All categories

3 years ago

How is the graph of y = (x minus 1) squared minus 3 transformed to produce the graph of y = one-half (x + 4) squared?

Mathematics

2 answers:

Valentin [98]3 years ago

6 0

Answer:

Step-by-step explanation:

edgenuity 2020

hope this helps!

san4es73 [151]3 years ago

3 0

Step-by-step explanation:

y = (x − 1)² − 3 → y = ½ (x + 4)²

The graph is shifted 3 units up, 5 units left, and scaled vertically by ½.

You might be interested in

I am pretty sure I know the answer but I need to be certain. Please answer and give explanation if possible.

creativ13 [48]

Answer:

Vertical opposite angles are the same so the answer is D.

8 0

3 years ago

Jimmy loaded three bags of sand into his wheelbarrow. The bags weighed 24

Rudik [331]

The whole numbers add up to 100. So now we are left with 1/2 or 0.5, 3/5 or .60, and 1/8 or .125. Those add up to 1.225 so the amount of sand in pounds in the wheelbarrow is 101.225. Hope I helped! If you have any questions just leave a comment.

3 0

3 years ago

Factor Зу2 - 4у.<br> y(4 - Зу)<br> y(Зу - 4)<br> y(4y-3)

gogolik [260]

Y(3y-4)
The y can be factored out of both equations

5 0

3 years ago

An automobile manufacturer would like to know what proportion of its customers are not satisfied with the service provided by th

butalik [34]

Answer:

a) A sample size of 5615 is needed.

b) 0.012

Step-by-step explanation:

In a sample with a number n of people surveyed with a probability of a success of $\pi$ , and a confidence level of $1-\alpha$ , we have the following confidence interval of proportions.

$\pi \pm z\sqrt{\frac{\pi(1-\pi)}{n}}$

In which

z is the zscore that has a pvalue of $1 - \frac{\alpha}{2}$ .

The margin of error is:

$M = z\sqrt{\frac{\pi(1-\pi)}{n}}$

99.5% confidence level

So $\alpha = 0.005$ , z is the value of Z that has a pvalue of $1 - \frac{0.005}{2} = 0.9975$ , so $Z = 2.81$ .

(a) Past studies suggest that this proportion will be about 0.2. Find the sample size needed if the margin of the error of the confidence interval is to be about 0.015.

This is n for which M = 0.015.

We have that $\pi = 0.2$

$M = z\sqrt{\frac{\pi(1-\pi)}{n}}$

$0.015 = 2.81\sqrt{\frac{0.2*0.8}{n}}$

$0.015\sqrt{n} = 2.81\sqrt{0.2*0.8}$

$\sqrt{n} = \frac{2.81\sqrt{0.2*0.8}}{0.015}$

$(\sqrt{n})^{2} = (\frac{2.81\sqrt{0.2*0.8}}{0.015})^{2}$

$n = 5615$

A sample size of 5615 is needed.

(b) Using the sample size above, when the sample is actually contacted, 12% of the sample say they are not satisfied. What is the margin of the error of the confidence interval?

Now $\pi = 0.12, n = 5615$ .