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notsponge [240]
3 years ago
13

Can someone please help me with this real quick

Mathematics
2 answers:
Nesterboy [21]3 years ago
7 0
I would say C, because there is 5 filled in out of 8, which 5/8=62.5%, which is not 75! Correct me if I am wrong.
iren [92.7K]3 years ago
5 0
It would be C

This is because..

A- 75 out of 100 boxes are filled

B- Represents 0.75 also known as 75%

C- Only has 8 boxes, 6 of which are filled in therefore showing that it is 60% of 80

D- is a replica to B but includes a bigger range of numbers, it still shows/represents 75%

Therefore, the answer is C
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Answer:

There are 14 problems worth 5 points and 15 problems worth 2 points.

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1. VIDEOTAPES Lyle is putting his
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12 divided by 1.5 = 8 tapes

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1) Fred Flint is paid $11.95 an hour with time-and-a-half pay for all hours he works over 40
allsm [11]

Fred's overtime rate and pay is $17.925 and $98.5875 respectively

<h3>Total payment</h3>

  • Amount paid per hour = $11.95

  • Amount paid overtime per hour = $11.95 × 1.5

= $17.925

  • Total hours worked last week = 45 1/2 hours

  • Overwork time = 45 1/2 hours - 40 hours

= 5 1/2 hours

Fred's overtime pay = 5 1/2 hours × $17.925

= 11/2 × 17.925

= $98.5875

Normal work rate = 40 hours × $11.95

= $478

Fred's total pay = Fred's overtime pay + Normal work rate

= $98.5875 + $478

= $576.5875

Therefore, Fred's total pay is $576.5875

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5 0
2 years ago
The failure of a circuit board interrupts work that utilizes a computing system until a new board is delivered. The delivery tim
Hatshy [7]

Answer:

The expected cost is 152

Step-by-step explanation:

Recall that since Y is uniformly distributed over the interval [1,5] we have the following probability density function for Y

f_Y(y ) = \frac{1}{5-1} = \frac{1}{4} if 1\leq y \leq 5 and 0 othewise. (To check this is the pdf, check the definition of an uniform random variable)

Recall that, by definition  

E(Y^k) = \int_{-\infty}^{\infty} y^kf_Y(y)dy

Also, we are given that C = 50+3Y+9Y^2. Recall the following properties of the expected value. If X,Y are random variables, then

E(a+bX+cY) = a+bE(X)+cE(Y)

Then, using this property we have that E[C] = 50+3E[Y]+ 9E[Y^2].

Thus, we must calculate E[Y] and E[Y^2].

Using the definition, we get that

E[Y] = \int_{1}^{5}\frac{y}{4} dy =\frac{1}{4}\left\frac{y^2}{2}\right|_{1}^{5} = \frac{25}{8}-\frac{1}{8} = 3

E[Y^2] = \int_{1}^{5}\frac{y^2}{4} dy =\frac{1}{4}\left\frac{y^3}{3}\right|_{1}^{5} = \frac{125}{12}-\frac{1}{12} = \frac{31}{3}

Then

E(C) = 50 + 3\cdot 3 + 9 \cdot \frac{31}{3} = 152

5 0
3 years ago
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