I'd say D just because if you can't rely on someone and I doesn't get there it's going to be really bad
Answer:
a. c+b≤360....equation 1
b. 2 c+1.5 b≥500.... equation 2, where c and b are the number of cans and bottles of water respectively.
c. The number of water bottles to be sold have to be equal to or more than 142 to cover the cost of renting costumes.
Explanation:
a.
<em>Step 1: Determine maximum number of cans and bottles</em>
As indicated, the number of cans and bottles can not exceed a certain value. This means that the number of cans and bottles can be either equal to or less than that value. The maximum number of cans and bottles can be represented in the following expression;
c+b≤m
where;
c=unknown
b=unknown
m=360
replacing;
c+b≤360....equation 1
b.
<em>Step 2: Determine total amount needed to raise $500</em>
Since $500 dollars is the minimum amount needed, the sales have to be $500 and more. This can be expressed as;
(C×c)+(B×b)≥T
where;
T=total amount needed
C=price per can of lemonade
c=number of cans sold
B=price per bottle of water
b=number of bottles sold
In our case;
T=$500
C=$2
c=unknown
B=$1.50
b=unknown
replacing;
(2×c)+(1.5×b)≥500
2 c+1.5 b≥500.... equation 2
c.
<em>Step 3: Determine least number of bottles of water that must be sold</em>
The least number of bottles of water that must be sold to cover the cost of renting costumes can be solved using equation 2 above;
2 c+1.5 b≥500
where;
c=144
b=unknown
replacing;
(2×144)+1.5 b≥500
288+1.5 b≥500
1.5 b≥500-288
1.5 b≥212
b≥212/1.5=141.33=142
b≥142, meaning the number of water bottles to be sold have to be equal or more than 142 to cover the cost of renting costumes.
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