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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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PLEASE HELP ASAP!! Represent the following expressions as a power of the number a where a≠0.
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After using the property of power, the expressions as a power of the number a is (a^3)^{-2} = a^{-6} , (a^{-1}\cdot a^{-2})^{-3} = a^{9} and ((a^2)^{-2})^2=a^{-8} .

In the given question we have to solve the given expressions as a power of the number a where a≠0.

The first given expression is (a^3)^{-2}.

Using the property of multiplication of power; (a^m)^n=a^{mn}.

Simplifying the expression.

(a^3)^{-2} = a^{3*(-2)}

(a^3)^{-2} = a^{-6}

The second expression is (a^{-1}\cdot a^{-2})^{-3}

Using the property of multiplication of power; (a^m)^n=a^{mn}.

(a^{-1}\cdot a^{-2})^{-3} = a^{(-1)\times(-3)}\cdot a^{(-2)\times(-3)}

(a^{-1}\cdot a^{-2})^{-3} = a^{3}\cdot a^{6}

Using the property of sum of power a^m\cdot a^n=a^{m+n}

(a^{-1}\cdot a^{-2})^{-3} = a^{(3+6)}

(a^{-1}\cdot a^{-2})^{-3} = a^{9}

The third expression is ((a^2)^{-2})^2.

Using the property of multiplication of power; (a^m)^n=a^{mn}.

((a^2)^{-2})^2=((a^2)^{(-2)\times2})

((a^2)^{-2})^2=(a^2)^{-4}

Again using the property of multiplication of power; (a^m)^n=a^{mn}.

((a^2)^{-2})^2=a^{2\times(-4)}

((a^2)^{-2})^2=a^{-8}

To learn more about property of power link is here

brainly.com/question/28747844

#SPJ1

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