The first 2 parts are correct
Third part is 40 %
Answer:
The overall probability of type II diabetes among 40- to 59-year-olds in Houston is 9.3%.
Step-by-step explanation:
We have these following rates of type II diabetes:
7% among Caucasians
10% among African-Americans
12% among Hispanics
5% among Asian-Americans.
The ethnic distribution of Houston is:
30% Caucasian
25% African-American
40% Hispanic
5% Asian-American
What is the overall probability of type II diabetes among 40- to 59-year-olds in Houston?

is the probability of finding a Caucasian with type II diabetes in Houston. So it is 7% of 30%.

is the probability of finding an African-American with type II diabetes in Houston. So it is 10% of 25%.

is the probability of finding a Hispanic with type II diabetes in Houston. So it is 12% of 40%.

is the probability of finding an Asian-American with type II diabetes in Houston. So it is 5% of 5%.


The overall probability of type II diabetes among 40- to 59-year-olds in Houston is 9.3%.
Answer:
a. V = 4/3
b. See attachment
Step-by-step explanation:
a.
Given
Z = x² + y²
V = ∫ ∫ (x² + y²) dxdy {0,1}{0,y} + ∫ ∫ (x² + y²) dxdy {0,1}{x,2-x}
V = ∫ ∫ (x² + y²) dxdy {0,1}{x,2-x}
Integrate with respect to y
V = ∫ x²y+ y³/3 dx {0,1}{x,2-x}
V = ∫ x²(2-x) + (2-x)³/3 - x²(2) - (2)³/3 dx {0,1}
V = ∫ 2x² -7x³/3 + (2-x)³/3 dx {0,1}
V = 2x³/3 - 7x⁴/12 + (2-x)⁴/12 {0,1}
V = (⅔ - 7/4 + 2/12) - (0-0+16/12)
V = 4/3
G would equal 2 because 8g-4g=4g so the equation would really be
4g=8
Divide by 4 to equal 2