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Zarrin [17]
3 years ago
13

What is the approximate area of a circle with a diameter of 10in

Mathematics
2 answers:
DanielleElmas [232]3 years ago
7 0

Answer:

78.54in²

Step-by-step explanation:

Nutka1998 [239]3 years ago
5 0

Answer:

A≈78.54in²

Step-by-step explanation:

Can I have brainliest pls?

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Jeff earns $27,800 per year. what is his bi-monthly gross pay
Anarel [89]
Below are the choices that can be found elsewhere:

<span>a. $1,158.33
b. $1,069.23
c. $2,316.66
d. $1,539.22
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Below is the solution:
$27,800 / 12 months = 2,316
$2,316 / 2 since it bi-monthly = $1,158.33

Thank you for posting your question here. I hope the answer helps. 

3 0
4 years ago
Answer correctly
lara31 [8.8K]

6w-10 represents the total amount of fence will need to complete the rectangle of fence in his yard. The fence will need two lengths and 2 widths to complete the rectangle. If the width is w, and the length is 2w-5, then the total for the rectangle would be w+w+2w-5+2w-5 which simplifies to 6w-10.

7 0
3 years ago
Read 2 more answers
Which of the following ( 15, 20, 30, 57 ) is the product of exactly three different prime numbers?
Andrei [34K]

Answer:

30=2*3*5, so it's 30

Step-by-step explanation:

15=3*5, so it's not 15

20=2*2*5, so it's not 20

30=2*3*5, so it's 30

57=19*3, so it's not 57

3 0
2 years ago
Which of these numbers is the greatest? No links
Ahat [919]

Answer:

c. 1.2 * 10^4

5 0
3 years ago
Read 2 more answers
Assume that a company hires employees on Mondays, Tuesdays, or Wednesdays with equal likelihood.a. If two different employees ar
horsena [70]

Answer:

P(A) = \frac{1}{9}

P(B) = \frac{1}{3}

P(C) = \frac{1}{729}

Step-by-step explanation:

We know that:

Only employees are hired during the first 3 days of the week with equal probability.

2 employees are selected at random.

So:

A. The probability that an employee has been hired on a Monday is:

P(M) = \frac{1}{3}.

If we call P(A) the probability that 2 employees have been hired on a Monday, then:

P(A) =P(M\ and\ M)\\\\P(A)=( \frac{1}{3})(\frac{1}{3})\\\\P(A) = \frac{1}{9}

B. We now look for the probability that two selected employees have been hired on the same day of the week.

The probability that both are hired on a Monday, for example, we know is P(A) = \frac{1}{9}. We also know that the probability of being hired on a Monday is equal to the probability of being hired on a Tuesday or on a Wednesday. But if both were hired on the same day, then it could be a Monday, a Tuesday or a Wednesday.

So

P(B) = \frac{1}{9} + \frac{1}{9} + \frac{1}{9}\\\\P(B) = \frac{1}{3}.

C. If the probability that two people have been hired on a specific day of the week is 3(\frac{1}{3}) ^ 2, then the probability that 7 people have been hired on the same day is:

P(C) = (\frac{1}{3}) ^ 7 + (\frac{1}{3}) ^ 7 + (\frac{1}{3}) ^ 7\\\\P(C) = \frac{1}{729}

D. The probability is \frac{1}{729}. This number is quite close to zero. Therefore it is an unlikely bastate event.

6 0
3 years ago
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