Mathematics: A circle is inscribed in a square with a side length of 144. If a point in the square is chosen at random, what is the probability that the point is inside the circle?

A circle is inscribed in a square with a side length

jimennacastillo15

09.05.2021

Given :

A circle is inscribed in a square with a side length of 144.

So, radius of circle, r = 144/2 = 72 units.

To Find :

The probability that the point is inside the circle.

Solution :

Area of circle,

$A_c = \pi r^2\\\\A_c = 3.14 \times 72^2\ units^2\\\\A_c = 16277.76 \ units^2$

Area of square,

$A_s = (2r)^2\\\\A_s = ( 2 \times 72)^2\ units^2\\\\A_s = 20736\ units^2$

Now, probability is given by :

$P = \dfrac{A_c}{A_s}\\\\P = \dfrac{16277.76}{20736}\\\\P = 0.785$

Therefore, the probability that the point is inside the circle is 0.785 .

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RioQNA

02.03.2023

See explanation below.

Step-by-step explanation:

Part A:

(i). They both correct because they could have a third degree polynomial that crosses the x-axis 3 times.

(ii). To find the zero's of the first one, set function equal to zero and use zero factor principle given as:

$\mathrm{Zero\:Factor\:Principle:\quad \:If}\:ab=0\:\mathrm{then}\:a=0\:\mathrm{or}\:b=0\:\left(\mathrm{or\:both}\:a=0\:\mathrm{and}\:b=0\right)$

So, we have:

$\mathrm{Solve\:}\:x+2=0:\quad x=-2\\\\\mathrm{Solve\:}\:x-1=0:\quad x=1\\\\\mathrm{Solve\:}\:x-2=0:\quad x=2$

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