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2 years ago

Someone help please, this is due in 10 minutes and I dont know the answer.

Mathematics

1 answer:

Greeley [361]2 years ago

8 0

Answer:

Step-by-step explanation:

Rate of change = the difference in y/the difference in x.

20 - 6.50/4 - 1 = 13.5/3 = 4.5/1

24.5 - 20/5 - 4 = 4.5/1

33.5 - 24.5/7 - 5 = 9/2 = 4.5/1

The rate of change is 4.5.

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3 years ago

What is the slope of the line 3y=6x-1

Allisa [31]

Answer:

Step-by-step explanation:

Simplify 3y=6x+1;

y=2x-1/3

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Slope; 2

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3 years ago

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Andrej [43]

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2 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:

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

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

$P(X \geq 4) = 0.145$

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

8 0

3 years ago

Brenda took out a personal loan for $12,000 at an interest rate of 12% compounded monthly. She made arrangements to pay the loan

iragen [17]

Answer: her monthly payments would be $267

Step-by-step explanation:

We would apply the periodic interest rate formula which is expressed as

P = a/[{(1+r)^n]-1}/{r(1+r)^n}]

Where

P represents the monthly payments.

a represents the amount of the loan

r represents the annual rate.

n represents number of monthly payments. Therefore

a = $12000

r = 0.12/12 = 0.01

n = 12 × 5 = 60

Therefore,

P = 12000/[{(1+0.01)^60]-1}/{0.01(1+0.01)^60}]

12000/[{(1.01)^60]-1}/{0.01(1.01)^60}]

P = 12000/{1.817 -1}/[0.01(1.817)]

P = 12000/(0.817/0.01817)

P = 12000/44.96

P = $267

7 0

2 years ago