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shtirl [24]
3 years ago
13

Help me please!..............

Mathematics
1 answer:
sladkih [1.3K]3 years ago
5 0
What's the problem you want to ask
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Miss watson runs a disatnce of 200 metres in 25 seconds.
iogann1982 [59]

Answer:

Step-by-step explanation:

ANSWER:

average = total distance / time

average = 200/25 = 8 m /s

8 0
2 years ago
(-102.75) ÷(-3) round to the nearest hundredth
astra-53 [7]

Answer:34.17

Step-by-step explanation:

(-102.5) ➗ (-3)=34.17

6 0
3 years ago
Read 2 more answers
8.6 mi<br> 18 mi<br> A) 53.2 mi?<br> C) 106.4 mi?<br> E) 212.8 mi?<br> B) 221 mi?<br> D) 110.5 mi?
GenaCL600 [577]

Answer:

C

Step-by-step explanation:

4 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
Please help ASAP! Will give BRAINLIEST! Please read the question THEN answer correctly! No guessing.
wlad13 [49]

Answer:

B

Step-by-step explanation:

7 0
3 years ago
Read 2 more answers
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