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alina1380 [7]
3 years ago
11

Two cars, a Porsche Boxster and a Toyota Scion XB, are traveling in the same direction, although the Boxster is 125.0 m behind t

he Scion. The speed of the Boxster is 26.4 m/s and the speed of the Scion is 15.6 m/s. How much time does it take for the Boxster to catch the Scion?
Mathematics
1 answer:
Gre4nikov [31]3 years ago
6 0

Answer: t = 11.57s

The time taken for the Boxster to catch the Scion is 11.57s

Step-by-step explanation:

Given;

Speed of the boxster = 26.4m/s

Speed of the Scion = 15.6m/s

Assuming the starting point of 0 for the boxter, the starting point of the Scion would be 125.0m.

Given that;

Distance = speed × time

For the Boxster the distance travelled is;

d = 0 + 26.4t

For the scion the distance travelled is ;

d = 125 + 15.6t

Where t is the time.

Equating the two distance;

26.4t = 125 + 15.6t

26.4t - 15.6t = 125

t = 125/(26.4-15.6)

t = 11.57s

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Answer:

1. P(X≥35) = 0.0183

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We have the proportion for women: pw=0.22, and the proportion for men: pm=0.19.

1. We have a sample of 140 woman and we have to calculate the probability of getting 35 or more who do volunteer work.

This is equivalent to a proportion of

p=X/n=35/140=0.25

The standard error of the proportion is:

\sigma_p=\sqrt{\dfrac{p(1-p)}{n}}=\sqrt{\dfrac{0.22*0.78}{300}}\\\\\\ \sigma_p=\sqrt{0.0006}=0.0239

We calculate the z-score as:

z=\dfrac{p-p_w}{\sigma_p}=\dfrac{0.25-0.22}{0.0239}=\dfrac{0.03}{0.0239}=0.8198

Then, the probability of having 35 women or more who do volunteer work in this sample of 140 women is:

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2. We have to calculate the probability of having 21 or fewer women in the group who do volunteer work.

The proportion is now:

p=X/n=21/140=0.15

We can calculate then the z-score as:

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Then, the probability of having 21 women or less who do volunteer work in this sample of 140 women is:

P(X

3. For the sample with men and women, we use the proportion for both, which is π=0.2.

The sample size is n=300.

Then, the standard error of the proportion is:

\sigma_p=\sqrt{\dfrac{p(1-p)}{n}}=\sqrt{\dfrac{0.2*0.8}{300}}\\\\\\ \sigma_p=\sqrt{0.0005}=0.0231

We can calculate the z-scores for p1=0.18 and p2=0.25:

z_1=\dfrac{p_1-\pi}{\sigma_p}=\dfrac{0.18-0.2}{0.0231}=\dfrac{-0.02}{0.0231}=-0.8660\\\\\\z_2=\dfrac{p_2-\pi}{\sigma_p}=\dfrac{0.25-0.2}{0.0231}=\dfrac{0.05}{0.0231}=2.1651

We can now calculate the probabilty of having a proportion within 0.18 and 0.25 as:

P=P(0.18

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