10q-30q is equivalent to

If a storm window has an area of 640 square inches, what is the equation to find the dimensions when the window is 20 inches hig

Lyrx [107]

Area = width * height
since the window is 20 inches higher than the width:
width +20 = height

640 = width * width+20
640 = width^2 +20*width
width^2 +20*width - 640 = 0
width = 17.203
**********************************************************
Double Check
if width = 17.203 then height = 37.203
area = 17.203 * 37.203 = 640 square inches Correct!!

8 0

3 years ago

Find the appropriate rejection regions for the large-sample test statistic z in these cases. (Round your answers to two decimal

Usimov [2.4K]

Answer:

a) We have that the significance is given by $\alpha =0.01$ and we know that we have a right tailed test.

So for this case we need to look in the normal standard dsitribution a critical value that accumulates 1% of the area on the right and 99% of the area on the left. This value can be founded with the following excel code:

"=NORM.INV(1-0.01,0,1)"

And we got for this case $z_{crit}=2.33$

So then the rejection region would be $z>2.33$

b) We have that the significance is given by $\alpha =0.05$ , $\alpha/2 =0.025$ and we know that we have a two tailed test.

So for this case we need to look in the normal standard dsitribution a critical value that accumulates 2.5% of the area on the right and 97.5% of the area on the left. This value can be founded with the following excel code:

"=NORM.INV(1-0.025,0,1)"

And we got for this case $z_{crit}=\pm 1.96$

So then the rejection region would be $z>1.96 \cup z$

Step-by-step explanation:

Part a

We have that the significance is given by $\alpha =0.01$ and we know that we have a right tailed test.

So for this case we need to look in the normal standard dsitribution a critical value that accumulates 1% of the area on the right and 99% of the area on the left. This value can be founded with the following excel code:

"=NORM.INV(1-0.01,0,1)"

And we got for this case $z_{crit}=2.33$

So then the rejection region would be $z>2.33$

Part b

We have that the significance is given by $\alpha =0.05$ , $\alpha/2 =0.025$ and we know that we have a two tailed test.

So for this case we need to look in the normal standard dsitribution a critical value that accumulates 2.5% of the area on the right and 97.5% of the area on the left. This value can be founded with the following excel code:

"=NORM.INV(1-0.025,0,1)"

And we got for this case $z_{crit}=\pm 1.96$

So then the rejection region would be $z>1.96 \cup z$

7 0

3 years ago

What is the common point between lines y = -x + 7 and y = 3x + 10?

neonofarm [45]

Did you figure it out yet? lol

4 0

2 years ago

Solve tan(x)(tanx+1) =0

Umnica [9.8K]

The answer is "A" for your question.

7 0

2 years ago

Given: (1/3)x + y = 15.

Kitty [74]

I believe its the second one.......the last one is eliminated based on the y intercept......
plug the x and y into the equations and see which side equals.....

3 0

3 years ago

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