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3 years ago

What is an equation for the line that passes through points (8, -3) and (9, 10)

Mathematics

1 answer:

34kurt3 years ago

3 0

Answer:

y=13x-107

Step-by-step explanation:

(10- -3)/(9-8)=13

10=13*9+b

b=10-117=-107

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For the write a story just make your own problem and answer it idk wut the the other one is

zavuch27 [327]

Answer:

1.) Alex and Jemma decided that they will separate the dozen doughnuts that they bought. Since Jemma loved doughnuts she got 7, and Alex got 5.

2.) 35:49

Divide 35 and 5.

35/5=7

Multiply 7 by 7.

7 x 7=49.

The missing side will be 35:49

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3 years ago

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dimaraw [331]

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3 years ago

Classify the following triangle. Check all that apply.

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3 years ago

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Type the correct answer in each box. Use numerals instead of words. Solve the equation by completing the square. x2 + 20x + 82 =

VikaD [51]

7=x^2+20x+82
0=x^2+20x+75

What adds to 20 and multiplies to 75? 5 and 15.

x^2+5x+15x+75
x(x+5) 15(x+5)
(x+15) (x+5)

Zeroes are x= -5 and x= -15

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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:

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

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

$P(X \geq 4) = 0.145$

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