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:
<h2>Per week: $1041.04</h2><h2>Per month: $4610.32</h2><h2>Per year: 55323.84</h2>
Answer:
See solutions below
Step-by-step explanation:
From the given diagram;
AC = opposite
AB = 7 = hypotenuse
Angle of elevation = 70 degrees
Using SOH CAH TOA
Sin theta = opp/hyp
Sin theta = AC/AB
Sin 70 = AC/7
AC = 7sin70
AC = 7(0.9397)
AC = 6.58
Similarly
tan 70 = AC/BC
tan 70 = 6.58/BC
BC = 6.58/tan70
BC = 6.58/2.7475
BC = 2.39
tan m<A = BC/AC
tanm<A = 2.39/6.58
tan m<A = 0.3632
m<A = 19.96degrees
you would change the denominators to the least common multiple, in this case, 12. then you would change the fractions to 8/12, 14/12, and 7/12. you would add those, and get 29/12. divide 29 by 12, and get 2. add the rest to get 2 5/12.
