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BlackZzzverrR [31]
3 years ago
11

A particular state has elected both a governor and a senator. Let A be the event that a randomly selected voter has a favorable

view of a certain party’s senatorial candidate, and let B be the corresponding event for that party’s gubernatorial candidate.
Suppose that P(A′) = .44, P(B′) = .57, and P(A ⋃ B) = .68 (these figures are suggested by the 2010 general election in California).
(a) What is the probability that a randomly selected voter has a favorable view of both candidates?
(b) What is the probability that a randomly selected voter has a favorable view of exactly one of these candidates?
(c) What is the probability that a randomly selected voter has an unfavorable view of at least one of these candidates?
Mathematics
1 answer:
Elodia [21]3 years ago
3 0

Answer:

Step-by-step explanation:

Given that A be the event that a randomly selected voter has a favorable view of a certain party’s senatorial candidate, and let B be the corresponding event for that party’s gubernatorial candidate.

Suppose that

P(A′) = .44, P(B′) = .57, and P(A ⋃ B) = .68

From the above we can find out

P(A) = 1-0.44 = 0.56

P(B) = 1-0.57 = 0.43

P(AUB) = 0.68 =

0.56+0.43-P(A\bigcap B)\\P(A\bigcap B)=0.30

a) the probability that a randomly selected voter has a favorable view of both candidates=P(AB) = 0.30

b) the probability that a randomly selected voter has a favorable view of exactly one of these candidates

= P(A)-P(AB)+P(B)-P(AB)

=0.99-0.30-0.30\\=0.39

c) the probability that a randomly selected voter has an unfavorable view of at least one of these candidates

=P(A'UB') = P(AB)'

=1-0.30\\=0.70

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Step-by-step explanation:

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Which ordered pair is a solution to the system of linear equations 1/2x-3/4y=11/60 and 2/5x+1/6y=3/10
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ANSWER

( \frac{2}{3} , \frac{1}{5} )

EXPLANATION

The first equation is

\frac{1}{2} x -  \frac{3}{4} y =  \frac{11}{60} ...(1)

The second equation is

\frac{2}{5} x  +  \frac{1}{6} y =  \frac{3}{10} ...(2)

We want to eliminate y, so we multiply the first equation by

\frac{4}{5}

\frac{4}{5}  \times \frac{1}{2} x - \frac{4}{5}    \times \frac{3}{4} y =  \frac{11}{60}  \times  \frac{4}{5}

\frac{2}{5} x - \frac{3}{5} y =  \frac{11}{75} ...(3)

We now subtract equation (3) from (2)

(\frac{2}{3} x  -  \frac{2}{3} x )+ ( \frac{1}{6} y -  -  \frac{3}{5}y ) =(  \frac{3}{10}  -  \frac{11}{75} )

\frac{1}{6} y  +  \frac{3}{5}y  =\frac{3}{10}  -  \frac{11}{75}

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Multiply both sides by

\frac{30}{23}

\implies \:  \frac{30}{23} \times  \frac{23}{30}y=  \frac{23}{150}  \times  \frac{30}{23}

\implies \: y =  \frac{1}{5}

Substitute into the first equation to solve for x .

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Multiply to obtain

\frac{1}{2} x -  \frac{3}{20} =  \frac{11}{60}

\frac{1}{2} x = \frac{11}{60} + \frac{3}{20}

\frac{1}{2} x = \frac{1}{3}

Multiply both sides by 2.

2 \times \frac{1}{2} x =2 \times  \frac{1}{3}

x = \frac{2}{3}

The solution is

( \frac{2}{3} , \frac{1}{5} )

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