I'll do the first two problems to get you started
Problem 1
1/2 = 3/6
1/3 = 2/6
For the first fraction, I multiplied top and bottom by 3. Note how 3 is the denominator of the second fraction.
For the second fraction, I multiplied top and bottom by 2. Note how 2 is the denominator of the first fraction.
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Problem 2
1/2 = 5/10
3/5 = 6/10
Like with problem 1, I multiplied top and bottom of the first fraction by the value 5 which is the denominator of the second fraction. Likewise, I did the same for the second fraction using the first denominator. The steps are essentially the same, but the numbers are different.
Problem 3 will be handled the same way. I'll let you try it out.
Answer:
(x+1) (x−6)
Step-by-step explanation:
Let events
A=Nathan has allergy
~A=Nathan does not have allergy
T=Nathan tests positive
~T=Nathan does not test positive
We are given
P(A)=0.75 [ probability that Nathan is allergic ]
P(T|A)=0.98 [probability of testing positive given Nathan is allergic to Penicillin]
We want to calculate probability that Nathan is allergic AND tests positive
P(T n A)
From definition of conditional probability,
P(T|A)=P(T n A)/P(A)
substitute known values,
0.98 = P(T n A) / 0.75
solving for P(T n A)
P(T n A) = 0.75*0.98 = 0.735
Hope this helps!!
