Answer:
The probability that the aircraft is overload = 0.9999
Yes , The pilot has to be take strict action .
Step-by-step explanation:
P.S - The exact question is -
Given - Before every flight, the pilot must verify that the total weight of the load is less than the maximum allowable load for the aircraft. The aircraft can carry 37 passengers, and a flight has fuel and baggage that allows for a total passenger load of 6,216 lb. The pilot sees that the plane is full and all passengers are men. The aircraft will be overloaded if the mean weight of the passengers is greater than 6216/37 = 168 lb. Assume that weight of men are normally distributed with a mean of 182.7 lb and a standard deviation of 39.6.
To find - What is the probability that the aircraft is overloaded ?
Should the pilot take any action to correct for an overloaded aircraft ?
Proof -
Given that,
Mean, μ = 182.7
Standard Deviation, σ = 39.6
Now,
Let X be the Weight of the men
Now,
Probability that the aircraft is loaded be
P(X > 168 ) = P(
)
= P( z >
)
= P( z > -0.371)
= 1 - P ( z ≤ -0.371 )
= 1 - P( z > 0.371)
= 1 - 0.00010363
= 0.9999
⇒P(X > 168) = 0.9999
As the probability of weight overload = 0.9999
So, The pilot has to be take strict action .
The is 12 I think hope that helps
Answer:
The 3rd option is correct.
Step-by-step explanation:
This is because of an exponent rule that goes:
(xy)^z = x^z(y^z)
The whole thing gets raised to the power. Therefore only the 3rd one works (ps. if you can mark as brainliest that would be amazing, but if not that's fine)
Answer:
17 for 1 point shots and 35 for two point shots.
Step-by-step explanation:
let the number of 1 point shots be x
let the number of 2 point shots be y
x + y = 52 (1)
x + 2y = 87 (2)
subtracting (1) from (2)
y = 35
from (1) x = 52 - y = 52 - 35 = 17
Hence number of 1 pointers = 17
number of 2 point shots = 35
checking; 17 + 35 =52 ; and 17+ 2*35 = 87
Answer:
42.3
Step-by-step explanation:
- Substitute in values
