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Marizza181 [45]
2 years ago
15

A simple random sample of size individuals who are currently employed is asked if they work at home at least once per week. Of t

he employed individuals​ surveyed, responded that they did work at home at least once per week. Construct a​ 99% confidence interval for the population proportion of employed individuals who work at home at least once per week.
Mathematics
1 answer:
Delvig [45]2 years ago
8 0

The confidence interval of employes individuals from home

Construction of hypothesis:

0.10 - 0.047 < p < 0.10 + 0.047

A simple random sample size n = 250 . Of the 250 employed individuals ​surveyed,42 responded that they did work home a minimum of once per week.

Construct 99% confidence interval for population

For proportion : 42 / 250 = 1/10 = 0.16

Mean = 2.5 * sqrt [ 0.1 * 0.9 / 250]

= 2.5 * 0.01

= 0.47

Construction of hypothesis:

0.10 - 0.047 < p < 0.10 + 0.047

To know more about confidence interval

brainly.com/question/24131141

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What are two factors to add to give negative 4 and multiply to give negative 10?
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Simplify each only using positive exponents:<br> 2x^-3 • 4x^2<br> 2x^4 • 4x^-3<br> 2x^3y^-3 • 2x
erica [24]
<h2>Answer:</h2>

\frac{2}{x}

\frac{x}{2}

\frac{4x^4}{y^3}

<h2>Step-by-step explanation:</h2>

a. 2x^-3 • 4x^2

To solve this using only positive exponents, follow these steps:

i. Rewrite the expression in a clearer form

2x⁻³ . 4x²

ii. The position of the term with negative exponent is changed from denominator to numerator or numerator to denominator depending on its initial position. If it is at the numerator, it is moved to the denominator. If otherwise it is at the denominator, it is moved to the numerator. When this is done, the negative exponent is changed to positive.

In our case, the first term has a negative exponent and it is at the numerator. We therefore move it to the denominator and change the negative exponent to  positive as follows;

\frac{1}{2x^3} . 4x^2

iii. We then solve the result as follows;

\frac{1}{2x^3} . 4x^2 = \frac{2}{x}

Therefore, 2x⁻³ . 4x² = \frac{2}{x}

b. 2x^4 • 4x^-3

i. Rewrite as follows;

2x⁴ . 4x⁻³

ii. The second term has a negative exponent, therefore swap its position and change the negative exponent to a positive one.

2x^4 . \frac{1}{4x^3}

iii. Now solve by cancelling out common terms in the numerator and denominator. So we have;

\frac{x}{2}

Therefore, 2x⁴ . 4x⁻³ = \frac{x}{2}

c. 2x^3y^-3 • 2x

i. Rewrite as follows;

2x³y⁻³ . 2x

ii. Change position of terms with negative exponents;

2x^3.\frac{1}{y^3} .2x

iii. Now solve;

\frac{4x^4}{y^3}

Therefore, 2x³y⁻³ . 2x = \frac{4x^4}{y^3}

8 0
3 years ago
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