Answer:
The radius, r₂, of the ball that uses one-half the amount of rubber coating used to cover the 16-inch ball is approximately 4.66 inches
Step-by-step explanation:
The dimension of the ball with known radius = 16-inch
The surface area of the ball with 16-inch radius = 4×π×r² = π·D² = π×16² = 804.24772 in.²
Given that the ball uses one-half the rubber material coating used to cover the 16-inch ball, we have the surface area of the ball = 804.24772 in.²/2 = 402.12386 in.²
The radius, r₂ of the new ball is found as follows;
402.12386 in.² = 4×π×r₂²
r₂² = 402.12386 in.² /(4×π) ≈ 32
r₂ = √32 = 4·√2 ≈ 4.66 inches
The radius, r₂, of the ball that uses one-half the amount of rubber coating used to cover the 16-inch ball ≈ 4.66 inches.
B = (2x+3)(4x^2-6x+9)-2(4^3-1)
B = 8x^3-99
Hope it helps : )
Answer: 100
5
Step-by-step explanation:
a) The mean of a normal distribution is also the median. Half the population will have values above the mean. Half of 200 is 100, so ...
... 100 students will have grades above 70%.
b) 84% is 14% above the mean. Each 7% is 1 standard deviation, so 14% is 2 standard deviations above the mean. The empirical rule tells you 95% of the population is within 2 standard deviations of the mean, so about 5% of students (10 students) got grades higher than 84% or lower than 56%. The normal distribution is symmetrical, so we expect about 5 students in each range.
... about 5 students will have grades above 84%.
