Hello,
Let a the first, b the second and c the third.
a+b+c=53
b=a/3
c=2a
==>a+a/3+2a=53
==>10a/3=53
==>a=15.9
c=2*a=2*15.9=31.8
b=a/3=15.9/3=5.3
Proof: a+b+c=15.9+5.3+31.8=53
The second number is 5.3
Answer:
129
Step-by-step explanation:
Considering the survey to be representative, you can simply multiply the share of students <em>p</em> preferring “Track & Field” with the whole school population at the same time to estimate the number of such students in the whole school.
First we need to find the relative share <em>p</em> of such answers in the study by dividing it by the sum of answers, assuming that the table is complete for that random sample:
<em>p</em> = 4/(8 + 5 + 4) = 4/17
Then for the whole school we get 550 <em>p</em> ≈ 129.4
18 is the answer im pretty sure if its wrong im sorry
Answer:
1 Adult ticket=$21.50 ; 1 Children ticket=$15.50
Step-by-step explanation:
3A+4C=126.50
2A+5C=120.50
We find the price of adult tickets 1st.
15A+20C=632.5
8A+20C=482
7A=632.5-482=150.5
1A=21.50
Now we find Children ticket
6A+8C=253
6A+15C=361.50
7C=108.50
1C=15.50
The answer is 2 3/10. how do you not know this
