ok so hi im elizabeth can i help you
15÷3 and 24÷3
= 5:8
hop u found this helpful
QUESTION:
The code for a lock consists of 5 digits (0-9). The last number cannot be 0 or 1. How many different codes are possible.
ANSWER:
Since in this particular scenario, the order of the numbers matter, we can use the Permutation Formula:–
- P(n,r) = n!/(n−r)! where n is the number of numbers in the set and r is the subset.
Since there are 10 digits to choose from, we can assume that n = 10.
Similarly, since there are 5 numbers that need to be chosen out of the ten, we can assume that r = 5.
Now, plug these values into the formula and solve:
= 10!(10−5)!
= 10!5!
= 10⋅9⋅8⋅7⋅6
= 30240.
Answer:
26 candy bars and 12 sodas
Step-by-step explanation:
This problem can be solved by a simple system of equations
I am going to call x the number of candy bars that Sally bought and y the number of sodas that Sally bought.
The fact that she bought 38 itens means the x + y = 38.
She spent $50.38. The value she spent is the sum of the number of candy bars multplied by the value of the candy bar and the number of sodas multiplied by value of a soda. So 1.25x + 1.49y = 50.38
So we have to solve the following system:
1)x + y = 38
2)1.25x + 1.49y = 50.38
From 1), we have that x = 38 - y. I am going to replace it in 2)
1.25(38-y) + 1.49y = 50.38
-1.25y + 1.49y = 50.38 - 47.5
0.24y = 2.88
y = 12.
Sally bought 12 sodas.
In 1: x = 38-y = 38-12 = 26
Sally bourght 26 candy bars
Answer:
It would be the third one hope i helped
Step-by-step explanation:
