Answer: nickels = 14, dimes = 42, quarters = 18
<u>Step-by-step explanation:</u>
Let n represent nickels <em>which has a value of $0.05 </em>
Let d represent dimes <em>which has a value of $0.10</em>
Let q represent quarters <em>which has a value of $0.25</em>
It is given that:
n = n
d = 3n
q = n + 4
and their total value is $9.40 ⇒ n + d + q = $9.40
Use substitution to find the quantity of nickels:
.05(n) + .10(3n) + .25(n + 4) = 9.40
5n + 10(3n) + 25(n+4) = 940 <em>multiplied by 100 </em>
5n + 30n + 25n + 100 = 940 <em>distributed</em>
60n + 100 = 940 <em>added like terms</em>
60n = 840 <em>subtracted 100</em>
n = 14 <em>divided by 60</em>
n = 14
d = 3n
= 3(14)
= 42
q = n + 4
= (14) + 4
= 18
Answer:

Step-by-step explanation:

Answer: 0.75
Step-by-step explanation:
Given : Interval for uniform distribution : [0 minute, 5 minutes]
The probability density function will be :-

The probability that a given class period runs between 50.75 and 51.25 minutes is given by :-
![P(x>1.25)=\int^{5}_{1.25}f(x)\ dx\\\\=(0.2)[x]^{5}_{1.25}\\\\=(0.2)(5-1.25)=0.75](https://tex.z-dn.net/?f=P%28x%3E1.25%29%3D%5Cint%5E%7B5%7D_%7B1.25%7Df%28x%29%5C%20dx%5C%5C%5C%5C%3D%280.2%29%5Bx%5D%5E%7B5%7D_%7B1.25%7D%5C%5C%5C%5C%3D%280.2%29%285-1.25%29%3D0.75)
Hence, the probability that a randomly selected passenger has a waiting time greater than 1.25 minutes = 0.75
Answer:
100 + 60 + 8 + 0.7
Step-by-step explanation:
Expanded form or expanded notation is a way of writing numbers to see the value of individual digits.
Remember property of exponents

add exponents of same base
ok so
50,000 is 5 times 10^4
so
50000 times 10^15=
5 times 10^4 times 10^15=
5 times

=
5 times 10^19