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.
The answer is 4/15, because we have to get the denominaters to be the same, so the LCM of 5 and 15 is 15, so 15/5 is 3 and 4*3 is 12. Now, its 12/15-8/15 which is 4/15. Hope this helps!
Answer:
Mike drew the following figure in his notebook. What is the value of ? Show all work
Step-by-step explanation:
Is 288
Answer:
C° = 71.6056
Step-by-step explanation:
Law of Cosines: c² = a² + b² - 2abcosC°
Step 1: Plug in known variables
29² = 30² + 15² - 2(30)(15)cosC°
Step 2: Evaluate
841 = 900 + 225 - 900cosC°
-59 = 225 - 900cosC°
-284 = -900cosC°
71/225 = cosC°
cos⁻¹(71/225) = C°
C° = 71.6056
And we have our answer!
Answer:
if your adding them together it would be -70
Step-by-step explanation:
4× -9 =-36
3×-8=-24
