I am not completely sure how to name faces, but I will name them using all four points.
Faces DCBA
Faces BFGC
Faces HGFE
Answer:
Option d
Step-by-step explanation:
We need 2 fundamental data to solve this problem.
1.- Average number of clients attended in 1 hour.
2.- Time in which it is expected that a client will be attended.
They tell us that the average number of clients served in an hour is m = 8
They tell us to calculate the probability that the client will be seen in less than 15 minutes.
But the time t in the formula is given in hours.
Therefore we must write 15 minutes according to hours.
We know that 
.
So:
![15\ minutes[\frac{1\ hour}{60\ minutes}] = 0.25\ hours](https://tex.z-dn.net/?f=15%5C%20minutes%5B%5Cfrac%7B1%5C%20hour%7D%7B60%5C%20minutes%7D%5D%20%3D%200.25%5C%20hours)
.
Now we substitute in the formula 
and 
to find P.
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:
x = 2 ; y = 3
Step-by-step explanation:
3x = 6
x = 2
4y = 12
y = 3
