Answer: Option 'c' is correct.
Step-by-step explanation:
Since we have given that
the optimized solution of a linear program to an integer as it does not affect the value of the objective function.
As if we round the optimized solution to the nearest integer, it does not change the objective function .
while it is not true that it always produces the most optimal integer solution or feasible solution.
Hence, Option 'c' is correct.
Given:
A right rectangular prism is packed with cubes of side length 
inch.
Prism is packed with 15 cubes along the length, 6 cubes along the width, and 5 cubes along the height.
To find:
The volume of the prism
Solution:
Side length of each cube 
inch.
15 cubes along the length, so length of prism is
6 cubes along the width, so width of prism is
5 cubes along the height, so height of prism is
Now,
Volume of rectangular prism is
Therefore, the correct option is B.
Y=6x-27...........(1)
y=4x-17...........(2)
By comparison
6x-27=4x-17
=> 2x=10 => x=5
Substitute x=5 into (1)
y=6(5)-27=3
Check:
y=4(5)-17=3 as well.
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
