Answer:
<u>Shelley did not budget enough. She will need additional US$ 16 for fulfilling the average sales of 760 sno-cones at the 4 events. With the budget she has, she just can supply 714 sno-cones (250 / 0.35).</u>
Step-by-step explanation:
1. Let's check all the information given for solving this question:
Cost of individual supply = US$ 0.35
Budget per week = US$ 250
Average sales of sno-cones per event = 190
Number of events per week = 4
2. Let's check if Shelley budgeted enough for the week events:
Total of sno-cones Shelley will sell = Average sales * Number of Events
Total of sno-cones Shelley will sell = 190 * 4 = 760
Cost of supplies for the week = Total of sno-cones * Cost of individual supply
Cost of supplies for the week = 760 * 0.35 = US$ 266
Budget per week = US$ 250
Budget per week - Cost of supplies for the week = 250 - 266 = - 16
<u>Shelley did not budget enough. She will need additional US$ 16 for fulfilling the average sales of 760 sno-cones at the 4 events. With the budget she has, she just can supply 714 sno-cones (250 / 0.35).</u>
Answer: The value of y is 
.
Explanation:
It is given that the graph of a proportional relationship passes through (12, 16)
and (1, y).
The graph of a proportional relationship means the x and y coordinates are in a proportion k. The equation of the graph is in the form of y=kx. Where k is the proportion factor.
It is given that the graph passing through (12,16).
So the equation of the line is,
put x=1.
Therefore, the value of y is .
The answer is f(x) = q * 1.025<span>x + 5, do you have that?
</span>
Take $190 and multiply it by 0.15 (28.5) and subtract that from $190= $161.50
Answer:
The fourth term of the expansion is -220 * x^9 * y^3
Step-by-step explanation:
Question:
Find the fourth term in (x-y)^12
Solution:
Notation: "n choose k", or combination of k objects from n objects,
C(n,k) = n! / ( k! (n-k)! )
For example, C(12,4) = 12! / (4! 8!) = 495
Using the binomial expansion formula
(a+b)^n
= C(n,0)a^n + C(n,1)a^(n-1)b + C(n,2)a^(n-2)b^2 + C(n,3)a^(n-3)b^3 + C(n,4)a^(n-4)b^4 +....+C(n,n)b^n
For (x-y)^12, n=12, k=3, a=x, b=-y, and the fourth term is
C(n,3)a^(n-3)b^3
=C(12,3) * x^(12-3) * (-y)^(3)
= 220*x^9*(-y)^3
= -220 * x^9 * y^3
