The claim is that for a smartphone carrier's data speeds atairports, the mean= 14.00Mbps. The sample size is equals=27 and the test statistic is t equals=1.386. What is the P-value=

Degree of freedom = sample size -1 = 27-1 = 26

Given t = 1.386

This is a two tailed test.

Using a TI83/84 calculator:

2 x tcdf(1e99,-1.386,26) = 0.177520

The p value = 0.177520 (round the answer as needed)

180

step-by-step explanation:

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answer: d.) the total number of protons and neutrons is 15

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