Let us formulate the independent equation that represents the problem. We let x be the cost for adult tickets and y be the cost for children tickets. All of the sales should equal to $20. Since each adult costs $4 and each child costs $2, the equation should be
4x + 2y = 20
There are two unknown but only one independent equation. We cannot solve an exact solution for this. One way to solve this is to state all the possibilities. Let's start by assigning values of x. The least value of x possible is 0. This is when no adults but only children bought the tickets.
When x=0,
4(0) + 2y = 20
y = 10
When x=1,
4(1) + 2y = 20
y = 8
When x=2,
4(2) + 2y = 20
y = 6
When x=3,
4(3) + 2y= 20
y = 4
When x = 4,
4(4) + 2y = 20
y = 2
When x = 5,
4(5) + 2y = 20
y = 0
When x = 6,
4(6) + 2y = 20
y = -2
A negative value for y is impossible. Therefore, the list of possible combination ends at x =5. To summarize, the combinations of adults and children tickets sold is tabulated below:
Number of adult tickets Number of children tickets
0 10
1 8
2 6
3 4
4 2
5 0
3+5=8 24÷8=3
3×3=9 muffins 5×3=15-pizza
I hope this helped!!!:)
Step-by-step explanation:
-2 = 0.75
-1 = 1.5
0 = 3
1 = 6
2 = 12
Answer:
Step-by-step explanation:
Given that the probability of a customer arrival at a grocery service counter in any one second is equal to 0.3
Assume that customers arrive in a random stream, so an arrival in any one second is independent of all others.
i.e. X the no of customers arriving is binomial with p = 0.3 and q = 1-0.3 =0.7
a) the probability that the first arrival will occur during the third one-second interval.
= Prob that customer did not arrive in first 2 seconds * prob customer arrive in 3rd sec
= 
b) the probability that the first arrival will not occur until at least the third one-second interval.
Prob that customer did not arrive in first two seconds *(Prob customer arrives in 3rd or 4th or 5th.....)
=
The term inside bracket is a geometric infinite progression with common ratio - 0.7 <1
Hence the series converges
Prob =