Q:

Identify the probability to the nearest hundredth that a point chosen randomly inside the rectangle is either in the circle or in the regular hexagon. HELP ASAP!!

Accepted Solution

A:
Answer:[tex]0.06[/tex]Step-by-step explanation:we know that The probability of an event is the ratio of the size of the event space to the size of the sample space.  The size of the sample space is the total number of possible outcomes  The event space is the number of outcomes in the event you are interested in.  so  Let x------> size of the event space y-----> size of the sample space  so [tex]P=\frac{x}{y}[/tex] step 1Find the probability that a point chosen randomly inside the rectangle is in the circleFind the area of the rectangle[tex]A=26.2*13=340.6\ in^{2}[/tex]Find the area of the circle[tex]A=3.14*(2)^{2} =12.56\ in^{2}[/tex]In this problem we have [tex]x=12.56\ in^{2}[/tex] [tex]y=340.6\ in^{2}[/tex] substitute [tex]P=\frac{12.56}{340.6}=0.037[/tex] step 2Find the probability that a point chosen randomly inside the rectangle is in the regular hexagonFind the area of the regular hexagon[tex]A=6[\frac{1}{2} (1.8)^{2}sin(60)]=8.42\ in^{2}[/tex]In this problem we have [tex]x=8.42\ in^{2}[/tex] [tex]y=340.6\ in^{2}[/tex] substitute [tex]P=\frac{8.42}{340.6}=0.025[/tex] step 3Find the probability that a point chosen randomly inside the rectangle is  either in the circle or in the regular hexagonIs the sum of the two probabilities [tex]0.037+0.025=0.062[/tex]Round to the nearest hundredth[tex]0.062=0.06[/tex]