Represent the number of days by x. With this representation, the variable cost of the rental is 31.67x. The total cost is the sum of the fixed and variable costs. This value should not be more than $500. The equation below shows the relationship.
130 + 31.67x ≤ 500
Solving for x gives x ≤ 11.68
Thus, the maximum number of days to rent the car is only 11 days.
Answer:
1/6
Step-by-step explanation:
As the purple die has already rolled and the number landed is 6, to find the probability of the sum of the dice being 7, there is only one option for the yellow die:
purple result + yellow result = 7
6 + yellow result = 7
yellow result = 1
So the yellow die needs to roll a 1, over a total of 6 possibilities, so the probability is 1/6
Answer:
The estimated probability that Ginger will eat a a pizza everyday of the week is;
D. 8/10 = 80%
Step-by-step explanation:
The given parameters are;
The frequency with which Ginger buys launch = Everyday
The percentage of the time the cafeteria has pizza out = 80%
The outcome of 0 and 1 = No pizza available
The outcome of 2, 3, 4, 5, 6, 7, 8, and 9 = Pizza available
Therefore, we have the;
Group number Percentage of time pizza available
1 80%
2 80%
3 80%
4 80%
5 40%
6 100%
7 80%
8 100%
9 80%
10 80%
Therefore, the sum of the percentages outcome the days Ginger eats pizza = 0.8 + 0.8 + 0.8 + 0.8 + 0.4 + 1 + 0.8 + 1 + 0.8 + 0.8 = 8
The number of runs of simulation = 10 runs
The estimated probability that Ginger will eat a a pizza everyday of the week = (The sum of the percentages outcome the days Ginger eats pizza)/(The number of runs of simulation)
∴ The estimated probability that Ginger will eat a a pizza everyday of the week = 8/10
Answer:
3yz² - 3z² - 5y + 7
Step-by-step explanation:
Sum of two polynomials = –yz² - 3z² – 4y + 4
One of the polynomial = y - 4yz²- 3
Find the other polynomial
The other polynomial = sum of the polynomials - one of the polynomial
= –yz² - 3z² – 4y + 4 - (y - 4yz² - 3)
= –yz² - 3z² – 4y + 4 - y + 4yz² + 3
= -yz² + 4yz² - 3z² - 4y - y + 4 + 3
= 3yz² - 3z² - 5y + 7
A. 0 -2yz?
B. – 4y + 7 01 - 2yz
C. – 3y + 1 0 -5yz² + 3z² – 3y + 1 D. 3yz² - 3z² – 5y + 7