One is 6 years old, one is 16 as per the rules. Now we have 5 unknowns, of which either the twins can be both a part of, or one can be either 6 or 16 as well.
Since the mode is the most frequently occurring number, let's assume that 15 years old appears twice to make it so.
Now we have one 6 y.o, two 15 y.os, and one 16 y.o.
Out of 7 kids, the middle child has to be 10 years old to secure the median.
Now we just have two missing, and we need to figure out the mean. 6+15+15+10+16 is 62, dividing by 5 is 12.4 The mean is currently 12.4 years old. For the mean to be 11, the combined total has to be 77 (since 77/7 kids is 11 mean) 77-62 is 15. But to make sure the median stays 10, both of those kids have to be less than 10 but more than 6. Therefore, their ages must be 7 and 8.
Their ages are 6, 7, 8, 10, 15, 15, 16
Let's double check to make sure we have all the qualifications. 7 kids, check. Youngest 6, check. Oldest 16, check. Twins, check. Median of 10, check. Mode of 15, check. Average of 11, well 6+7+8+10+15+15+16 is 77, 77/7 is 11, so check. Phew.
29
I hope this helps!
Thank you!!
Answer:
63.5
Step-by-step explanation:
Answer:
3:5 <em>the colon is needed to represent the ratio</em>
Step-by-step explanation:
r = 10
g = 4
y = 30 - 14 = 6
y = 6
so ratio of total balls is 10:4:6
and ratio of y:r is
6:10
simply by dividing by 2
3:5
