mike charges 100 dollars service charge plus 65 dollars per hour.
amy charges 40 dollars service charge plus 80 dollars per hour.
the cost to repair from mike is 100 + 65 * x.
the cost to repair from amy is 40 + 80 * x.
let x = 3 and you get:
the cost to repair from mike is 100 + 65 * 3 = 295 dollars.
the cost to repaid from amy is 40 + 80 * 3 = 280 dollars.
as long as the time to repair is 3 hours or less, the cost to repair will be cheaper from amy.
Answer:
(-5,-3)
Step-by-step explanation:
y values minus y values over x values minus x values
Answer:
is this weird but in all my years of school I've never learn something like this
Step-by-step explanation:
Total tomato plants = 24 + 30 = 54
Total green bean plants = 18 + 25 = 43
The ratio of tomato plants to green bean plants


a. 9:00 AM is the 60 minute mark:

b. 8:15 and 8:30 AM are the 15 and 30 minute marks, respectively. The probability of arriving at some point between them is

c. The probability of arriving on any given day before 8:40 AM (the 40 minute mark) is

The probability of doing so for at least 2 of 5 days is

i.e. you're virtually guaranteed to arrive within the first 40 minutes at least twice.
d. Integrate the PDF to obtain the CDF:

Then the desired probability is
