;

98% C.I. for (\mu_1-\mu_2) = (\bar X_1-\bar X_2) \pm (t_(_\frac{\alpha}{2}_) \times s_p \times  \sqrt{\frac{1}{n_1}+\frac{1}{n_2} })

                                 = (8-5.3) \pm (2.326 \times 3.641 \times  \sqrt{\frac{1}{110}+\frac{1}{75} })

                                 = [2.7 \pm 1.268]

                                 = [1.432, 3.968]

Here, the critical value of t at a 1% level of significance is 2.326.

Hence, the 98% confidence interval for the difference between the two population means is [1.432 years, 3.968 years].

7 0
3 years ago
A line has _____
VMariaS [17]
Two fold rotational symmetry
6 0
4 years ago
GUYS QUICK! what is the surface area of a sphere that has a radius of 5 cm
alisha [4.7K]

Answer:

100 pi cm^2

or approximately 314 cm^2

Step-by-step explanation:

The surface area of a sphere is given by

SA = 4 pi r^2

The radius is 5

SA = 4 *pi * 5^2

SA = 4*pi(25

SA = 100 pi cm^2

If pi is 3.14

SA = 100 *3.14 = 314 cm^2

4 0
3 years ago
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PLS HELP NOW!
Lena [83]

Answer:

The candidate should get 5829 votes.

Step-by-step explanation:

This is a proportion question. The ratio should hold for both the pole and the actual vote.

Equation

x /9570 = 67/110          Multiply both sides by 9570

x = 67 * 9570 / 110       Multiply the two numerators.

x = 641190 / 110            Divide by 110

x = 5829

8 0
4 years ago
Ian invests $250 each month for 9 months into an account that pays 6% APR compounded monthly. What is the future value of Ian's
Feliz [49]

Answer:

B

Step-by-step explanation:

6 0
3 years ago
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