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:
195
Step-by-step explanation:
87+38+40=165
360-165=195
Hope this helps!
If not, I am sorry.
D. Three less than five times the quantity of a number plus two.
Answer:
Step-by-step explanation:
-2x + 3y – 4z = 8 ----------------(I)
5x – 3y + 5z = -8 -----------------(II)
7x – 3y + 3z = 8 ------------------(III)
Add equation (I) & (II) and thus y will be eliminated
(I) -2x + 3y – 4z = 8
(II) <u>5x – 3y + 5z = -8</u> {Add}
3x + z = 0 ------------------------(A)
Multiply equation (II) by (-1) and then add with equation (III). Thus y will be eliminated.
(II) * (-1) -5x + 3y - 5z = +8
<u>7x – 3y + 3z = 8</u> {Add}
2x -2z = 16 ---------------(B)
Multiply equation (A) by 2 and then add. Thus z will be eliminated and we will get the value of x
(A) * 2 6x + 2z = 0
(B) <u>2x - 2z = 16</u> {Add}
8x = 16
Divide both sides by 8
x = 16/8
x = 2
Plugin x = 2 in equation (A)
3x + z = 0
3*2 + z = 0
6 + z = 0
z = -6
Plug in x = 2 and z = - 6 in equation (I)
-2x +3y - 4z = 8
-2*2 + 3y - 4*(-6) = 8
-4 + 3y + 24 = 8
3y + 20 = 8
3y = 8 - 20
3y = -12
y = -12/3
y = -4
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You will add them all together, then multiply the sum by 6.
713/6 = 118.833333
this rounded to the nearest hundredth
$118.83 per month
