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Assoli18 [71]
3 years ago
12

A computer room has 12 computers. The room is open for 4 hours each day. 25 students sigh up for computer time. Each student get

s the same number of minutes. What is the graetest whole number of minutes each student can get?
Mathematics
1 answer:
GuDViN [60]3 years ago
8 0

Answer:

155 minutes

Step-by-step explanation:

1. Find the total available "computer time": 4 hours * 12 computers= 48 hours

2. Split this among all of the students: 48 hours/25 students= 1.92 hours

3. Convert to minutes: 1.92 hr/student * 60 minutes ≈ 155 minutes/student

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The mean annual tuition and fees for a sample of 15 private colleges was with a standard deviation of . A dotplot shows that it
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Answer:

Step-by-step explanation:

The question is incomplete. The complete question is:

The mean annual tuition and fees for a sample of 15 private colleges was $35,500 with a standard deviation of $6500. A dotplot shows that it is reasonable to assume that the population is approximately normal. You wish to test whether the mean tuition and fees for private colleges is different from $32,500. State the null and alternate hypotheses. A) H0: 4 = 32,500, H:4=35,500 C) H: 4 = 35,500, H7:35,500 B) H: 4 = 32,500, H : 4 # 32,500 D) H0:41 # 32,500, H : 4 = 32,500

Solution

We would set up the hypothesis test. This is a test of a single population mean since we are dealing with mean

For the null hypothesis,

H0: µ = 32500

For the alternative hypothesis,

Ha: µ ≠ 32500

This is a two tailed test.

Since the number of samples is small and the population standard deviation is not given, the distribution is a student's t.

Since n = 15,

Degrees of freedom, df = n - 1 = 15 - 1 = 14

t = (x - µ)/(s/√n)

Where

x = sample mean = 35500

µ = population mean = 32500

s = samples standard deviation = 6500

t = (35500 - 32500)/(6500/√15) = 1.79

We would determine the p value using the t test calculator. It becomes

p = 0.095

Assuming alpha = 0.05

Since alpha, 0.05 < than the p value, 0.095, then we would fail to reject the null hypothesis.

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