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baherus [9]
3 years ago
9

The probability that a randomly selected elementary or secondary school teacher from a city is a female is , holds a second job

is , and is a female and holds a second job is . Find the probability that an elementary or secondary school teacher selected at random from this city is a female or holds a second job.
Mathematics
1 answer:
julsineya [31]3 years ago
4 0

Answer:

The probability that an elementary or secondary school teacher selected at random from this city is a female or holds a second job is 0.90.

Step-by-step explanation:

Denote the events as follows:

<em>X</em> = an elementary or secondary school teacher from a city is a female

<em>Y</em> = an elementary or secondary school teacher holds a second job

The information provided is:

P (X) = 0.66

P (Y) = 0.46

P (X ∩ Y) = 0.22

The addition rule of probability is:

P(A\cup B)=P(A)+P(B)-P(A\cap B)

Use this formula to compute the probability that an elementary or secondary school teacher selected at random from this city is a female or holds a second job as follows:

P(X\cup Y)=P(X)+P(Y)-P(X\cap Y)\\=0.46+0.66-0.22\\=0.90

Thus, the probability that an elementary or secondary school teacher selected at random from this city is a female or holds a second job is 0.90.

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If BTS=GHD BS=25 TS=14 BT=31 GD=4x-11 S=56 B=21 and H=(7y+5) find the values of x and y
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Given:

Consider the completer question is "If ∆BTS≅∆GHD, BS=25, TS=14, BT=31, GD=4x-11, m∠S=56, m∠B=21 and m∠H=(7y+5), find the values of x and y.

To find:

The values of x and y.

Solution:

We have,

\Delta BTS\cong \Delta GHD       (Given)

BS=GD                       (CPCTC)

25=4x-11

25+11=4x

36=4x

Divide both sides by 4.

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\angle B+\angle T+\angle S=180^\circ    (Angle sum property)

21^\circ+\angle T+56^\circ=180^\circ

77^\circ+\angle T=180^\circ

\angle T=180^\circ-77^\circ

\angle T=103^\circ

Now,

\angle T=\angle H            (CPCTC)

103=7y+5

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Divide both sides by 7.

14=y

Therefore, the value of x is 9 and value of y is 14.

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