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3 years ago

I’m trying to correct this problem but I don’t know how can someone help me

Mathematics

1 answer:

BartSMP [9]3 years ago

6 0

Remark

It is not a straight line distance from the park to the mall. None of the answers give you that result. And if you know what displacement is, none of the answers are really displacement either. The distance is sort of a "as the crow flies." distance. There's a stop off in the middle of town.

Method

You need to use the Pythagorean Formula twice -- once from the park to the city Center and once from the city center to the mall.

Distance from the Park to the city center.

a = 3 [distance east]

b = 4 [distance south]

c = ??

c^2 = 3^2 + 4^2 Take the square root of both sides.

c = sqrt(3^2 + 4^2)

c = sqrt(9 + 16) Add

c = sqrt(25)

c = 5

So the distance from the park to the city center is 5 miles

Distance from City center to the mall

a = 2 miles [distance east]

b = 2 miles [distance north]

c = ??

c^2 = a^2 + b^2 Substitute

c^2 = 2^2 + 2^2 Expand this.

c^2 = 4 + 4

c^2 = 8 Take the square root of both sides.

sqrt(c^2) = sqrt(8)

c = sqrt(8) This is the result

c = 2.8

Answer

Total distance = 5 + 2.8 = 7.8

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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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