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3 years ago

Determine the equation of the graph and select the correct answer below.

Mathematics

1 answer:

Assoli18 [71]3 years ago

5 0

Answer:

y = (x -1)² -3

Step-by-step explanation:

A quadratic with a vertex at (h, k) will have an equation of the form ...

y = a(x -h)² +k

You have (h, k) = (1, -3), and a vertical scale factor* of 1. So, the equation of the graphed curve is ...

y = (x -1)² -3

_____

* One way to determine the value of "a" in the form shown is to look at the vertical difference between the vertex and the points 1 unit right or left of the vertex. Here, those points are 1 unit above the vertex, so the vertical scale factor "a" is 1.

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The crew of a spaceship search for a missing shuttle craft for 1000 miles in every direction from their position how many cubic

Semmy [17]

Answer:

4,188,790,200 cubic miles to the nearest hundred.

Step-by-step explanation:

That will be the volume of an imagined sphere with radius 1000 miles.

= 4/3 * pi * (1000)^3

= 4,188,790,200 cubic miles.

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4 years ago

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3 years ago

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Solve for x. 36x+2=110. show work

BartSMP [9]

Answer:

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4 years ago

Read 2 more answers

Eight less than 7times a number is -29

Alona [7]

7n-8=-29
You got the equation correct.
I'm not sure if you want me to solve n but I'll do it in case you need help on that.
First, add 8 on both sides which would equal 7n=-21.
Now, divide 7 on both sides.
n=-3
Hope this helps

3 0

3 years ago

Refer to the random sample of customer order totals with an average of $78.25 and a population standard deviation of $22.50. a.

zysi [14]

Answer:

a) $78.25- 1.64 \frac{22.50}{\sqrt{40}}= 72.416$

$78.25+ 1.64 \frac{22.50}{\sqrt{40}}= 84.084$

b) $78.25- 1.64 \frac{22.50}{\sqrt{75}}= 73.989$

$78.25+ 1.64 \frac{22.50}{\sqrt{75}}= 82.511$

c) For this case when we increase the sample size the margin of error would be lower and then the interval would be narrower

d) $ME=z_{\alpha/2}\frac{\sigma}{\sqrt{n}}$ (a)

Solving for n we got:

$n=(\frac{z_{\alpha/2} \sigma}{ME})^2$ (b)

And replacing the info we have:

$n=(\frac{1.640(22.50)}{5})^2 =54.46 \approx 55$

Step-by-step explanation:

Part a

For this case we have the following data given

$\bar X = 78.25$ represent the sample mean for the customer order totals

$\sigma =22.50$ represent the population deviation

$n= 40$ represent the sample size selected

The confidence level is 90% or 0.90 and the significance level would be $\alpha=0.1$ and $\alpha/2 = 0.05$ and the critical value from the normal standard distirbution would be given by:

$z_{\alpha/2}=1.64$

And the confidence interval is given by:

$\bar X -z_{\alpha/2} \frac{\sigma}{\sqrt{n}}$

And replacing we got:

$78.25- 1.64 \frac{22.50}{\sqrt{40}}= 72.416$

$78.25+ 1.64 \frac{22.50}{\sqrt{40}}= 84.084$

Part b

The sample size is now n = 75, but the same confidence so the new interval would be:

$78.25- 1.64 \frac{22.50}{\sqrt{75}}= 73.989$

$78.25+ 1.64 \frac{22.50}{\sqrt{75}}= 82.511$

Part c

For this case when we increase the sample size the margin of error would be lower and then the interval would be narrower

Part d

The margin of error is given by:

$ME=z_{\alpha/2}\frac{\sigma}{\sqrt{n}}$ (a)

Solving for n we got:

$n=(\frac{z_{\alpha/2} \sigma}{ME})^2$ (b)

And replacing the info we have:

$n=(\frac{1.640(22.50)}{5})^2 =54.46 \approx 55$

3 0

3 years ago