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klio [65]
3 years ago
9

How many terms are in the expression below? 17 – (5n = 4) + 8%

Mathematics
1 answer:
tankabanditka [31]3 years ago
7 0

Answer: 4

Step-by-step explanation:

(Can someone please answer my math question)

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Some day help me with this 2 question?
gregori [183]

Answer:

As x approaches 0, f(x)=2

Step-by-step explanation:

First plug in f(x) and f(0) into the limit.

\lim_{x \to 0\0} 2x^2+1-(2(0^2)+1)/x^2\\\\, simplify things a bit

\lim_{x \to \ 0} 2x^2/x^2,   divide by x^2

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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)

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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