The answer is 1432 x 54 = 77,328
Answer:
200 new students
Step-by-step explanation:
Let the number of students enrolled 2 years ago be X
Then last year because of the increase of 25% it became 500
Increase of 25% = 0.25X
Total student population <u>after </u> increase = X + 0.25X = 1.25X
We know that after the first increase, the number became 500
So 1.25X = 500, X = 500/1.25 = 400 (number 2 years ago)
Last year it increased from 500 by 20%. So current number of students = 500 x 1.20 = 600
Difference between current number and number two years ago gives the number of new students as 600-400 = 200
Answer:
Step-by-step explanation:
25 + 4x = 4x + 25
0 = 0
infinitely many solutions
<span>We are not told how often the interest is compounded, so assuming it is <em /><u><em>compounded yearly</em></u>, you need to keep $9.99 in the account to pay the fee.
<u><em>Explanation: </em></u>
Compound interest follows the formula A=p(1+r)^t,
where:
A is the total amount in the account,
p is the amount of principal,
r is the interest rate as a decimal number,
and t is the number of years.
<u>For our problem: </u>
A = 9.99,
p is unknown,
r = 0.018% = 0.00018,
and t=1.
<u>This gives us: </u>
9.99=p(1+0.00018)^1;
9.99=p(1.00018).
<u>Divide both sides by 1.00018: </u>
9.99=p.</span>
The permutation equation
is true for all integers n ≥ 3
<h3>How to prove the equation?</h3>
We have:
![^{n+1}P_3 - ^nP_3= 3 * ^nP_2](https://tex.z-dn.net/?f=%5E%7Bn%2B1%7DP_3%20-%20%5EnP_3%3D%203%20%2A%20%5EnP_2)
Apply the following permutation formula
![^{n}P_r = \frac{n!}{(n- r)!}](https://tex.z-dn.net/?f=%5E%7Bn%7DP_r%20%3D%20%5Cfrac%7Bn%21%7D%7B%28n-%20r%29%21%7D)
So, we have:
![\frac{(n + 1)!}{(n + 1 - 3)!} - \frac{n!}{(n - 3)!} = 3 * \frac{n!}{(n - 2)!}](https://tex.z-dn.net/?f=%5Cfrac%7B%28n%20%2B%201%29%21%7D%7B%28n%20%2B%201%20-%203%29%21%7D%20-%20%5Cfrac%7Bn%21%7D%7B%28n%20-%203%29%21%7D%20%3D%203%20%2A%20%5Cfrac%7Bn%21%7D%7B%28n%20-%202%29%21%7D)
Evaluate the difference
![\frac{(n + 1)!}{(n - 2)!} - \frac{n!}{(n - 3)!} = 3 * \frac{n!}{(n - 2)!}](https://tex.z-dn.net/?f=%5Cfrac%7B%28n%20%2B%201%29%21%7D%7B%28n%20-%202%29%21%7D%20-%20%5Cfrac%7Bn%21%7D%7B%28n%20-%203%29%21%7D%20%3D%203%20%2A%20%5Cfrac%7Bn%21%7D%7B%28n%20-%202%29%21%7D)
Expand
![\frac{(n + 1)!}{(n - 2)(n - 3)!} - \frac{n!}{(n - 3)!} = 3 * \frac{n!}{(n - 2)(n - 3)!}](https://tex.z-dn.net/?f=%5Cfrac%7B%28n%20%2B%201%29%21%7D%7B%28n%20-%202%29%28n%20-%203%29%21%7D%20-%20%5Cfrac%7Bn%21%7D%7B%28n%20-%203%29%21%7D%20%3D%203%20%2A%20%5Cfrac%7Bn%21%7D%7B%28n%20-%202%29%28n%20-%203%29%21%7D)
Multiply through by (n - 3)!
![\frac{(n + 1)!}{(n - 2)} - n! = 3 * \frac{n!}{(n - 2)}](https://tex.z-dn.net/?f=%5Cfrac%7B%28n%20%2B%201%29%21%7D%7B%28n%20-%202%29%7D%20-%20n%21%20%3D%203%20%2A%20%5Cfrac%7Bn%21%7D%7B%28n%20-%202%29%7D)
Expand
![\frac{(n + 1) * n!}{(n - 2)} - n! = 3 * \frac{n!}{(n - 2)}](https://tex.z-dn.net/?f=%5Cfrac%7B%28n%20%2B%201%29%20%2A%20n%21%7D%7B%28n%20-%202%29%7D%20-%20n%21%20%3D%203%20%2A%20%5Cfrac%7Bn%21%7D%7B%28n%20-%202%29%7D)
Divide through by n!
![\frac{(n + 1)}{(n - 2)} - 1 = \frac{3}{(n - 2)}](https://tex.z-dn.net/?f=%5Cfrac%7B%28n%20%2B%201%29%7D%7B%28n%20-%202%29%7D%20-%201%20%3D%20%20%5Cfrac%7B3%7D%7B%28n%20-%202%29%7D)
Take the LCM
![\frac{n + 1 - n + 2}{(n - 2)} = \frac{3}{(n - 2)}](https://tex.z-dn.net/?f=%5Cfrac%7Bn%20%2B%201%20-%20n%20%2B%202%7D%7B%28n%20-%202%29%7D%20%20%3D%20%20%5Cfrac%7B3%7D%7B%28n%20-%202%29%7D)
Evaluate the like terms
![\frac{3}{(n - 2)} = \frac{3}{(n - 2)}](https://tex.z-dn.net/?f=%5Cfrac%7B3%7D%7B%28n%20-%202%29%7D%20%3D%20%20%5Cfrac%7B3%7D%7B%28n%20-%202%29%7D)
Both sides of the equations are the same.
Hence, the permutation equation
is true for all integers n ≥ 3
Read more about permutation at:
brainly.com/question/11732255
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