You have $15 and admission is $5, so you know have $10 before doing any games. If you wanted to play 12 games, you would need $10.20, which you do not have. But, if you wanted to play 11 games, then you would need $9.35, which is just under $10.

So the maximum amount of games that you could play would be 11 games.

8 0

2 years ago

Write the least common denominator for 2/3, 4/5, and 5/6.

Lunna [17]

Answer:

30 I think but I'm probably wrong

5 0

2 years ago

Read 2 more answers

This table shows the number of points two teams scored in five games

mezya [45]

Answer:

2.16

Step-by-step explanation:

The question is on mean absolute deviation

The general formula ,

Mean deviation = sum║x-μ║/N where x is the each individual value, μ is the mean and N is number of values

Team 1

Finding the mean ;

$mean= \frac{51+47+35+48+64}{5} =49$

Points Absolute Deviation from mean

51 2

47 2

35 14

48 1

64 15

Sum 34

Absolute mean deviation = 34/5= 6.8

Team 2

Finding the mean

$mean=\frac{27+55+53+38+41}{5} = 42.8$

Points Absolute deviation from the mean

27 15.8

55 12.2

53 10.2

38 4.8

41 1.8

Sum 44.8

Absolute deviation from the mean = 44.8/5 =8.96

Solution

Difference in mean absolute deviation of the two teams = 8.96-6.8 = 2.16

5 0

2 years ago