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amid [387]
3 years ago
10

Yumi and Juan pick up litter along the highway at the same rate. Yumi picks up litter for 3 miles and takes two hours longer tha

n Juan, who picks up litter for 2 miles. If it took Juan x hours, how long did it take Yumi? A. 1 hour 20 minutes B. 3 hours 20 minutes C. 4 hours D. 6 hours
Mathematics
2 answers:
MA_775_DIABLO [31]3 years ago
5 0

Answer:

6 HOURS:) ed quiz!

Step-by-step explanation:

oksano4ka [1.4K]3 years ago
3 0
Juan's time = x
Yumi's time = x + 2
Yumi's distance = 3 miles
Juan's distance = 2 miles
Rate = distance/time
Juan's rate = 2/x
Yumi's rate = 3/(x + 2)
The rates are equal so:
2/x = 3/(x + 2)
2x + 4 = 3x
x = 4 hours
Yumi's time = 4 + 2 = 6 hours
The answer is D.
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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:

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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:

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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

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Answer:

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Step-by-step explanation:

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