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gogolik [260]
3 years ago
13

Someone help me with this

Mathematics
2 answers:
Alex3 years ago
7 0

Answer:

y=7.5x, y=9.25x, y=6x

Step-by-step explanation:

on the first table we have for 2 tikets $ 15 so the price for one ticket (also known as unit price) is 15/2= $ 7.5

the equation for the price of ticket must be

y= 7.5 x (where y is price you pay for x number of tikets)

on the second table you can pick that you have $37 for 4 tikets so unit price is 37/4= $9.25 per one ticket

y=9.25x

last table $12 for 2 tikets so for one ticket is 12/2=$6

y=6x

nikitadnepr [17]3 years ago
5 0
I agree with the person above
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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
Fudgin [204]

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)

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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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