PLEASE HELP ASAP!! RATIONAL EXPRESSION MULTIPLE CHOICE QUESTION!! PLS

madam [21]

Well the slope is -3 and I'm assuming that the y-intercept is 250 because that's where it touches it at all on the graph

Hope this helps. If it does, please rate it as Brainliest

3 0

3 years ago

Suppose we are interested in analyzing the weights of NFL players. We know that on average, NFL players weigh 247 pounds with a

Diano4ka-milaya [45]

Answer:

a) The standard error would be of 8.58 pounds.

b) The margin of error is 14.11 pounds.

c) The 90% confidence interval for the population mean is between 232.89 pounds and 261.11 pounds

Step-by-step explanation:

a.What is the standard error?

The standard error is

$s = \frac{\sigma}{\sqrt{n}}$

In which $\sigma$ is the standard deviation of the population and n is the size of the sample. So

$s = \frac{47}{\sqrt{30}} = 8.58$

The standard error would be of 8.58 pounds.

b.What is the margin of error at 90% confidence?

We have that to find our $\alpha$ level, that is the subtraction of 1 by the confidence interval divided by 2. So:

$\alpha = \frac{1-0.9}{2} = 0.05$

Now, we have to find z in the Ztable as such z has a pvalue of $1-\alpha$ .

So it is z with a pvalue of $1-0.05 = 0.95$ , so $z = 1.645$

Now, find the margin of error M as such

$M = z*s = 1.645*8.58 = 14.11$

The margin of error is 14.11 pounds.

c. Using my sample of 30, what would be the 90% confidence interval for the population mean?

Lower bound: Sample mean subtracted by the margin of error.

247 - 14.11 = 232.89 pounds

Upper bound

247 + 14.11 = 261.11 pounds

The 90% confidence interval for the population mean is between 232.89 pounds and 261.11 pounds

4 0

3 years ago

What are the vertices of A'B'C if ABC is dilated by a scale factor of 2?

malfutka [58]

The vertices of A'B'C' if A(0, 10), B(8, 6), C(4, 2) is dilated by a scale factor of 2 are;

C. A'(0, 20), B'(16, 12), C'(8, 4)

<h3>Which method can be used to find the vertices of A'B'C'?</h3>

Given that triangle ABC is dilated by a scale factor of 2, we have;

A'B' = 2 × AB

A'C' = 2 × AC

B'C' = 2 × BC

Length of AB = √((8-0)^2 + (6-10)^2) = 4•√5

Length of AC = √((4-0)^2 + (2-10)^2) = 4•√5

Length of BC = √((8-4)^2 + (6-2)^2) = 4•√2

By multiplying the given coordinates by the scale factor, we have;

A' = 2 × (0, 10) = (0, 20)

B' = 2 × (8, 6) = (16, 12)

C' = 2 × (4, 2) = (8, 4)

A'B' = √((16-0)^2 + (12-20)^2) = 8•√5

A'C' = √((8-0)^2 + (4-20)^2) = 8•√5

B'C' = √((16-8)^2 + (12-4)^2) = 8•√2

Therefore when we have;

A'(0, 20), B'(16, 12), C'(8, 4), we get;

A'B' = 2 × AB

A'C' = 2 × AC
B'C' = 2 × BC

The correct option is therefore;

C. A'(0, 20), B'(16, 12), C'(8, 4)

Learn more about finding the distance between points on the coordinate plane here:

brainly.com/question/7243416

#SPJ1

5 0

1 year ago

Volume of the cylinder with diameter of 8mm and height of 8mm

alexdok [17]

Answer:

V = 402.12385mm

Step-by-step explanation:

When the diameter is 8mm, the radius is 4mm.

V = (π(4^2))8

V = (16π)8

V = (50.2654)8

V = 402.12385

V ≈ 402, 402.1, 402.12, 402.124

4 0

2 years ago

Since multiplication is implied algebra we often don’t need to actually write the

KengaRu [80]

Ab = 4c
Just drop the ‘x’

8 0

3 years ago