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Vinvika [58]
3 years ago
5

A store offers 20% off all items. If x is the original purchase price, which expression represents the final price with the disc

ount? Select all that apply.
A. 1.2x
B. 0.8x
C. x
D. x – 0.2x
E. x + 0.2x
Mathematics
2 answers:
Verizon [17]3 years ago
8 0

Answer:

B & D

Step-by-step explanation:

We use percents in decimal form to multiply it with the price. We convert percents into decimals by dividing the percent number by 100. For example, 78% divided by 100 becomes 0.78.

There are two ways to look at it:

For finding the price we pay during a sale, we focus on the percent we pay. If 22% off is the sale, then we spend 78% or 100-22=78. If 20% off is the sale, then we pay 80% or 0.80. Multiply that by x an unknown price and we have 0.8x.

We can find the percent off by multiplying the price by the percent conversion. So 20% is 0.20. Then subtract it from the original price to find the leftover that we pay. This is x-0.2x.

ivanzaharov [21]3 years ago
5 0

Answer:

B and D

Step-by-step explanation:

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Answer:

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When a linear equation is of the form ...

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Step-by-step explanation:

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6 0
3 years ago
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
How would the factor pairs change if nancy had only 5 posters to arrange?
Romashka [77]

Answer:

Part A) Nancy can arrange the posters in 4 ways

Part B) Nancy can arrange the posters in 2 ways

Step-by-step explanation:

<u><em>The complete question is</em></u>

Nancy has 10 movie posters. She wants to hang them on a wall in equals rows.

A) Find all the ways that she can arrange the posters.

B) How would the factor pairs change if Nancy had only 5 posters to arrange?

Part A) Find all the ways that she can arrange the posters.

we know that

The factors of 10 are

1x10 -----> 1 row of 10 posters

2x5 -----> 2 rows of 5 posters

5x2 -----> 5 rows of 2 posters

10x1 -----> 10 rows of 1 poster

therefore

Nancy can arrange the posters in 4 ways

Part B) How would the factor pairs change if Nancy had only 5 posters to arrange?

we know that

if Nancy had only 5 posters to arrange

then

The factors of 5 are

1x5 -----> 1 row of 5 posters

5x1 -----> 5 rows of 1 poster

therefore

Nancy can arrange the posters in 2 ways

3 0
3 years ago
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