Answer:
u=10
Step-by-step explanation:
Answer:
Step-by-step explanation:
Given the volume of the sphere expressed as V = 4/3πr³
If the balloon is being inflated so that the radius increases at a constant rate r(t) = 12t + 2, the function that represents the volume of the weather balloon in terms of time can be derived by simply substituting r = 12t+2 into the formula for finding the volume of the balloon as shown;
V(t) = 4/3 π(12t+2)³
V(t) = 4π(12t+2)³/3
b) To find the volume of the balloon 12 seconds after inflation begins, we will substitute t = 12 into the resulting equation in (a)
V(12) = 4π(12(12)+2)³/3
V(12) = 4π(144+2)³/3
V(12) = 4π(146)³/3
V(12) = 4π(3,112,136)/3
V(12) = 39,108,254.38/3
V(12) = 13,036,084.79
Volume of the balloon to neatest cube metres is 13,036,085m³
Normalizing the bounds,
y=186, z = (186-200)/12 = -1.17
y = 210, z = (210 - 200)/12 = 0.83
We want the area of the standard normal between those. Looking it up in the cumulative table,
Ф(1.17) = 0.87900
Ф(0.83) = 0.79673
We need Ф(-1.17) which is 1 - Ф(1.17)
Then the probability we seek is
p = Ф(0.83) - Ф(-1.17) = Ф(0.83) + Ф(1.17) - 1 = 0.87900 + 0.79673 - 1 = 0.67573
Answer: 67.6%
Answer:
i need more information to answer this
Step-by-step explanation:
