Solve the Equation (NO LINKS AND ONLY IF YOU KNOW PLS)

Anton [14]

Answer:

y < 4

y > 2x - 4

y > -2x - 4

Step-by-step explanation:

For each line, figure out the equation, and then turn the = into a < or > depending on which side of the line you want to include.

Example, line B has equation y = 2x - 4, and you want to be above it, so y has to be larger than 2x - 4.

The second graph works the same way, try it.

5 0

3 years ago

A magazine provided results from a poll of 500 adults who were asked to identify their favorite pie. Among the 500 respondents,

nataly862011 [7]

Answer:

n=500 represent the random sample taken

$\hat p=0.12$ estimated proportion of people that chose chocolate pie

$\hat q =1-\hat p=1-0.12=0.88$ represent the people that NOT chose chocolate pie

E=0.05 represent the error or margin of error given by the following formula:

$ME=z_{\alpha/2} \sqrt{\frac{\hat p(1-\hat p)}{n}}$

p= true population proportion of people that chose chocolate pie

If the confidence level is 90 %, what is the value of alpha ?

$\alpha=1-0.9 =0.1$ and the value of $\alpha/2 =0.05$ ,

$z_{\alpha/2}=-1.64$ and $z_{1-\alpha/2}=1.64$

$ME=1.64 \sqrt{\frac{0.12(1-0.12)}{500}}=0.0238$

Step-by-step explanation:

Data given and notation

What values do ModifyingAbove p with caret , ModifyingAbove q with caret , n, E, and p represent?

n=500 represent the random sample taken

X represent the people that chose chocolate pie

$\hat p=0.12$ estimated proportion of people that chose chocolate pie

$\hat q =1-\hat p=1-0.12=0.88$ represent the people that NOT chose chocolate pie

E=0.05 represent the error or margin of error given by the following formula:

$ME=z_{\alpha/2} \sqrt{\frac{\hat p(1-\hat p)}{n}}$

z would represent the quantile of the normal standard distribution

p= true population proportion of people that chose chocolate pie

The confidence interval for the population proportion is given by this formula :

$\hat p \pm z_{\alpha/2} \sqrt{\frac{\hat p(1-\hat p)}{n}}$

If the confidence level is 90 %, what is the value of alpha ?

On this case the value for the significance would be $\alpha=1-0.9 =0.1$ and the value of $\alpha/2 =0.05$ , we can find the quantiles of the normal standard distribution given by:

$z_{\alpha/2}=-1.64$ and $z_{1-\alpha/2}=1.64$

And with the following excel codes:

"=NORM.INV(0.05,0,1)" "=NORM.INV(1-0.05,0,1)"

And we can find the margin of error like this:

$ME=1.64 \sqrt{\frac{0.12(1-0.12)}{500}}=0.0238$

8 0

3 years ago

Determine the average rate of change of f(x)=2x2−11 over the interval [−3, 3] .

Kazeer [188]

The average rate of change of the function f(x) = $2x^2-11$ over the interval [-3, 3] is 0

The function is

f(x) = $2x^2-11$

The average rate of change = $\frac{f(b)-f(a)}{b-a}$

First we have to find each values and substitute the values in the equation.

The interval = [-3, 3]

The value of a = -3

The value of b = 3

To find the value of f(-3), substitute the value of x = -3 in the given function

f(-3) = $2(-3)^2-11$

= 2×9 - 11

= 18-11

= 7

The value of f(b) = $2(3)^2-11$

= 2×9 - 11

= 18-11

= 7

The average rate of change = $\frac{7-7}{3--3}$

= 0/6

= 0

Hence, the average rate of change of the function f(x) = $2x^2-11$ over the interval [-3, 3] is 0

Learn more about average rate of change here

brainly.com/question/23715190

#SPJ1

8 0

1 year ago

Morgan makes and sells bread at a local market. Her

il63 [147K]

0.80x-30=58 x=110 the answer

7 0

3 years ago

What do you do when a dog is ceying

Ugo [173]

You mean crying?

Well, there's a wide range of reasons why the dog is whimpering.
I'd personally contact a vet or do some research.
Although, it may be anxiety or somethiing.

6 0

3 years ago