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Paraphin [41]
3 years ago
13

In a random sample of 63 women at a company, the mean salary is $48,902 with a standard deviation of $5270. in a rando sample of

50 men at the company the mean salary is $53,454 with a standard deviation of $4677. Which interval in the 95% confidence interval for the difference between the mean salaries of all women and men at the company?
A. 3614.81 , 5489.19
B. 2715.11 , 6388.89
C. 4083,40 , 5020.60
D. 2134.05 , 6969.95
Mathematics
1 answer:
sdas [7]3 years ago
4 0
So u would make at least 101,000 per year
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Answer:

The answer is 225

Step-by-step explanation:

Setup a ratio saying 150:2/3

Next divide 2/3 by 1 to see how many times it can go into it which is 1 1/2

After that multiply 1 1/2 by 150

Answer is 225.

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f the price charged for a candy bar is​ p(x) cents, where p (x )equals 162 minus StartFraction x Over 10 EndFraction ​, then x t
andreev551 [17]

Answer:

a. 1620-x^2

b. x=810

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Step-by-step explanation:

(a) Total revenue from sale of x thousand candy bars

P(x)=162 - x/10

Price of a candy bar=p(x)/100 in dollars

1000 candy bars will be sold for

=1000×p(x)/100

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x thousand candy bars will be

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=10( 1620-x/10) * x

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=1620x-x^2

R(x)=1620x-x^2

(b) Value of x that leads to maximum revenue

R(x)=1620x-x^2

R'(x)=1620-2x

If R'(x)=0

Then,

1620-2x=0

1620=2x

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810=x

x=810

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R(810)=1620x-x^2

=1620(810)-810^2

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4 years ago
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B: m = -6 or m = 5/2

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<u>breakdown</u>

\rightarrow \sf 2m^2 +12m -5m- 30 = 0

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\rightarrow \sf 2m(m +6) -5(m+ 6) = 0

<u>collect into groups</u>

\rightarrow \sf (2m -5)(m+ 6) = 0

<u>set to zero</u>

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<u>change sides</u>

\rightarrow \sf m=2.5, \ m= -6

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