How much greater is the area of a square with a side length of 9 inches than the area of a circle with a radius of 3 inches?

1 answer:

7 0

The area of the square is 81 sq in, and the area of the circule is (3.14)(3 in)^2, or approx 3.14(9) sq in, or approx 28.27 sq in.

Subtract: 81 sq in - 28.27 sq in = approx. 52.73 sq in.

The difference is 52.73 sq in; the square is larger, the circle smaller.

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Suppose you just received a shipment of nine televisions. Three of the televisions are defective. If two televisions are randoml

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The probability that both televisions work is: 0.42

The probability at least one of the two televisions does not work is:

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<h2>Step-by-step explanation:</h2>

There are a total of 9 televisions.

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Three of the televisions are defective.

This means that the number of televisions which are non-defective are:

9-3=6

The probability that both televisions work is calculated by:

$=\dfrac{6_C_2}{9_C_2}$

( Since 6 televisions are in working conditions and out of these 6 2 are to be selected.

and the total outcome is the selection of 2 televisions from a total of 9 televisions)

Hence, we get:

$=\dfrac{\dfrac{6!}{2!\times (6-2)!}}{\dfrac{9!}{2!\times (9-2)!}}\\\\\\=\dfrac{\dfrac{6!}{2!\times 4!}}{\dfrac{9!}{2!\times 7!}}\\\\\\=\dfrac{5}{12}\\\\\\=0.42$

The probability at least one of the two televisions does not work:

Is equal to the probability that one does not work+probability both do not work.

Probability one does not work is calculated by:

$=\dfrac{3_C_1\times 6_C_1}{9_C_2}\\\\\\=\dfrac{\dfrac{3!}{1!\times (3-1)!}\times \dfrac{6!}{1!\times (6-1)!}}{\dfrac{9!}{2!\times (9-2)!}}\\\\\\=\dfrac{3\times 6}{36}\\\\\\=\dfrac{1}{2}\\\\\\=0.5$