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Maurinko [17]
3 years ago
10

The population of Oak is increasing at a rate of 2% per year. If the population is 64,242 what will it be in three years answer

Mathematics
1 answer:
Alekssandra [29.7K]3 years ago
4 0

Answer: 68096.52; rounded to 68097

Step-by-step explanation:

64242  x 2/100 (because 2% = 0.02 = 2/100) = 1284.84

So the population of Oak is increasing at a rate of 1284.84 people per year.

But in three years?

1284.84 x 3 = 3854.52.

64242 + 3854.52 = 68096.52.

The population in three years will be 68096.52

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Question 10 of 20
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After a college football team once again lost a game to their archrival, the alumni association conducted a survey to see if alu
valentina_108 [34]

Answer:

P-value for this hypothesis test is 0.00175.

Step-by-step explanation:

We are given that the alumni association conducted a survey to see if alumni were in favor of firing the coach.

A simple random sample of 100 alumni from the population of all living alumni was taken. Sixty-four of the alumni in the sample were in favor of firing the coach.

<u><em>Let p = proportion of all living alumni who favored firing the coach</em></u>

SO, Null Hypothesis, H_0 : p = 0.50   {means that the majority of alumni are not in favor of firing the coach}

Alternate Hypothesis, H_A : p > 0.50   {means that the majority of alumni are in favor of firing the coach}

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  = sample proportion of the alumni in the sample who were in favor of firing the coach = \frac{64}{100} = 0.64

            n = sample of alumni = 100

So, <em><u>test statistics</u></em>  =  \frac{0.64-0.50}{{\sqrt{\frac{0.64(1-0.64)}{100} } } } }

                               =  2.92

<u>Now, P-value of the hypothesis test is given by ;</u>

         P-value = P(Z > 2.92) = 1 - P(Z \leq 2.92)

                                             = 1 - 0.99825 = 0.00175

Therefore, the P-value for this hypothesis test is 0.00175.

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