So i'm assuming 33^a is saying any number that is a square of 33 inside of 1,313 so first, to find how many even numbers divide 1,313 by two to get 656.5 so if I am not mistaken 656 numbers in 1,313 will be even. Now find all squares of 33 that don't pass 1,313 which is 33 itself and 1,089 because 33^3 is 35,937 so add 2 to 656 to get 668. so the probability is 658/1,313 i believe.
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Answer:</h3>
- B. f(x) = 3,000(0.85)^x
- $1566.02
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Step-by-step explanation:</h3>
Part A
At the end of the year, the value of the computer system is ...
... (beginning value) - 15% · (beginning value) = (beginning value) · (1 - 0.15)
... = 0.85 · (beginning value)
Since the same is true for the next year and the next, the multiplier after x years will be 0.85^x. Then the value after x years is ...
... f(x) = (beginning value) · 0.85^x
The beginning value is given as $3000, so this is ...
... f(x) = 3000·0.85^x
____
Part B
For x=4, this is ...
... f(4) = 3000·0.85^4 = 3000·0.52200625 ≈ 1566.02
The value after 4 years is $1566.02.
Answer:
Sorry im not good at math
Answer:
Multiply by Four. x4
12/3
16/4
24/6
36/9
40/10
Step-by-step explanation:
