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 this person doesn't use any facility?

59 over 130

Step-by-step 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

answer:

sometimes

*i think

d

step-by-step explanation:

hello and welcome!

so, when we see the words "proportional" along with a table, we need to know that we're looking numbers that divide by the same thing.

looking at this table, we simply have to do a trial and error, and divide to see which numbers have the same quotient. if you look at choice a & c, they both have equal quotients for three of the numbers, with one exception, so those can't be the answer. looking at b, there is no similarity, so that can't be it. then, we see d; if you divide all the numbers in the table, you'll see that they all divide by two, which is a proportional relationship, and our answer!

also, always remember that the x will always be the number that goes into y, and somewhere behind this table, is a formula. for this specific table, the formula would be:

since our x-value is being divided by two to get the y-value. and if they were to ask for more numbers in the series, this is the formula you would use to find them.