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2 years ago

My teacher says everything is there to answer this question but I don't know. Someone please explain.

1 answer:

5 0

We can start with the basics. What we know is that this entire trip took 6 hours. So the travelers were able to walk on the level, uphill, and downhill in a total of 6 hours.

A level mile takes 1/4 of an hour (15 minutes). An uphill mile takes 1/3 of an hour (20 minutes). A downhill mile takes 1/6 of an hour (10 minutes). So, to go and return over the same mile takes half an hour. This means that in 6 hours, they went 12 miles and came 12 miles back. So they went a total of 24 hours.

If the 12 miles had been all level, it would have taken approximately 3 hours. It the 12 miles had been all uphill, it would have taken approximately 4 hours. When reaching the peak, 3.5 hours would have been the approximate time it took them to reach the top since it is in between. <span />

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Pls help with my geometry!(question 3)

diamong [38]

It’s either 30° or 45° bc of the quality I can’t really see, the trick is look at the corner and see if it’s a right angle, and every right angle is 90° if the line is straight across, it’s 45° if it’s uneven, it’s 30°

7 0

3 years ago

Please answer this question now

shusha [124]

Hi there! :)

Answer:

$\huge\boxed{V = 359.01 mm^{3} }$

Use the formula V = 1/3(bh) to solve for the volume of the cone where b = πr² where π ≈ 3.14:

Find the area of the base:

b = π(7)²

b = 49π

b = 153.86 mm²

Find the volume:

V = 1/3(153.86 · 7)

V = 1/3(1077.02)

V = 359.006 ≈ 359.01 mm³.

4 0

3 years ago

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

AnnZ [28]

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:

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