The answer is sixteen thousand four hundred ninety
Let x = no. of 10 oz cups sold
Let y = no. of 14 oz cups sold
Let z = no. of 20 oz cups sold
:
Equation 1: total number of cups sold:
x + y + z = 24
:
Equation 2: amt of coffee consumed:
10x + 14y + 20z = 384
:
Equation 3: total revenue from cups sold
.95x + 1.15y + 1.50z = 30.60
:
Mult the 1st equation by 20 and subtract the 2nd equation from it:
20x + 20y + 20z = 480
10x + 14y + 20z = 384
------------------------ subtracting eliminates z
10x + 6y = 96; (eq 4)
Mult the 1st equation by 1.5 and subtract the 3rd equation from it:
1.5x + 1.5y + 1.5z = 36.00
.95x + 1.15y+ 1.5z = 30.60
---------------------------subtracting eliminates z again
.55x + .35y = 5.40; (eq 5)
Multiply eq 4 by .055 and subtract from eq 5:
.55x + .35y = 5.40
.55x + .33y = 5.28
--------------------eliminates x
0x + .02y = .12
y = .12/.02
y = 6 ea 14 oz cups sold
Substitute 6 for y for in eq 4
10x + 6(6) = 96
10x = 96 - 36
x = 60/10
x = 6 ea 10 oz cups
That would leave 12 ea 20 oz cups (24 - 6 - 6 = 12)
Check our solutions in eq 2:
10(6) + 14(6) + 20(12) =
60 + 84 + 240 = 384 oz
A lot steps, hope it made some sense! I hope this helps!! ;D
Move all terms to the left side and set equal to zero. Then set each factor equal to zero.
x=10,−1
Answer:
The answer is "Option A and Option B".
Step-by-step explanation:
In question 1:
In all cases, the entire population is measured so that the actual medium discrepancy could be measured as well as an interval of trust cannot be used.
This issue would be that she calculated the ages with all representatives of both classes, such that she measured a whole population. It's not necessary.
In question 2:
When the p-value is 0.042. At 90% trust and 92% trust level 11 (p-value below 0.10 and 0.08) are not included. however the biggest confidence level of 92%. Consequently, the largest trust level where the 11 is Not included in the trust interval is 92% trust.
