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valina [46]
3 years ago
12

A used car dealer sells SUVs and cars. Of all the vehicles, 70% are cars. Of all the vehicles, 10% are black SUVs. What is the p

robability that an SUV chosen at random is black? Round your answer to the nearest tenth of a percent.
Mathematics
1 answer:
grin007 [14]3 years ago
5 0

Answer: 0.3

Step-by-step explanation:

From the question,

A used car dealer sells SUVs and cars. Of all the vehicles, 70% are cars. Since 70% are cars, SUV will be (100% - 70%) = 30%. Of all the vehicles, 10% are black SUVs. To calculate the probability than an SUV chosen is black goes thus:

P(A|B) = P(A and B) / P(B)

P(SUV | black) = P(SUV and black) / P(suv)

P(suv | black) = 10%/ 30%

P(suv | black) = 0.1/0.3

P(suv | black) = 0.3

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4 years ago
tickets for a harlem globetrotter show cost $28 general admission, $43 courtside, or $173 bench seats. Nine times as many genera
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Answer:

general admission tickets  1170

courtside tickets  985

tickets bench seats.  130

Step-by-step explanation:

To solve the problem, it is necessary to generate a system of equations with the information provided by the statement.

First be

x = # general admission tickets

y = # courtside tickets

z = # tickets bench seats.

The first equation would be that they sold nine times more general admission tickets than bench seats tickets.

That is: x = 9 * z (1)

And the second equation is that the number of general admission tickets sold was 55 more than the sum of the number of courtside tickets and bench seats tickets.

That is: x = 55 + y + z (2)

Now the third equation would be the money raised.

28 * x + 43 * y + 173 * z = 97605 (3)

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x = 55 + y + x / 9, rearranging:

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Now replacing (4) and (5) in 3, we have:

28 * x + 43 * ((8/9) * x - 55) + 173 * (x / 9) = 97605

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We can confirm this with equation (3)

28 * x + 43 * y + 173 * z = 97605

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