Write a linear Write a linear function f with f(−9)=10 and f(−1)=−2.function f with f(−9)=10 and f(−1)=−2.

Find the value of x for m (arc AB)=23 and m(arc CD)=32 (the answer will be a decimal )

Marta_Voda [28]

Ok Ok Ok Ok Ok Ok Ok Ok Ok

5 0

3 years ago

For the equation, complete the solution. 8x + y = −7 <br> (x, y) = , 1

butalik [34]

Answer:

(x, y) = (-1, 1)

Step-by-step explanation:

8x + y = −7

for y = 1

8x + 1 = -7

subtract 1 from boh sides

8x = -8

divide bot sides by 8

x = -1

(x, y) = (-1, 1)

7 0

3 years ago

Item 16 A sphere has a radius of 8 centimeters. A second sphere has a radius of 2 centimeters. What is the difference of the vol

Zigmanuir [339]

Answer:

$672\pi \text{ cm}^3$ .

Step-by-step explanation:

We have been given that a sphere has a radius of 8 centimeters. A second sphere has a radius of 2 centimeters. We are asked to find the difference of the volumes of the spheres.

We will use volume formula of sphere to solve our given problem.

$\text{Volume of sphere}=\frac{4}{3}\pi r^3$ , where r is radius of sphere.

The difference of volumes would be volume of larger sphere minus volume of smaller sphere.

$\text{Difference of volumes}=\frac{4}{3}\pi(\text{8 cm})^3-\frac{4}{3}\pi(\text{2 cm})^3$

$\text{Difference of volumes}=\frac{4}{3}\pi(512)\text{ cm}^3-\frac{4}{3}\pi(8)\text{ cm}^3$

$\text{Difference of volumes}=\frac{4}{3}\pi(512-8)\text{ cm}^3$

$\text{Difference of volumes}=4\pi(168)\text{ cm}^3$

$\text{Difference of volumes}=672\pi\text{ cm}^3$

Therefore, the difference between volumes of the spheres is $672\pi \text{ cm}^3$ .

3 0

3 years ago

Michael opens a bag of Skittles and organizes them by color. He has 5 yellow, 6 green, 8

ELEN [110]

Answer:

I think 8:32 because there are 8 oranges and 32 in total

5 0

3 years ago

Read 2 more answers

Individuals filing federal income tax returns prior to March 31 had an average refund of $1102. Consider the population of "last

lions [1.4K]

Answer:

a) The null hypothesis states that the last-minute filers average refund is equal to the early filers refund. The alternative hypothesis states that the last-minute filers average refund is less than the early filers refund.

$H_0: \mu=1102\\\\H_a:\mu < 1102$

b) P-value = 0.0055

c) The null hypothesis is rejected.

There is enough statistical evidence to support the claim that that the refunds of the individuals that wait until the last five days to file their returns is on average lower than the early filers refund.

d) Critical value tc=-1.96.

As t=-2.55, the null hypothesis is rejected.

Step-by-step explanation:

We have to perform a hypothesis test on the mean.

The claim is that the refunds of the individuals that wait until the last five days to file their returns is on average lower than the early filers refund ($1102).

a) The null hypothesis states that the last-minute filers average refund is equal to the early filers refund. The alternative hypothesis states that the last-minute filers average refund is less than the early filers refund.

$H_0: \mu=1102\\\\H_a:\mu < 1102$

b) The sample has a size n=600, with a sample refund of $1050 and a standard deviation of $500.

We can calculate the z-statistic as:

$t=\dfrac{\bar x-\mu}{s/\sqrt{n}}=\dfrac{1050-1102}{500/\sqrt{600}}=\dfrac{-52}{20.41}=-2.55$

The degrees of freedom are df=599

$df=n-1=600-1=599$

The P-value for this test statistic is:

$P-value=P(t$

c) Using a significance level α=0.05, the P-value is lower than the significance level, so the effect is significant. The null hypothesis is rejected.

There is enough statistical evidence to support the claim that that the refunds of the individuals that wait until the last five days to file their returns is on average lower than the early filers refund.

d) If the significance level is α=0.025, the critical value for the test statistic is t=-1.96. If the test statistic is below t=-1.96, then the null hypothesis should be rejected.

This is the case, as the test statistic is t=-2.55 and falls in the rejection region.

4 0

3 years ago