Someone Help Pls . idkvthis

Joe and maxine are playing a game where they flip a fair coin four times and try to predict the outcomes. Maxine thinks that the

dolphi86 [110]

H,t,h,h they are both equal portability because you could get eithed

8 0

3 years ago

How does an earthquake of magnitude 7.9 compare in intensity with an earthquake of magnitude of 3.7 on the Richter scale? The l

Andre45 [30]

We are required to compare the magnitude for two earthquake intensity:
the larger intensity is 7.9
the smaller intensity is 3.7
thus the ratio of the intensity will be:
(larger intensity)/(smaller intensity)
=7.9/3.7
=2.135
~2
Thus we conclude that:
The larger earthquake's intensity was 2 times as great as the smaller earthquake's intensity.

5 0

4 years ago

Aubrey can run at a pace of six miles per hour. running at the same rate how many miles can she run in ninety minutes

Eddi Din [679]

Aubrey would be able to run 9 miles in ninety minutes.

90 minutes=1.5 hours

6*1.5=9

Hope this helps

7 0

3 years ago

While conducting a test of modems being manufactured, it is found that 10 modems were faulty out of a random sample of 367 modem

Kitty [74]

Answer:

We conclude that this is an unusually high number of faulty modems.

Step-by-step explanation:

We are given that while conducting a test of modems being manufactured, it is found that 10 modems were faulty out of a random sample of 367 modems.

The probability of obtaining this many bad modems (or more), under the assumptions of typical manufacturing flaws would be 0.013.

Let p = population proportion.

So, Null Hypothesis, $H_0$ : p = 0.013 {means that this is an unusually 0.013 proportion of faulty modems}

Alternate Hypothesis, $H_A$ : p > 0.013 {means that this is an unusually high number of faulty modems}

The test statistics that would be used here One-sample z-test for proportions;

T.S. = $\frac{\hat p-p}{\sqrt{\frac{p(1-p)}{n} } }$ ~ N(0,1)

where, $\hat p$ = sample proportion faulty modems= $\frac{10}{367}$ = 0.027

n = sample of modems = 367

So, the test statistics = $\frac{0.027-0.013}{\sqrt{\frac{0.013(1-0.013)}{367} } }$

= 2.367

The value of z-test statistics is 2.367.

Since, we are not given with the level of significance so we assume it to be 5%. Now at 5% level of significance, the z table gives a critical value of 1.645 for the right-tailed test.

Since our test statistics is more than the critical value of z as 2.367 > 1.645, so we have sufficient evidence to reject our null hypothesis as it will fall in the rejection region due to which we reject our null hypothesis.

Therefore, we conclude that this is an unusually high number of faulty modems.

6 0

4 years ago

Which fraction is equal to 2/3? 4/7,8/12,6/9

xeze [42]

The answer to the question is 6/9.

3 0

4 years ago