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spayn [35]
2 years ago
7

A marketing research company is estimating the average total compensation of CEOs in the service industry. Data were randomly co

llected from 18 CEOs and the 99 % confidence interval for the mean was calculated to be Explain what the phrase " 99 % confident" means.
Mathematics
1 answer:
daser333 [38]2 years ago
8 0

Answer:

Step-by-step explanation:

Confidence interval measures the degree of uncertainty and certainty and comprises a range of values that might contain an unknown population parameter. A 99% confidence interval means we can expect that 99% of the 18 samples taken contains the parameter mean values between the upper and the lower bound values from the data collected from the 18 CEOs.

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If f(x) = 3 – 2x and 1/x+5, what is the value of (f/g)(8)? –169 –1 13 104
Basile [38]
I'm assuming that this is the complete question.
If f(x) = 3 – 2x and g(x)=1/(x+5), what is the value of (f/g)(8)? a) –169       b) –1      c) 13     d) 104
x = 8
f(x) = 3 -2xf(8) = 3 - 2(8) = 3 - 16 = -13
g(x) = 1/(x+5)g(8) = 1/(8+5) = 1/13
(f/g)(8)f(8)/g(8) = -13/ (1/13) = -13 * 13 = -169  Choice A :)
8 0
2 years ago
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which of the following is equivalent to 3 sqrt 32x^3y^6 / 3 sqrt 2x^9y^2 where x is greater than or equal to 0 and y is greater
Nutka1998 [239]

Answer:

\frac{\sqrt[3]{16y^4}}{x^2}

Step-by-step explanation:

The options are missing; However, I'll simplify the given expression.

Given

\frac{\sqrt[3]{32x^3y^6}}{\sqrt[3]{2x^9y^2} }

Required

Write Equivalent Expression

To solve this expression, we'll make use of laws of indices throughout.

From laws of indices \sqrt[n]{a}  = a^{\frac{1}{n}}

So,

\frac{\sqrt[3]{32x^3y^6}}{\sqrt[3]{2x^9y^2} } gives

\frac{(32x^3y^6)^{\frac{1}{3}}}{(2x^9y^2)^\frac{1}{3}}

Also from laws of indices

(ab)^n = a^nb^n

So, the above expression can be further simplified to

\frac{(32^\frac{1}{3}x^{3*\frac{1}{3}}y^{6*\frac{1}{3}})}{(2^\frac{1}{3}x^{9*\frac{1}{3}}y^{2*\frac{1}{3}})}

Multiply the exponents gives

\frac{(32^\frac{1}{3}x*y^{2})}{(2^\frac{1}{3}x^{3}*y^{\frac{2}{3}})}

Substitute 2^5 for 32

\frac{(2^{5*\frac{1}{3}}x*y^{2})}{(2^\frac{1}{3}x^{3}*y^{\frac{2}{3}})}

\frac{(2^{\frac{5}{3}}x*y^{2})}{(2^\frac{1}{3}x^{3}*y^{\frac{2}{3}})}

From laws of indices

\frac{a^m}{a^n} = a^{m-n}

This law can be applied to the expression above;

\frac{(2^{\frac{5}{3}}x*y^{2})}{(2^\frac{1}{3}x^{3}*y^{\frac{2}{3}})} becomes

2^{\frac{5}{3}-\frac{1}{3}}x^{1-3}*y^{2-\frac{2}{3}}

Solve exponents

2^{\frac{5-1}{3}}*x^{-2}*y^{\frac{6-2}{3}}

2^{\frac{4}{3}}*x^{-2}*y^{\frac{4}{3}}

From laws of indices,

a^{-n} = \frac{1}{a^n}; So,

2^{\frac{4}{3}}*x^{-2}*y^{\frac{4}{3}} gives

\frac{2^{\frac{4}{3}}*y^{\frac{4}{3}}}{x^2}

The expression at the numerator can be combined to give

\frac{(2y)^{\frac{4}{3}}}{x^2}

Lastly, From laws of indices,

a^{\frac{m}{n} = \sqrt[n]{a^m}; So,

\frac{(2y)^{\frac{4}{3}}}{x^2} becomes

\frac{\sqrt[3]{(2y)}^{4}}{x^2}

\frac{\sqrt[3]{16y^4}}{x^2}

Hence,

\frac{\sqrt[3]{32x^3y^6}}{\sqrt[3]{2x^9y^2} } is equivalent to \frac{\sqrt[3]{16y^4}}{x^2}

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77julia77 [94]

Answer:

$0.85 per tomato

Step-by-step explanation:

A unit rate is the cost for one item. Like $0.99 for this one bag of skittles.

So $3.40 divided by four is $0.85.

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Answer:

A. Given

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D.2 times the measure of angle CBD=90

Step-by-step explanation:

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