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Marianna [84]
3 years ago
14

PLEASE ANSWER QUICK A manufacturing facility pays its employees an average wage of $4.50 an hour with a standard deviation of 50

cents. If the wages are normally distributed, what is the percentage of workers getting paid between #3.75 and $5.00 an hour? A. 80.4% B.77.4% C.70.5% D.65.4%
Mathematics
1 answer:
evablogger [386]3 years ago
8 0

Answer:

B.77.4%

Step-by-step explanation:

Mean wage (μ) = $4.50

Standard deviation (σ) = $0.50

For nay given salary X, the z-score is given by:

z=\frac{X-\mu}{\sigma}

For X = $3.75, the z-score is:

z=\frac{3.75-4.50}{0.50}\\z=-1.5

For X = $5.00, the z-score is:

z=\frac{5.00-4.50}{0.50}\\z=1

A z-score of -1.5 corresponds to the 6.68th percentile, while a score of 1 corresponds to the 84.13th percentile. Therefore, the percentage of workers getting paid between $3.75 and $5.00 an hour is:

P=84.13-6.68\\P=77.45\%

The answer is alternative B.77.4%

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A Web music store offers two versions of a popular song. The size of the standard version is 2.8 megabytes (MB). The size of the
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Answer:

Number of high quality version downloads = 880

Step-by-step explanation:

Let,

x be the number of standard version songs downloaded

y be the number of high quality version songs downloaded

According to given statement,

x+y=1600        Eqn 1

2.8x+4.1y=5624     Eqn 2

Multiplying Eqn 1 by 2.8 to eliminate x

2.8(x+y=1600)

2.8x+2.8y=4480     Eqn 3

Subtracting Eqn 3 from Eqn 2

(2.8x+4.1y)-(2.8x+2.8y)=5624-4480

2.8x+4.1y-2.8x-2.8y=1144

1.3y=1144

Dividing both sides by 1.3

\frac{1.3y}{1.3}=\frac{1144}{1.3}\\y=880

Hence,

Number of high quality version downloads = 880

3 0
3 years ago
A rancher wishes to build a fence to enclose a 2250 square yard rectangular field. Along one side the fence is to be made of hea
Bess [88]

Answer:

The least cost of fencing for the rancher is $1200

Step-by-step explanation:

Let <em>x</em> be the width and <em>y </em>the length of the rectangular field.

Let <em>C </em>the total cost of the rectangular field.

The side made of heavy duty material of length of <em>x </em>costs 16 dollars a yard. The three sides not made of heavy duty material cost $4 per yard, their side lengths are <em>x, y, y</em>.  Thus

C=4x+4y+4y+16x\\C=20x+8y

We know that the total area of rectangular field should be 2250 square yards,

x\cdot y=2250

We can say that y=\frac{2250}{x}

Substituting into the total cost of the rectangular field, we get

C=20x+8(\frac{2250}{x})\\\\C=20x+\frac{18000}{x}

We have to figure out where the function is increasing and decreasing. Differentiating,

\frac{d}{dx}C=\frac{d}{dx}\left(20x+\frac{18000}{x}\right)\\\\C'=20-\frac{18000}{x^2}

Next, we find the critical points of the derivative

20-\frac{18000}{x^2}=0\\\\20x^2-\frac{18000}{x^2}x^2=0\cdot \:x^2\\\\20x^2-18000=0\\\\20x^2-18000+18000=0+18000\\\\20x^2=18000\\\\\frac{20x^2}{20}=\frac{18000}{20}\\\\x^2=900\\\\\mathrm{For\:}x^2=f\left(a\right)\mathrm{\:the\:solutions\:are\:}x=\sqrt{f\left(a\right)},\:\:-\sqrt{f\left(a\right)}\\\\x=\sqrt{900},\:x=-\sqrt{900}\\\\x=30,\:x=-30

Because the length is always positive the only point we take is x=30. We thus test the intervals (0, 30) and (30, \infty)

C'(20)=20-\frac{18000}{20^2} = -25 < 0\\\\C'(40)= 20-\frac{18000}{20^2} = 8.75 >0

we see that total cost function is decreasing on (0, 30) and increasing on (30, \infty). Therefore, the minimum is attained at x=30, so the minimal cost is

C(30)=20(30)+\frac{18000}{30}\\C(30)=1200

The least cost of fencing for the rancher is $1200

Here’s the diagram:

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