Answer: ...
Step-by-step explanation: I will rather let you use a variable and solve (for example: x!
Answer:
Pi
Step-by-step explanation:
We will get the number of possible selections, and then subtract the number less than 25 cents.
We can choose the number of dimes 5 ways 0,1,2,3 or 4.
We can choose the number of nickels 4 ways 0,1,2 or 3.
We can choose the number of quarters 3 ways 0,1, or 2.
That's 5*4*3 = 60 selections
Now we must subtract from the 60 the number of selections of coins that are less than 25 cents. These will involve only dimes and nickels.
To get a selection of coin worth less than 25 cents:
If we use no dimes, we can use 0,1,2 on all 3 nickels.
That's 4 selections less than 25 cents. (that includes the choice of No coins at all in the 60, which we must subtract).
If we use exactly 1 dime , we can use 0,1,2, or all 3 nickels.
That's the 3 combinations less than 25 cents.
And there is 1 other selection less than 25 cents, 2 dimes and no nickels.
So that's 4+3+1 = 8 selections which we must subtract from the 60.
Answer 60-8 = 52 selections of coins worth 25 cents or more.
Answer:
5040,56
Step-by-step explanation:
We have to construct pass words of 4 digits
a) None of the digits can be repeated
We have total digits as 0 to 9.
4 digits can be selected form these 10 in 10P4 ways (since order matters in numbers)
No of passwords = 10P4
= 
b) start with 5 and end in even digit
Here we restrain the choices by putting conditions
I digit is compulsorily 5 and hence only one way
Last digit can be any one of 0,2,4,6,8 hence 5 ways
Once first and last selected remaining 2 digits can be selected from remaining 8 digits in 8P2 ways (order counts here)
=56
Answer:
Step-by-step explanation:
perp. -5
y + 7 = -5(x - 4)
y + 7 = -5x + 20
y = -5x + 13
