# Which equation represents a line that passes through -2,4 and has a slope of 2/5 need help Now

y = 2/5x + 24/5

Step-by-step explanation:

y - 4 = 2/5(x - (-2))

y - 4 = 2/5(x +2)

y - 4 = 2/5x + 4/5

y = 2/5x + 4/5 + 4

y = 2/5x + 24/5

## Related Questions

Find a generating function for the sequence of squares: g(x) = P∞ n=0 n 2x n . Then for fun, as above evaluate your expression g(1/100) or g(1/1000) to get a fraction that contains the squares in its decimal expansion.

The sequence of squares, , has generating function

Recall that for ,

Taking the derivative, we have

and taking the derivative again, we have

From this we can get an expression for in terms of the derivatives of :

Then

let d equal the distance in meters and t equal the time in seconds. Which is a direct variation equation for this relationship

d = s x t

Step-by-step explanation:

The formula for distance.

EXAMPLE: If the function f(x) = 2x==x+4 is restricted to a domain interval of
-4 sxs 4, is the domain over the interval continuous or discrete,
and what is the range of the function?

The range of the given function in interval notation is .

The given function is and the domain interval is .

The domain and range are defined for a relation and they are the sets of all the x-coordinates and all the y-coordinates of ordered pairs respectively.

Substitute x=4 in the given function, we get

Substitute x=-4 in the given function, we get

So, the range is

Therefore, the range of the given function is .

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Step-by-step explanation:

"The municipal transportation authority determined that 58% of all drivers were speeding along a busy street. In an attempt to reduce this percentage, the city put up electronic speed monitors so that drivers would be warned if they were driving over the speed limit. A follow-up study is now planned to see if the speed monitors were effective. The null and alternative hypotheses of the test are H0 : π = 0.58 versus Ha : π < 0.58. It is planned to use a sample of 150 drivers taken at random times of the week and the test will be conducted at the 5% significance level. (a) What is the most number of drivers in the sample that can be speeding and still have them conclude that the alternative hypothesis is true? (b) Suppose the true value of π is 0.52. What is the power of the test? (c) How could the researchers modify the test in order to increase its power without increasing the probability of a Type I Error?"

a) X=77 drivers

b) Power of the test = 0.404

c) Increasing the sample size.

Step-by-step explanation:

This is a hypothesis test of proportions. As the claim is that the speed monitors were effective in reducing the speeding, this is a left-tail test.

For a left-tail test at a 5% significance level, we have a critical value of z that is zc=-1.645. This value is the limit of the rejection region. That means that if the test statistic z is smaller than zc=-1.645, the null hypothesis is rejected.

The proportion that would have a test statistic equal to this critical value can be expressed as:

The standard error of the proportion is:

Then, the proportion is:

This proportion, with a sample size of n=150, correspond to

The power of the test is the probability of correctly rejecting the null hypothesis.

The true proportion is 0.52, but we don't know at the time of the test, so the critical value to make a decision about rejecting the null hypothesis is still zc=-1.645 corresponding to a critical proportion of 0.51.

Then, we can say that the probability of rejecting the null hypothesis is still the probability of getting a sample of size n=150 with a proportion of 0.51 or smaller, but within a population with a proportion of 0.52.

The standard error has to be re-calculated for the new true proportion:

Then, we calculate the z-value for this proportion with the true proportion:

The probability of getting a sample of size n=150 with a proportion of 0.51 or lower is:

Then, the power of the test is β=0.404.

The only variable left to change in the test in order to increase the power of the test is the sample size, as the significance level can not be changed (it is related to the probability of a Type I error).

It the sample size is increased, the standard error of the proprotion decreases. As the standard error tends to zero, the critical proportion tend to 0.58, as we can see in its equation:

Then, if the critical proportion increases, the z-score increases, and also the probability of rejecting the null hypothesis.

F(x)=3x-7 and g(x)=(1/3)x+7 are inverses of each other.

.True
.False

False

Step-by-step explanation:

Sorry for the lat reply hopefully you still have that question ready. But basically in order for these equations to be considered inverses of one another it has to map its domain value and switch it to the range value and in this case it does not match the inverse when graphed.

The functions f(x) = 3x - 7 and g(x) = (1/3)x + 7 are inverses of each other.

### Explanation:

Two functions are inverses of each other if the composition of the functions results in the identity function. To check if f(x) = 3x - 7 and g(x) = (1/3)x + 7 are inverses, we need to find their composition.

Let's substitute g(x) into f(x) and simplify: f(g(x)) = f((1/3)x + 7) = 3((1/3)x + 7) - 7 = x + 7 - 7 = x.

Since f(g(x)) = x, it means that f(x) and g(x) are inverses of each other, and therefore the statement is True.

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In tilapia, an important freshwater food fish from Africa, the males actively court females. They have more incentive to court a female who has already laid all of her eggs, but can they tell the difference? an experiment was done to measure the male tilapia's response to the smell of female fish. Water containing feces from females that were either pre-ovulatory (they still had eggs) or post-ovulatory (they had already laid their eggs) was washed over the gills of males hooked up to an electro-olfactogram machine which measured when the senses of the males were excited. The amplitude of the electro-olfactogram was used as a measure of the excitability of the males in the two different circumstances. Six males were exposed to the scent of pre-ovulatory females; their readings average 1.51 with a standard deviation of .25. Six different males were exposed to post-ovulatory females; their average readings of 0.87 with standard deviation is .31. Assume that the electro-olfactogram readings were approximately normally distributed within the groups.(A) test for a difference in the excitability of the males with exposure to these two types of females
(B) what is the estimated average difference in electro-olfactogram readings between the two groups? What is the 95% confidnece limit for the difference between population means?

a)

"=T.INV(1-0.025,10)", and we got

Statistical decision

Since our calculated value is higher than our critical value,, we have enough evidence to reject the null hypothesis at 5% of significance.

b)

The degrees of freedom are given:

Step-by-step explanation:

Part a

Data given and notation

represent the mean for scent of pre ovulatory

represent the mean for post ovolatory

represent the sample standard deviation for preovulatory

represent the sample standard deviation for postovulatory

sample size for the group preovulatory

sample size for the group postovulatory

z would represent the statistic (variable of interest)

represent the p value

Concepts and formulas to use

We need to conduct a hypothesis in order to check if the mean's are different, the system of hypothesis would be:

H0:

H1:

If we analyze the size for the samples both are lower than 30, so for this case is better apply a t test to compare means, and the statistic is given by:

(1)

z-test: Is used to compare group means. Is one of the most common tests and is used to determine whether the means of two groups are equal to each other.

Calculate the statistic

We have all in order to replace in formula (1) like this:

Find the critical value

We find the degrees of freedom:

In order to find the critical value we need to take in count that we are conducting a two tailed test, so we are looking for thwo values on the t distribution with df =10 that accumulates 0.025 of the area on each tail. We can us excel or a table to find it, for example the code in Excel is:

"=T.INV(1-0.025,10)", and we got

Statistical decision

Since our calculated value is higher than our critical value,, we have enough evidence to reject the null hypothesis at 5% of significance.

Part b

For this case the confidence interval is given by:

The degrees of freedom are given: