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.
Yes they can be greater than one no they can't be grater than 2
Answer:
.76 meters
Step-by-step explanation:
The area of a circle is pi*r^2
3.14*.85^2 = 2.26865
Dividing this into thirds gives us approximately .75621666666
Round to the nearest hundredth.
.76
Label the answer.
.76 meters
Answer:
AC =4DB
Step-by-step explanation:
D is midpoint of AB .So, AD = DB
AB = 2DB ---------- (i)
B is the midpoint of AC.So, AB = BC
AC = 2AB
= 2*2DB {FROM 1}
=4DB