Answer:
The fraction of the area of ACIG represented by the shaped region is 7/18
Step-by-step explanation:
see the attached figure to better understand the problem
step 1
In the square ABED find the length side of the square
we know that
AB=BE=ED=AD
The area of s square is

where b is the length side of the square
we have

substitute


therefore

step 2
Find the area of ACIG
The area of rectangle ACIG is equal to

substitute the given values

step 3
Find the area of shaded rectangle DEHG
The area of rectangle DEHG is equal to

we have


substitute

step 4
Find the area of shaded rectangle BCFE
The area of rectangle BCFE is equal to

we have


substitute

step 5
sum the shaded areas

step 6
Divide the area of of the shaded region by the area of ACIG

Simplify
Divide by 5 both numerator and denominator

therefore
The fraction of the area of ACIG represented by the shaped region is 7/18
Answer:
a) P(B'|A) = 0.042
b) P(B|A') = 0.625
Step-by-step explanation:
Given that:
80% of the light aircraft that disappear while in flight in a certain country are subsequently discovered
Of the aircraft that are discovered, 63% have an emergency locator,
whereas 89% of the aircraft not discovered do not have such a locator.
From the given information; it is suitable we define the events in order to calculate the probabilities.
So, Let :
A = Locator
B = Discovered
A' = No Locator
B' = No Discovered
So; P(B) = 0.8
P(B') = 1 - P(B)
P(B') = 1- 0.8
P(B') = 0.2
P(A|B) = 0.63
P(A'|B) = 1 - P(A|B)
P(A'|B) = 1- 0.63
P(A'|B) = 0.37
P(A'|B') = 0.89
P(A|B') = 1 - P(A'|B')
P(A|B') = 1 - 0.89
P(A|B') = 0.11
Also;
P(B ∩ A) = P(A|B) P(B)
P(B ∩ A) = 0.63 × 0.8
P(B ∩ A) = 0.504
P(B ∩ A') = P(A'|B) P(B)
P(B ∩ A') = 0.37 × 0.8
P(B ∩ A') = 0.296
P(B' ∩ A) = P(A|B') P(B')
P(B' ∩ A) = 0.11 × 0.2
P(B' ∩ A) = 0.022
P(B' ∩ A') = P(A'|B') P(B')
P(B' ∩ A') = 0.89 × 0.2
P(B' ∩ A') = 0.178
Similarly:
P(A) = P(B ∩ A ) + P(B' ∩ A)
P(A) = 0.504 + 0.022
P(A) = 0.526
P(A') = 1 - P(A)
P(A') = 1 - 0.526
P(A') = 0.474
The probability that it will not be discovered given that it has an emergency locator is,
P(B'|A) = P(B' ∩ A)/P(A)
P(B'|A) = 0.022/0.526
P(B'|A) = 0.042
(b) If it does not have an emergency locator, what is the probability that it will be discovered?
The probability that it will be discovered given that it does not have an emergency locator is:
P(B|A') = P(B ∩ A')/P(A')
P(B|A') = 0.296/0.474
P(B|A') = 0.625
Answer:
a = 4
Step-by-step explanation:
STEP 1) 4a - 16 = 4 • (a - 4)
Answer: The p-value is 0.154.
Step-by-step explanation:
Since we have given that
We claim that
Null hypothesis :

Alternate hypothesis :

Population mean = 20 hours
Sample mean = 18.5 hours
Sample standard deviation = 4.3 hours
Sample size n = 35
So, test statistic would be

So, the p value would be 0.154.
Hence, the p-value is 0.154.