There are 130 people in a sport centre. 61 people use the gym. 68 people use the swimming pool. 60 people use the track. 25 people use the gym and the pool. 30 people use the pool and the track. 19 people use the gym and the track. 6 people use all three facilities. a person is selected at random. what is the probability that the person uses exactly one of the facilities?

Explanation:

As shown in the Venn diagram,

Let

U=[The total number of people in the sport center ]

This implies that

n(U)=130

G=[Those who use the gym]

This implies that

n(G)=61

P=[Those who use the pool]

This implies that

n(P)=68

T=[Those who use the track]

This implies that

n(T)=60

a=those who use only the gym

b=those who use only the pool

c=those who use only the track

From the digram,

a=61-(13+6+19)=61-38=23

b=68-(6+24+19)=68-49=19

c=60-(13+6+24)=60-43=17

The number of people who use exactly one facilities=a+b+c=23+19+17=59

Hence, the probability that a person selected at random uses exactly one the facilities

-3

step-by-step explanation:

a drawing really does here:

so i draw the horizontal line y=2 and the vertical line x=-2 at this intersection i put b (this is where my right angle is going to be).   so point b is at (-2,2).   point a is 5 units long and is also on y=2 so this point has to be (3,2) or (-7,2) -doesn't matter i will go with (3,2).

now we know c is at (-2,y) we don't know the y yet bet we know that c is on x=-2.

we have the area of the triangle is .5bh=12.5

we know the base length is 5 (ab)

we have the height is (2-y) if you look at your

plug in

.5(5)(2-y)=12.5

solve for y to get the y-coordinate for c

2.5(2-y)=12.5

divide both sides by 2.5 giving:

2-y=5

now subtract 2 on both sides

-y=3

y=-3

answer

106 emergencies

step-by-step explanation: