Answer:
A = 0.8 litres
B = 0.7 litres
C = 0.5 litres
D = 0.2 litres
Step-by-step explanation
Here's what we know:
1. Jug A = B + .1 litres
2. Jug C = B - 200 (or 0.2 litres)
3. Jug D = .25 x A
4. Jug A + Jug B = 1.5 litres
In problem 1, we learned that Jug A has .1 litres more than Jug B and in problem 4, the two of them added together are 1.5 litres. To solve this we can combine the problems.
B + .1 litres + B = 1.5 litres
2B + .1 = 1.5
Subtract .01 from each side and you have 2B = 1.4
Divide each side by 2 and you have B = 0.7 litres
Plug this info into problem 1 and you can solve for A. (0.7 + 0.1 = 0.8)
Plug this info into problem 2 and you can solve for C. (0.7 - 0.2 = 0.5)
Since you have A, you can use that info to solve problem 3 (0.25 x 0.8 = 0.2)
Answer:
Step-by-step explanation:
5 can cancel
Answer:
Step-by-step explanation:
Assuming that all of the 255 sold seats were filled, then
[tex]\frac{sold}{total} *100\\[percent filled]
(225/260)*100=86.53%
100%-86.53%=13.4%
13.4% of seats are empty!
Answer:
9ms
Step-by-step explanation:
Hi,
x + y = 350 (equation 1)
3x + 2y = 950 (equation 2)
Use Substitution Method
Isolate variable x in equation 1
x = 350 – y (equation 3)
Substitute equation 3 into equation 2
3(350 – y) + 2y = 950
1050 – 3y + 2y = 950
3y – 2y = 1050 – 950
y = 100
Substitute y = 100 into equation 1
x + 100 = 350
x = 250
Answer: 250 $3 tickets and 100 $2 tickets were sold.
