Answer:
a) the probability that the minimum of the three is between 75 and 90 is 0.00072
b) the probability that the second smallest of the three is between 75 and 90 is 0.396
Step-by-step explanation:
Given that;
fx(x) = { 1/5 ; 50 < x < 100
0, otherwise}
Fx(x) = { x-50 / 50 ; 50 < x < 100
1 ; x > 100
a)
n = 3
F(1) (x) = nf(x) ( 1-F(x)^n-1
= 3 × 1/50 ( 1 - ((x-50)/50)²
= 3/50 (( 100 - x)/50)²
=3/50³ ( 100 - x)²
Therefore P ( 75 < (x) < 90) = ⁹⁰∫₇₅ 3/50³ ( 100 - x)² dx
= 3/50³ [ -2 (100 - x ]₇₅⁹⁰
= (3 ( -20 + 50)) / 50₃
= 9 / 12500 = 0.00072
b)
f(k) (x) = nf(x) ( ⁿ⁻¹_k₋ ₁) ( F(x) )^k-1 ; ( 1 - F(x) )^n-k
Now for n = 3, k = 2
f(2) (x) = 3f(x) × 2 × (x-50 / 50) ( 1 - (x-50 / 50))
= 6 × 1/50 × ( x-50 / 50) ( 100-x / 50)
= 6/50³ ( 150x - x² - 5000 )
therefore
P( 75 < x2 < 90 ) = 6/50³ ⁹⁰∫₇₅ ( 150x - x² - 5000 ) dx
= 99 / 250 = 0.396
Answer: Therefore, Option 'D' is correct.
Step-by-step explanation:
Since we have given that
Number of population = 132000
Number of voting districts = 11
According to question, no district is to have a population that is more than 10 percent greater than the population of any other district.
Let the number of population that the least populated district could have be 'x'.
So, it becomes,
Hence, there are 11000 population that the least populated district could have.
Therefore, Option 'D' is correct.
We have to determine the value of a house after 10 years. We know that the present value of the house is $125,000 and that the value increases 3.5% each year. Formula is: FV = PV * ( 1 + r )^t, where FV is the future value, PV is the present value, r = 3.5% = 0.035 and t = 10 years. FV = 125,000 *( 1.035 )^10 = 125,000 * 1.4105; FV = $176,324.85 ( $ 176,325 to the nearest dollar ). Answer: The value of the house after 10 years will be <span>$176,325.</span>
Least to greatest: -10, -3, 5, 16
Closet to farthest from 0: -3, 5, -10, 16
