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Jlenok [28]
3 years ago
11

The time between a flash of lightning and the sound of its thunder can be used to estimate the distance you are from a lightning

strike. The distance, in miles, from the strike is approximately the number of seconds between seeing the flash and hearing the thunder divided by 5. Write an equation to show this relationship using t for time in seconds and d for distance in miles.
Mathematics
1 answer:
julia-pushkina [17]3 years ago
3 0

If you count the number of seconds between the flash of lightning and the sound of thunder, and then divide by 5, you'll get the distance in miles to the lightning: 5 seconds = 1 mile, 15 seconds = 3 miles, 0 seconds = very close. Keep in mind that you should be in a safe place while counting.

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Help with 9 and 10 and show your work plz and thanks
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Step-by-step explanation:

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salantis [7]

Answer:

No, there is no strong evidence that the percentage of the fleet out of compliance is different from their initial thought.

Step-by-step explanation:

We are given that a company with a fleet of 150 cars found that the emissions systems of only 4 out of the 25 they tested failed to meet pollution control guidelines.

The company initially believed that 30% of the fleet was out of compliance.

<u><em>Let p = percentage of the fleet that was out of compliance.</em></u>

SO, Null Hypothesis, H_0 : p = 30%   {means that the percentage of the fleet out of compliance is same as their initial thought}

Alternate Hypothesis, H_A : p \neq 30%   {means that the percentage of the fleet out of compliance is different from their initial thought}

The test statistics that will be used here is <u>One-sample z proportion</u> <u>statistics</u>;

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

where, \hat p = percentage of the fleet out of compliance = \frac{4}{25} = 16%

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So, <u><em>test statistics</em></u>  =  \frac{0.16-0.30}{{\sqrt{\frac{0.16(1-0.16)}{25} } } } }

                              =  -1.909

The value of the test statistics is -1.909.

<em>Since in the question we are not given with the level of significance at which hypothesis can be tested, so we assume it to be 5%. Now at 5% significance level, </em><u><em>the z table gives critical values between -1.96 and 1.96 for two-tailed test.</em></u><em> </em>

<em>Since our test statistics lies within the range of critical values of z, so we have insufficient evidence to reject our null hypothesis as it will not fall in the rejection region due to which we fail to reject our null hypothesis.</em>

Therefore, we conclude that the percentage of the fleet out of compliance is same as their initial thought.

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