A) The peak of the curve has coordinates (3, 45). Thus the maximum is 45 hundred dollars. It occurs when 3 hundred dollars are spent on advertisiing.
The maximum possible monthly profit of the shop is about $4500.
b) The x interval over which P(x) is greater than zero appears to be 0 to 10 hundred dollars. What is not clear from the question is whether the expected answer is (0, 10) or (0, 1000).
c) The maximum profit is earned when $300 is spent on advertising.
Answer:
Simplifying
f(r) = 5 + 1.75r
Multiply f * r
fr = 5 + 1.75r
Solving
fr = 5 + 1.75r
Solving for variable 'f'.
Move all terms containing f to the left, all other terms to the right.
Divide each side by 'r'.
f = 5r-1 + 1.75
Simplifying
f = 5r-1 + 1.75
Reorder the terms:
f = 1.75 + 5r-1
Step-by-step explanation:
tada i think
Answer:
Option E is correct.
The expected number of meals expected to be served on Wednesday in week 5 = 74.2
Step-by-step Explanation:
We will use the data from the four weeks to obtain the fraction of total days that number of meals served at lunch on a Wednesday take and then subsequently check the expected number of meals served at lunch the next Wednesday.
Week
Day 1 2 3 4 | Total
Sunday 40 35 39 43 | 157
Monday 54 55 51 59 | 219
Tuesday 61 60 65 64 | 250
Wednesday 72 77 78 69 | 296
Thursday 89 80 81 79 | 329
Friday 91 90 99 95 | 375
Saturday 80 82 81 83 | 326
Total number of meals served at lunch over the 4 weeks = (157+219+250+296+329+375+326) = 1952
Total number of meals served at lunch on Wednesdays over the 4 weeks = 296
Fraction of total number of meals served at lunch over four weeks that were served on Wednesdays = (296/1952) = 0.1516393443
Total number of meals expected to be served in week 5 = 490
Number of meals expected to be served on Wednesday in week 5 = 0.1516393443 × 490 = 74.3
Checking the options,
74.3 ≈ 74.2
Hence, the expected number of meals expected to be served on Wednesday in week 5 = 74.2
Hope this Helps!!!
Answer: C. -102 + (-25.5)
