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tia_tia [17]
3 years ago
11

Translate this sentence into an equation.

Mathematics
1 answer:
Andrew [12]3 years ago
3 0

Step-by-step explanation:

"product" means multiplication. the same as "times".

so,

135 = r×9

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Suppose you are working in an insurance company as a statistician. Your manager asked you to check police records of car acciden
pochemuha

Answer:

(a) 95% confidence interval for the percentage of all car accidents that involve teenage drivers is [0.177 , 0.243].

(b) We are 95% confident that the percentage of all car accidents that involve teenage drivers will lie between 17.7% and 24.3%.

(c) We conclude that the the percentage of teenagers has not changed since you join the company.

(d) We conclude that the the percentage of teenagers has changed since you join the company.

Step-by-step explanation:

We are given that your manager asked you to check police records of car accidents and out of 576 accidents you selected randomly, teenagers were at the wheel in 120 of them.

(a) Firstly, the pivotal quantity for 95% confidence interval for the population proportion is given by;

                        P.Q. = \frac{\hat p-p}{\sqrt{\frac{\hat p(1-\hat p)}{n} }}  ~ N(0,1)

where, \hat p = sample proportion teenage drivers = \frac{120}{576} = 0.21

           n = sample of accidents = 576

           p = population percentage of all car accidents

<em>Here for constructing 95% confidence interval we have used One-sample z proportion statistics.</em>

So, 95% confidence interval for the population population, p is ;

P(-1.96 < N(0,1) < 1.96) = 0.95  {As the critical value of z at 2.5% level

                                                     of significance are -1.96 & 1.96}  

P(-1.96 < \frac{\hat p-p}{\sqrt{\frac{\hat p(1-\hat p)}{n} }} < 1.96) = 0.95

P( -1.96 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} }} < {\hat p-p} < 1.96 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} }} ) = 0.95

P( \hat p-1.96 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} }} < p < \hat p+1.96 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} }} ) = 0.95

<u>95% confidence interval for p</u> = [\hat p-1.96 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} }} , \hat p+1.96 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} }}]

  = [ 0.21-1.96 \times {\sqrt{\frac{0.21(1-0.21)}{576} }} , 0.21+1.96 \times {\sqrt{\frac{0.21(1-0.21)}{576} }} ]

  = [0.177 , 0.243]

Therefore, 95% confidence interval for the percentage of all car accidents that involve teenage drivers is [0.177 , 0.243].

(b) We are 95% confident that the percentage of all car accidents that involve teenage drivers will lie between 17.7% and 24.3%.

(c) We are also provided that before you were hired in the company, the percentage of teenagers who where involved in car accidents was 18%.

The manager wants to see if the percentage of teenagers has changed since you join the company.

<u><em>Let p = percentage of teenagers who where involved in car accidents</em></u>

So, Null Hypothesis, H_0 : p = 18%    {means that the percentage of teenagers has not changed since you join the company}

Alternate Hypothesis, H_A : p \neq 18%    {means that the percentage of teenagers has changed since you join the company}

The test statistics that will be used here is <u>One-sample z proportion statistics</u>;

                              T.S.  =  \frac{\hat p-p}{\sqrt{\frac{\hat p(1-\hat p)}{n} }}  ~ N(0,1)

where, \hat p = sample proportion teenage drivers = \frac{120}{576} = 0.21

           n = sample of accidents = 576

So, <u><em>test statistics</em></u>  =  \frac{0.21-0.18}{\sqrt{\frac{0.21(1-0.21)}{576} }}  

                              =  1.768

The value of the sample test statistics is 1.768.

Now at 0.05 significance level, the z table gives critical value of -1.96 and 1.96 for two-tailed test. Since our test statistics lies within the range of critical values of z, so we have insufficient evidence to reject our null hypothesis as it will not fall in the rejection region due to which <u>we fail to reject our null hypothesis</u>.

Therefore, we conclude that the the percentage of teenagers has not changed since you join the company.

(d) Now at 0.1 significance level, the z table gives critical value of -1.6449 and 1.6449 for two-tailed test. Since our test statistics does not lie within the range of critical values of z, so we have sufficient evidence to reject our null hypothesis as it will fall in the rejection region due to which <u>we reject our null hypothesis</u>.

Therefore, we conclude that the the percentage of teenagers has changed since you join the company.

4 0
3 years ago
Please Help Me! I will give brainly!!
DerKrebs [107]

Answer:

5 1/10 meters

Step-by-step explanation:

18 4/5 can be changed to 18 8/10 so

18 8/10 - 13 7/10 = 5 1/10 meters

8 0
2 years ago
Read 2 more answers
A fruit seller purchased 20 kg of apples at Rs 80 per kg and sold them at Rs 85 per kg. find profit and profit percent​
viktelen [127]

Answer:

Profit = Rs 100

Profit %age = 6.25%

Step-by-step explanation:

Cost Price = Rs 80

Cost Price for 20 kg = Rs 1600

Selling Price = Rs 85

<u><em>Profit for 1 kg</em></u> = Rs 85 - Rs 80

=> Rs 5 (For 1 Kg)

<u><em>Profit for 20 kg:</em></u>

=> Rs 5*20

=> Rs 100

<u><em>Now Profit %age:</em></u>

=> \frac{100}{1600}* 100

=> 0.0625 * 100

=> 6.25%

7 0
3 years ago
Read 2 more answers
Clara scored a 79 on her history test what grade does she have to get on her next test so she will have a average of a 83
Airida [17]
79 is 4 below 83 so her next grade needs to be 4 above in order to get an average of 83. The answer is 87.
6 0
3 years ago
1. PART A: Determine the missing lengths:
Otrada [13]

#1

Divide into two rectangles

\\ \tt\hookrightarrow "1"=14-12=2

\\ \tt\hookrightarrow "2"=3+1=4

#2

\\ \tt\hookrightarrow P=14+1+12+3+2+4=36

#3

\\ \tt\hookrightarrow A=14(1)+3(2)=14+6=20units^2

3 0
2 years ago
Read 2 more answers
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