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Liono4ka [1.6K]
3 years ago
14

True or False: The shape and standard deviation of a population distribution of a variable (such as income) can be estimated wit

h a distribution of a sample of sufficient size
Mathematics
1 answer:
nasty-shy [4]3 years ago
8 0
I belv it is  true. do not ask why.

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What is the least common denominator for the rational expressions 1 / x² -5 x-6} and 1/x²-12 x+36 ? Show your work.
enot [183]

The least common denominator for the rational expressions 1/x²-5x-6 and 1/x²-12 x+36 is  (x-6)(x+1)(x-6).

In the given question

=\frac{1}{x^{2} -5x-6} , \frac{1}{x^{2} -12x+36}  ...(i)

First we will factor each denominator

So,

x²-5x-6

On splitting the factors of the above expression , we get

(x+1)(x-6)   ....(ii)

x²-12x+36

On splitting the factors of the above expression , we get

(x-6)(x-6)    ...(iii)

Substituting the value of (ii) and (iii) in (i) ,we get ;

=\frac{1}{(x+1)(x-6)} , \frac{1}{(x-6)(x-6)}

As the LCM of denominator is the Least common denominator , we have

LCM of (x+1)(x-6) and (x-6)(x-6)

LCM=(x-6)(x+1)(x-6)

Hence the LCD is (x-6)(x+1)(x-6).

Therefore , the least common denominator for the rational expressions 1/x²-5x-6 and 1/x²-12 x+36 is  (x-6)(x+1)(x-6).

Learn more about Least Common Denominator here brainly.com/question/11879136

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7 0
1 year ago
A physical therapist wants to determine the difference in the proportion of men and women who participate in regular sustained p
Alekssandra [29.7K]

Using the z-distribution and the formula for the margin of error, it is found that:

a) A sample size of 54 is needed.

b) A sample size of 752 is needed.

In a sample with a number n of people surveyed with a probability of a success of \pi, and a confidence level of \alpha, we have the following confidence interval of proportions.

\pi \pm z\sqrt{\frac{\pi(1-\pi)}{n}}

In which z is the z-score that has a p-value of \frac{1+\alpha}{2}.

The margin of error is of:

M = z\sqrt{\frac{\pi(1-\pi)}{n}}

90% confidence level, hence\alpha = 0.9, z is the value of Z that has a p-value of \frac{1+0.9}{2} = 0.95, so z = 1.645.

Item a:

The estimate is \pi = 0.213 - 0.195 = 0.018.

The sample size is <u>n for which M = 0.03</u>, hence:

M = z\sqrt{\frac{\pi(1-\pi)}{n}}

0.03 = 1.645\sqrt{\frac{0.018(0.982)}{n}}

0.03\sqrt{n} = 1.645\sqrt{0.018(0.982)}

\sqrt{n} = \frac{1.645\sqrt{0.018(0.982)}}{0.03}

(\sqrt{n})^2 = \left(\frac{1.645\sqrt{0.018(0.982)}}{0.03}\right)^2

n = 53.1

Rounding up, a sample size of 54 is needed.

Item b:

No prior estimate, hence \pi = 0.05

M = z\sqrt{\frac{\pi(1-\pi)}{n}}

0.03 = 1.645\sqrt{\frac{0.5(0.5)}{n}}

0.03\sqrt{n} = 1.645\sqrt{0.5(0.5)}

\sqrt{n} = \frac{1.645\sqrt{0.5(0.5)}}{0.03}

(\sqrt{n})^2 = \left(\frac{1.645\sqrt{0.5(0.5)}}{0.03}\right)^2

n = 751.7

Rounding up, a sample of 752 should be taken.

A similar problem is given at brainly.com/question/25694087

5 0
3 years ago
What is the simplified form of the expression √1/121
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\sqrt{a/b} =  \sqrt{a}/\sqrt{b}
B. 1/11
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The ticket office at orchestra center estimates
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Answer:

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Least likely - lemon
Middle - cherry
Most likely - grape
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