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3 years ago

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HELP ME PLEASE!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!1 WILL GIVE BRAINLIEST AND WILL HELP U BACK WITH QUESTIONS!!! I NEED HELP ITS URG

ENT

1 answer:

FrozenT [24]3 years ago

3 0

Shareholders is the answer.

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Given the graph above find an equation for the graph and write your anwser in slope intercept form

lys-0071 [83]

<u>Given:</u>

Given that the graph of a linear function.

We need to determine the equation of the graph in slope intercept form.

<u>Slope</u>:

Consider any two coordinates from the equation of the line to determine the slope.

Let the two coordinates are (-4,3) and (4,-1)

The slope can be determined using the formula,

$m=\frac{y_2-y_1}{x_2-x_1}$

Substituting the coordinates in the above formula, we get;

$m=\frac{-1-3}{4+4}$

$m=\frac{-4}{8}$

$m=-\frac{1}{2}$

Thus, the slope of the equation is $m=-\frac{1}{2}$

<u>Value of y - intercept:</u>

The value of y - intercept is the value of y when x = 0.

Hence, from the graph, it is obvious that the value of y when x = 0 is 1.

Hence, the value of y - intercept is 1.

Thus, y - intercept is $b=1$

<u>Equation of the line:</u>

The equation of the line can be determined using the formula,

$y=mx+b$

Substituting the slope $m=-\frac{1}{2}$ and the y - intercept $b=1$

Thus, we get;

$y=-\frac{1}{2}x+1$

Thus, the equation of the line is $y=-\frac{1}{2}x+1$

5 0

3 years ago

If f(x) varies directly with x and f(x) = 48 when x = 8 find the value of f(x) when x = 2

const2013 [10]

Here, f(x) = kx

k = f(x)/x
k = 48/8
k = 6

So, when x = 2,
f(x) = 6 * 2
z = 12

In short, Your Answer would be: 12

Hope this helps!

6 0

3 years ago

The lengths of pregnancies are normally distributed with a mean of days and a standard deviation of days. a. Find the probabilit

Alik [6]

Answer:

a) The probability of a pregnancy lasting X days or longer is given by 1 subtracted by the p-value of $Z = \frac{X - \mu}{\sigma}$ , in which $\mu$ is the mean and $\sigma$ is the standard deviation.

b) We have to find X when Z has a p-value of $\frac{a}{100}$ , and X is given by: $X = \mu - Z\sigma$ , in which $\mu$ is the mean and $\sigma$ is the standard deviation.

Step-by-step explanation:

Normal Probability Distribution

Problems of normal distributions can be solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$ , the z-score of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the p-value, we get the probability that the value of the measure is greater than X.

In this question:

Mean $\mu$ , standard deviation $\sigma$

a. Find the probability of a pregnancy lasting X days or longer.

The probability of a pregnancy lasting X days or longer is given by 1 subtracted by the p-value of $Z = \frac{X - \mu}{\sigma}$ , in which $\mu$ is the mean and $\sigma$ is the standard deviation.

b. If the length of pregnancy is in the lowest a%, then the baby is premature. Find the length that separates premature babies from those who are not premature.

We have to find X when Z has a p-value of $\frac{a}{100}$ , and X is given by: $X = \mu - Z\sigma$ , in which $\mu$ is the mean and $\sigma$ is the standard deviation.

8 0

3 years ago

Which of the following rational functions is graphed below?

DanielleElmas [232]

Answer:

Step-by-step explanation:

hello :

answer C the odd and positif function

4 0

3 years ago

You can mow one acre in 126 minutes. i can mow one acre in 117 minutes. how long will it take you to mow a 0.17-acre lawn? 13 in

Morgarella [4.7K]

a) (0.17 ac)×(126 min/ac) = 21.42 min

b) I can mow 1/126 ac/min; you can mow 1/117 ac/min. Together, we can mow at a rate that is the sum of these:

... (1/126 + 1/117) ac/min = (117 + 126)/(126×117) = 243/14,742 ac/min

Then it will take 14,742/243 min/ac × 0.17 ac/lawn ≈ 10.31 min/lawn

5 0

3 years ago