He had to put $65 bc. he had to balance his account then added $25
75
60+7.20= 67.20
67.20+7.80= 75
Answer:
How many drinks should be sold to get a maximal profit? 468
Sales of the first one = 345 cups
Sales of the second one = 123 cups
Step-by-step explanation:
maximize 1.2F + 0.7S
where:
F = first type of drink
S = second type of drink
constraints:
sugar ⇒ 3F + 10S ≤ 3000
juice ⇒ 9F + 4S ≤ 3600
coffee ⇒ 4F + 5S ≤ 2000
using solver the maximum profit is $500.10
and the optimal solution is 345F + 123S
Since we're doing compound interest for four years it will be simpler than daily.
In case you didn't know the formula, here it is: A = P (1 +r/n)^(nt)
now lets substitute.
A = 500 (1 + 0.03)^4
A = 500 (1.03)^4
A = 500 (1.12550881)
A = 562.754405
Since that decimal is too big for cash, let's round it. Notice that the number after 5 is lower than 5. Thus the first two numbers stay the same and the final answer will be..
A = 562.75
Hope this helps :3
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