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3 years ago

Find the GCF of the monomials: 72x^3 y^2 and 210x^2y^5

Mathematics

2 answers:

denpristay [2]3 years ago

4 0

Answer: The GCF of the monomials is 1.

Elden [556K]3 years ago

3 0

For this case we have by definition, that the Greatest Common Factor or GFC, of two or more whole numbers is the largest integer that divides them without leaving a residue.

Now, we look for the factors of 72 and 210:

72: 1,2,3,4,6,8,9,12,16,24,36,72

210: 1,2,3, 5,6,7 ...

Thus, the GFC of both is 6.

Then, the GFC of $72x ^ 3y ^ 2$ and $210x ^ 2y ^ 5$ is given by:

$6x ^ 2y ^ 2$

ANswer:

$6x ^ 2y ^ 2$

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Part A

Given info:

xbar = sample mean = 4.90 ppm
s = sample standard deviation = 1.12 ppm
n = 23 = sample size

Because n > 30 is not true, and we don't know the population standard deviation (sigma), this means we must use a T distribution.

The degrees of freedom here are n-1 = 23-1 = 22.

At 90% confidence and the degrees of freedom mentioned, the t critical value is roughly t = 1.717

Use a T distribution table or calculator to determine this. If you don't have a calculator for the task, then you can search out "inverse T calculator" and there are tons of free options to pick from.

The margin of error E is

E = t*s/sqrt(n)

E = 1.717*1.12/sqrt(23)

E = 0.400982

This is approximate and accurate to 6 decimal places.

The confidence interval is going to be xbar plus or minus that E value

L = lower bound = xbar - E = 4.90 - 0.400982 = 4.499018 = 4.50

U = upper bound = xbar + E = 4.90 + 0.400982 = 5.300982 = 5.30

The confidence interval in the format of (L, U) is (4.50, 5.30)

You could also express it as the format L < mu < U and it would be 4.50 < mu < 5.30; however, I'll stick to the first method.

<h3>Answer: (4.50, 5.30)</h3>

=====================================================

Part B

Since we know sigma = 2.6 is the population standard deviation, we can use a Z distribution now.

At 90% confidence, the z critical value is roughly 1.645; use a table or calculator to determine this.

$n = \left(\frac{z*\sigma}{E}\right)^2\\\\n \approx \left(\frac{1.645*2.6}{0.03}\right)^2\\\\n \approx 20325.254444 \\\\$

Round this up to the nearest integer to get 20326. For min sample size problems, <u>always</u> round up.

<h3>Answer: 20326</h3>

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