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liraira [26]
3 years ago
11

a store has a 20% off sale on pants with this discount, the price of one pair of pants before tax is $15.20. what was the origin

al price of the pants
Mathematics
1 answer:
anastassius [24]3 years ago
7 0

Answer:

19$

Step-by-step explanation:

In this problem, we have to apply the rule of three. Also, we must have present that $15.20 is the 80% of the price, because the store offers a 20% of discount. So, if $15.20 is 80%, how much would be the 100%?

Therefore, the original price of the pants is $19.

Remember that we must identified the right proportion to apply the rule of three, in this case, that proportion is the 80% with 15.20, that's the reason or fraction center of the whole operation.

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Answer:

.324 or 18.9 degrees

Step-by-step explanation:

rule is

sine equals opposite over hypotenuse,

cosine equals adjacent over hypotenuse, and

tangent equals opposite over adjacent

OR

soh cah toa

sine = soh = 12/37 = .324

arcsin(.324) or sin^-1(.324) in DEGREES not radians =

18.9 degrees

6 0
3 years ago
Simplify. 5(x - y) - (x - y) <br> a. x - y <br> b. x y <br> c. 4x - 4y <br> d. 4x 4y
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C is the answer hope this helps. 
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Ricardo has read two thirds of a book and Lynn has read 2/4 of the same book who read more
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manufacturing company produces digital cameras and claim that their products maybe 3% defective. A video company, when purchasin
alexdok [17]

Answer:

P(X>17) = 0.979

Step-by-step explanation:

Probability that a camera is defective, p = 3% = 3/100 = 0.03

20 cameras were randomly selected.i.e sample size, n = 20

Probability that a camera is working, q = 1 - p = 1 - 0.03 = 0.97

Probability that more than 17 cameras are working P ( X > 17)

This is a binomial distribution P(X = r) nCr q^{r} p^{n-r}

nCr = \frac{n!}{(n-r)!r!}

P(X>17) = P(X=18) + P(X=19) + P(X=20)

P(X=18) = 20C18 * 0.97^{18} * 0.03^{20-18}

P(X=18) = 20C18 * 0.97^{18} * 0.03^{2}

P(X=18) = 0.0988

P(X=19) = 20C19 * 0.97^{19} * 0.03^{20-19}

P(X=19) = 20C19 * 0.97^{19} * 0.03^{1}

P(X=19) = 0.3364

P(X=20) = 20C20 * 0.97^{20} * 0.03^{20-20}

P(X=20) = 20C20 * 0.97^{20} * 0.03^{0}

P(X=20) = 0.5438

P(X>17) = 0.0988 + 0.3364 + 0.5438

P(X>17) = 0.979

The probability that there are more than 17 working cameras should be 0.979 for the company to accept the whole batch

6 0
3 years ago
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