What are the answers to 2 and 4
1 answer:
You might be interested in
Answer:
ds/dt=69743.35m^2/min
Step-by-step explanation:
The radius r(t)r(t)r, (, t, )of a sphere is increasing at a rate of 7.57.57, point, 5 meters per minute. At a certain instant t_0t 0 t, start subscript, 0, end subscript, the radius is 555 meters. What is the rate of change of the surface area S(t)S(t)S, (, t, )of the sphere at that instant?
If i could understand the question correctly, the typos notwithstanding
r=radius, 555m
dr/dt=rate of change in radius 5m/min
ds/dt=rate of change in surface area ?
S=area of a sphere 4
ds/dt=ds/dr*dr/dt.................................1
ds/dr=differentiation of the area of a sphere=
ds/dt=*5
ds/dt=69743.35m^2/min
The surface area of the sphere will change by 69733.35 square meters for every minute.
Answer:
0.5818
Step-by-step explanation:
Suppose the average teenage romantic relationship is normally distributed with a mean number of 100 days with a standard deviation of 30 days.
Answer:
The z score is used to determine how many standard deviations that the raw score is above or below the mean. If the z score is positive then the raw score is above the mean and if it is negative then it is below the mean. It is given by:
Given that:
Therefore, from the normal distribution table, P(90 < x < 150) = P(-0.33 < z < 1.67) = P(z < 1.67) - P(z < -0.33) = 0.9525 - 0.3707 = 0.5818