Geometric Probability Find the probability that a point chosen randomly inside the larger rectangle is in each given smaller shape. Round to the nearest percent. PLEASE HELP!
1) The circle
2) The smaller rectangle
3) Not the circle or smaller rectangle

Step-by-step explanation:

1) P= Area of Circle/ Area of large rectangle

Area of the circle = pi·r² = pi·2²=4 pi ft.²

Area of large rectangle=  l·w -12·10 =120 ft.²

P = 4pi/120 rewrite 120 as 4·30

P= 4 pi/4*30 = pi/30 = 3.14/40 ≈ .1047 ≈10% (because .1047·100 =10.47≅10)

2) P = Area of smaller rectangle/ Area of large rectangle

Area of smaller rectangle = l·w = 2·4 =8 ft.²

Area of large rectangle=l·w = 12·10=120 ft²

P= 8/120 ≅ .0666≅ 7% (because .0666·100 =6.66≅7)

3) P= Not the circle or smaller rectangle/ Area of large rectangle

Not the circle or smaller rectangle area

= Area of large rectangle - Area of circle -Area of smaller rectangle

= 120 -4·pi -8 = 120 - (4· 3.14) -8 = 99.4362939 ft²

Area of large rectangle = l·w = 12·10 =120 ft²

P = 99.4362939 /120 ≅ .8286 ≅83% (because .8286·100 =82.86≅83)

idk

step-by-step explanation:

9.55 degrees.

step-by-step explanation:

the given angle is

1 cycle is equal to 2π radians. so 4π will be equal to two cycles.

5.166667π-4π = 1.166667π

hence, this angle goes a little more than negative x axis.

1.166667π-π= 0.166667π

hence, the reference angle is 0.166667π radians.

1 radian = 57.2958 degrees

now, in degrees, this should be equals to 9.55 degrees.