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3 years ago

Two buses leave a station at the same time and travel in opposite directions. One bus travels

Mathematics

1 answer:

natima [27]3 years ago

4 0

Answer:

63 miles per hour

54 miles per hour

Step-by-step explanation:

Let the two buses be represented as bus A and bus B.

bus A and bus B leave a station at the same time and travel in opposite directions.

Let the speed of bus A be A miles per hour.

Let the speed of bus B be B miles per hour.

One bus travels 9 miles/hour slower than the other.

Let bus A be the faster bus.

This means that the speed of bus B is B = A - 9

If the two buses are 585 miles apart after 5hours, this means that after 5hours, the total distance travelled by bus A and bus B is 585 miles.

Distance travelled by bus A in 5 hours = speed × time = 5A

Distance travelled by bus B in 5 hours = speed × time = 5B

5A + 5B = 585 - - - - - -1

Substituting B = A - 9 in equation 1

5A + 5(A - 9) = 585

5A + 5A - 45 = 585

10A = 585 + 45 = 630

A = 630/10 = 63 miles per hour

B = A - 9 = 63 - 9 = 54 miles per hour

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Based on historical data, your manager believes that 26% of the company's orders come from first-time customers. A random sample

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Answer:

$\hat p \sim N( p, \sqrt{\frac{p (1-p)}{n}})$

And we can use the z score formula given by:

$z = \frac{\hat p -\mu_p}{\sigma_p}$

And if we find the parameters we got:

$\mu_p = 0.26$

$\sigma_p = \sqrt{\frac{0.26(1-0.26)}{158}} = 0.0349$

And we can find the z score for the value of 0.4 and we got:

$z = \frac{0.4-0.26}{0.0349}= 4.0119$

And we can find this probability:

$P(z>4.0119) = 1-P(z$

And if we use the normal standard table or excel we got:

$P(z>4.0119) = 1-P(z$

Step-by-step explanation:

For this case we have the following info given:

$p = 0.26$ represent the proportion of the company's orders come from first-time customers

$n=158$ represent the sample size

And we want to find the following probability:

$p(\hat p >0.4)$

And we can use the normal approximation since we have the following two conditions:

1) np = 158*0.26 = 41.08>10

2) n(1-p) = 158*(1-0.26) = 116.92>10

And for this case the distribution for the sample proportion is given by:

$\hat p \sim N( p, \sqrt{\frac{p (1-p)}{n}})$

And we can use the z score formula given by:

$z = \frac{\hat p -\mu_p}{\sigma_p}$

And if we find the parameters we got:

$\mu_p = 0.26$

$\sigma_p = \sqrt{\frac{0.26(1-0.26)}{158}} = 0.0349$

And we can find the z score for the value of 0.4 and we got:

$z = \frac{0.4-0.26}{0.0349}= 4.0119$

And we can find this probability:

$P(z>4.0119) = 1-P(z$

And if we use the normal standard table or excel we got:

$P(z>4.0119) = 1-P(z$

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