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3 years ago

Will mark brainiest

2 answers:

5 0

Ime to walk uphill : t1 = d/3
time to walk downhill: t2 = d/6
total distance = 2d ( up hill and down ill)
2d = (d/3 + d/6)r
2/(1/3 +1/6) = r
r = 4
average speed up and down the hill was 4 mph

total distance = (2/3 x 6 x 4) + (1/3 x 6 x 4)
total distance = 16 + 8 = 24 miles total

Average speed to top of hill = 2/(1/4 +1/3) = 3.43 mph

to top of hill = 24/2 = 12 miles

12 / 3.43 = 3.5 hours
12 oclock + 3.5 hours = 3:30

total miles was 24
arrived at top of hill at 3:30

Sonja [21]3 years ago

3 0

12 oclock to 6 oclock is 6 hours

d = distance
r = rate of speed

time to walk uphill : t1 = d/3
time to walk downhill: t2 = d/6
total distance = 2d ( up hill and down ill)
2d = (d/3 + d/6)r
2/(1/3 +1/6) = r
r = 4
average speed up and down the hill was 4 mph

total distance = (2/3 x 6 x 4) + (1/3 x 6 x 4)
total distance = 16 + 8 = 24 miles total

Average speed to top of hill = 2/(1/4 +1/3) = 3.43 mph

to top of hill = 24/2 = 12 miles

12 / 3.43 = 3.5 hours
12 oclock + 3.5 hours = 3:30

total miles was 24
arrived at top of hill at 3:30

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For a certain river, suppose the drought length Y is the number of consecutive time intervals in which the water supply remains

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Answer:

a) There is a 9% probability that a drought lasts exactly 3 intervals.

There is an 85.5% probability that a drought lasts at most 3 intervals.

b)There is a 14.5% probability that the length of a drought exceeds its mean value by at least one standard deviation

Step-by-step explanation:

The geometric distribution is the number of failures expected before you get a success in a series of Bernoulli trials.

It has the following probability density formula:

$f(x) = (1-p)^{x}p$

In which p is the probability of a success.

The mean of the geometric distribution is given by the following formula:

$\mu = \frac{1-p}{p}$

The standard deviation of the geometric distribution is given by the following formula:

$\sigma = \sqrt{\frac{1-p}{p^{2}}$

In this problem, we have that:

$p = 0.383$

$\mu = \frac{1-p}{p} = \frac{1-0.383}{0.383} = 1.61$

$\sigma = \sqrt{\frac{1-p}{p^{2}}} = \sqrt{\frac{1-0.383}{(0.383)^{2}}} = 2.05$

(a) What is the probability that a drought lasts exactly 3 intervals?

This is $f(3)$

$f(x) = (1-p)^{x}p$

$f(3) = (1-0.383)^{3}*(0.383)$

$f(3) = 0.09$

There is a 9% probability that a drought lasts exactly 3 intervals.

At most 3 intervals?

This is $P = f(0) + f(1) + f(2) + f(3)$

$f(x) = (1-p)^{x}p$

$f(0) = (1-0.383)^{0}*(0.383) = 0.383$

$f(1) = (1-0.383)^{1}*(0.383) = 0.236$

$f(2) = (1-0.383)^{2}*(0.383) = 0.146$

Previously in this exercise, we found that $f(3) = 0.09$

$P = f(0) + f(1) + f(2) + f(3) = 0.383 + 0.236 + 0.146 + 0.09 = 0.855$

There is an 85.5% probability that a drought lasts at most 3 intervals.

(b) What is the probability that the length of a drought exceeds its mean value by at least one standard deviation?

This is $P(X \geq \mu+\sigma) = P(X \geq 1.61 + 2.05) = P(X \geq 3.66) = P(X \geq 4)$ .

We are working with discrete data, so 3.66 is rounded up to 4.

Either a drought lasts at least four months, or it lasts at most thee. In a), we found that the probability that it lasts at most 3 months is 0.855. The sum of these probabilities is decimal 1. So:

$P(X \leq 3) + P(X \geq 4) = 1$

$0.855 + P(X \geq 4) = 1$

$P(X \geq 4) = 0.145$

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