Either way. The probability of hitting the circle is:
P(C)=Area of circle divided by area of square
P(W)=(area of square minus area of circle divided by area of square
P(C)=(πr^2)/s^2
P(W)=(s^2-πr^2)/s^2
...
Okay with know dimensions, r=1 (because r=d/2 and d=2 so r=1), s=11 we have:
P(inside circle)=π/121 (≈0.0259 or 2.6%)
P(outside circel)=(121-π)/121 (≈0.9744 or 97.4%)
Answer:
A) A(t) = 4500*π - 1600*t
B) A(4) = 7730 in³
C) t = 8,8 sec
Step-by-step explanation:
The volume of the sphere is:
d max = 30 r max = 15 in
V(s) = (4/3)*π*r³ V(s) = (4/3)*π* (15)³
V(s) = 4500*π
A) Amount of air needed to fill the ball A(t)
A(t) = Total max. volume of the sphere - rate of flux of air * time
A(t) = 4500*π - 1600*t in³
B) After 4 minutes
A(4) = 4500*π - 6400
A(4) = 14130 - 6400
A(4) = 7730 in³
C) A(t) = 4500*π - 1600*t
when A(t) = 0 the ball got its maximum volume then:
4500*π - 1600*t = 0
t = 14130/1600
t = 8,8 sec
Answer:
c) 380 m²
Step-by-step explanation:
A = π·r²
= 3.14×121
= 379.94
≈ 380 m²
