Answer:
The number of customers needed to break even is
Step-by-step explanation:
Rent per month = $ 7200
Rent per year, $ 
= $ 86400 -------------(1)
Insurance per year = $ 4200 ---------(2)
Total direct cost for 2 servers (withoout tax) per year,
= $ 
= $ 60000
Total cost for helper ( without tax) per year,
= $ 
= $ 15000
Total cost for employees per year (without tax),
= $ (60000 +15000) = $ 75000
Total cost per year for employees including tax,
$ 
= $ 
= $ 80737.5 -----------------------(3)
So,
total cost for running the restaurant (except food)
= (1) + (2) + (3) = $ 
= $ 171337.5
Average total profit per customer,
= ( Average per customer payment - average cost of food per customer)
= $ (45.47 - 20.18)
= $ 25.29
So, total number of customer needed to break even,
= 
\simeq 6775
Answer:
Equivalent
Step-by-step explanation:
Look at 6/18. 18 divided by 6 is 3, which is the denominator. 6 divided by 6 is 1, so the numerator. So 1/3.
Answer:
1hr : 10min
Step-by-step explanation:
2 hrs = 120 min
1 hr : 60 min
so
1hr : 10min
Answer:
Does not factor; domain = -∞<x<∞, range = f(x) ≤ 2.
Step-by-step explanation:
Assuming your end goal is to factor the polynomial
If we use the guess and check method we can see that there is no way to factor this polynomial.
However, if you are looking for a domain and range we can graph it on a graphing calculator. Just by looking at the graph, we can see that the domain is all real numbers, or -∞<x<∞, and the range is f(x) ≤ 2.
I hope this helped!
Answer: 0.171887
Step-by-step explanation:
Given that S0 and S1 are binary symbol of equal probabilty;
P(S0) = P(S1) = 0.5
This probability is a Binomial random variable of sequence Sn, where Sn counting the number of success in a repeated trials.
P(Sn =X) = nCx p^x (1-p)^(n-1)
Pr(at most 3) = P(0<= x <=3) = P(X=0) + 0) + P(X=1) + P(X=2) + P(X=3)
Since there are only 2 values that occur in sequence 0 and 1 ( or S0 and S1).Let the distribution be given by the sequence (0111111111),(1011111111),(11011111111),...(1111111110) for Sn= 1 is the sequence for 1 error.
10C0, 10C1, 10C2, and 10C3 is the number of sequences in value for X= 0, 1, 2, 3 having value 0 and others are 1. Let the success be p(S0)=0.5 and p(S1)= 0.5
P(0<= X <=3) = 10C0 × (0.5)^0 × (0.5)^10 + 10C1× (0.5)¹ × (0.5)^9 + 10C2 × (0.5)² × (0.5)^8 + 10C3(0.5)³(0.5)^7
= 1 × (0.5)^10 + 10 × (0.5)^10 + 45 × (0.5)^10 + 120 × (0.5)^10
=0.000977 + 0.00977 + 0.04395 + 0.11719
= 0.171887
