Answer:
The relative frequency of the middle school students who watch soccer the most is 33.3%.
Step-by-step explanation:
We are given the table,
Grade Basketball Baseball Football Soccer Tennis Other Total
Elementary 51 18 26 32 4 6 137
Middle 50 24 44 34 7 18 177
High 42 17 58 36 11 24 188
Total 143 59 128 102 22 48 502
Now, we are required to find the relative frequency of the middle school students who watch soccer.
Since, we know,
Total number of students who watch soccer = 102
Number of middle school students who watch soccer = 34
Thus, the relative frequency of the middle school students = 
= 0.333 i.e. 33.3%
Hence, the relative frequency of the middle school students who watch soccer the most is 33.3%.
The value of y is 29
Since x=4, put 4 in 9x
y = 9(4) - 7
y = 36 - 7
y = 29
Answer: 8 + 2/3 batches of brownies you can bake.
Step-by-step explanation:
6 1/2 butter = 6 + 1/2 = (2·6 + 1)/2 = 13/2 sticks of butter, we have this
Since we have 13/2 sticks of butter and each batch of brownies needs 3/4 of a stick of butter, we have to divide the amount of butter we have between the amount of butter we use to bake a batch of brownies.
13/2 butter ÷ 3/4 butter/batch = 13·4/2·3 batches = 52/6 batches = 8 + 2/3 batches
Answer: 8 + 2/3 batches of brownies you can bake.
how many 3 element subsets of {1, 2, 3, 4, 5, 6, 7, 8, ,9, 10, 11} are there for which the sum of the elements in the subset is
AURORKA [14]
Answer:
There are 155 ways in which these elements casn occur.
Step-by-step explanation:
We want 3 element subsets whose sum are multiples of 3
1+2+3= 6
1+2+6= 9
1+2+9= 12
1+9+11=21
1+3+5=9
1+4+8=12
1+5+6=12
1+6+8=15
1+7+10=18
1+8+9=18
1+9+11=21
2+3+7=12
2+4+6=12
2+4+9=15
2+5+11=18
2+6+7=15
2+7+9=18
2+8+5=15
2+8+11=21
2+9+10=21
3+6+9= 18
3+9+11=21
3+10+11=24
6+9+10=27
6+8+11=27
6+7+11=24
7+8+9= 24
8+9+10=27
7+9+11=27 .........
We have 11 elements
We need a combination of 3
The combinations can be in the form
even+ even+ odd
odd+odd+odd
even + odd+odd
So there are 3 ways in which these elements can occur
Total number of combinations with 3 elements =11C3= 165
There are 6 odd numbers and 5 even numbers.
Number of subsets with 3 odd numbers = 6C3= 20
Number of two even numbers and 1 odd number = 5C2*6C1=10*6= 60
Number of 2 odd and 1 even number = 6C2* 5C1= 5*15= 75
So 20+60+75=155
There are 155 ways in which this combination can occur
this is a picture from a question in a math class last year :>
