On the Ford Aquatics swim team, 11 year olds and 12 year olds swim together. The Gold team has a 2:3 ratio of 11 year olds to 12 year olds. The Silver team has a 3:8 ratio of 11 year olds to the total swimmers. If the two teams have the same number of 11 year olds what could the size of the teams be?

The Gold team has 8 swimmers and the silver team also has 8 swimmers.

Step-by-step explanation:

Let T be the total number of swimmers on the Gold team. Since, the Gold team has a 2:3 ratio of 11 year olds to 12 year olds, the number of 11 year old is T' = 2/(2 + 3) × T = 2T/5 and the number of 12 year olds is T" = 3/(2 + 3) × T = 3T/5.

The silver team has a 3:8 ratio of 11 year olds to the total swimmers. Let T₁ be the number of 11 year old swimmers in the silver team. So, from the ratio 3:8, the number of 11 year old swimmers in the silver team is 3 and the total number of swimmers in the silver team is 8.

Since the number of 11 year old swimmers in both teams are the same, then

2T/5 = 3

T = 3 × 5 ÷ 2

T = 15/2

T = 7.5

T≅ 8

So, the Gold team also has 8 swimmers.

So, the Gold team has 8 swimmers and the silver team also has 8 swimmers.

$450

step-by-step explanation:

first, you have to multiply 150x2=300

next, you have to add 150+300= 450

x -

median = 1.25

range = 12-14.5

interquartile = q12-q14.5

y -

median = 1

range = 15-17

interquartile = q15-q17