At the movies, we had 286 people buy tickets. the adult tickets were $12 and the children's tickets were $4. the total sold was $1968. how many adults and kids attended?

( explain the answer algebraically)

103 adults and 183 children's attended

Solution:

Let "a" be the number of adults attended

Let "c" be the number of children attended

Cost of 1 adult ticket = $ 12

Cost of 1 children ticket = $ 4

At the movies, we had 286 people buy tickets

Therefore,

number of adults attended + number of children attended = 286

a + c = 286 ---- eqn 1

The total sold was $1968

Therefore,

number of adults attended x Cost of 1 adult ticket + number of children attended x Cost of 1 children ticket = 1968

12a + 4c = 1968 ------- eqn 2

Let us solve eqn 1 and eqn 2

From eqn 1,

a = 286 - c -- eqn 3

Substitute eqn 3 in eqn 2

12(286 - c) + 4c = 1968

3432 -12c + 4c = 1968

-8c = 1968 - 3432

-8c = -1464

8c = 1464

Substitute c = 183 in eqn 3

a = 286 - 183

Thus 103 adults and 183 children's attended

an advantage of the standard deviation is that it increases as the dispersion of the data increases. e. the interquartile range is preferred when the data are not skewed or no have outliers. an advantage of the standard deviation is that it uses all the observations in its computation.