7/10 + 1/2
Find common denominator will be 10
7/10+ 1×5/2×5
7/10+5/10
Add numerators keeping same denominator
7+5/10
Simplify
12/10
Reduce fraction
6/5
Answer:
Step-by-step explanation:
The given equation is expressed as
x1 + 2x2 = -24- - - - - - - - --1
x1 + 7x2 = -11- - - - - - - --2
We would eliminate x1 by subtracting equation 2 from equation 1. It becomes
- 5x2 = - 13
Dividing both sides of the equation by - 5, it becomes
- 5x2/- 5 = - 13/- 5
x2 = 13/5
Substituting x2 = 13/5 into equation 2, it becomes
x1 + 7 × 13/5 = -11
x1 + 91/5 = - 11
x1 = - 11 - 91/5
x1 = - 146/5
Answer:
(f+g)(x) = 7x -1 (Answer C)
Step-by-step explanation:
Rewrite these two functions, one above the other, for ease in adding:
f(x) = 5x - 2
+g(x) =2x + 1
------------------
(f+g)(x) = 7x -1
Answer:
20
Explanation:
To estimate, we round. 102.3 rounds to 100 and 4.7 rounds to 5; this means we divide 100/5, which is 20.
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
