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ICE Princess25 [194]
3 years ago
15

On wensday 72% of the customers who bought gas at a gas station made a additional purchases. There were 250 customers who bought

gas. How msny of these 250 customerd made additional purchases?
Mathematics
1 answer:
worty [1.4K]3 years ago
5 0

Answer:

180 people made additional purchases

Step-by-step explanation:

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Dafna1 [17]

Hello :D

The answer is 8.432

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Two less than 5 times a number is the same as the number plus 6
Alex777 [14]

Answer:

3

Step-by-step explanation:

from the question,you get an expresion of

6n-5= n + 10

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3 years ago
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Varvara68 [4.7K]

Answer:

1) 8n-3

2) 3+(8-n)

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4) 3(n+8)

6 0
3 years ago
The first three steps in determining the solution set of the system of equations algebraically are shown.
jolli1 [7]
Since both are equivalent to y, the equations must be equivalent.
x^2-x-3= -3x+5
x^2+2x-8=0

(x+4)(x-2)=0
x=-4, x=2

Plug the values of x in to either equation
y=-3(-4)+5
y= 12+5
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y= -3(2)+5
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Final answer: (-4,17) and (2,-1)
8 0
3 years ago
Read 2 more answers
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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