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jekas [21]
2 years ago
14

There are 12 candidates running for any of 6 distinct positions in a school's student council. In how many different ways could

the 6 positions be filled?
Mathematics
1 answer:
deff fn [24]2 years ago
7 0

Answer:

Ways = 665280\ ways

Step-by-step explanation:

Given

Candidates = 12

Positions = 6

Required

Determine the number of ways to fill the vacant position

The first position can be any of the 12 candidates;

The next candidate can be any of the remaining 11

The next candidate can be any of the remaining 10

The next candidate can be any of the remaining 9

The next candidate can be any of the remaining 8

The next candidate can be any of the remaining 7

So: we have;

Ways= 12 * 11 * 10 * 9 * 8 * 7

Ways = 665280\ ways

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Joey buys a home for $263,000. His home is predicted to increase in value 5% each year.
ipn [44]

Answer:

$890,611.35   :)

Step-by-step explanation:

First, we determine the equation.

Y=263,000(1+0.05)^{x}

890,611.35

7 0
2 years ago
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Simplified product ?
mel-nik [20]

Answer:

Last choice is correct.

Step-by-step explanation:

\left(\sqrt{10x^4}-x\sqrt{5x^2}\right)\left(2\sqrt{15x^4}+\sqrt{3x^3}\right)

\left(x^2\sqrt{10}-x\cdot x\sqrt{5}\right)\left(2\cdot x^2\sqrt{15}+x\sqrt{3x}\right)

\left(x^2\sqrt{10}-x^2\sqrt{5}\right)\left(2x^2\sqrt{15}+x\sqrt{3x}\right)

x^2\sqrt{10}\left(2x^2\sqrt{15}+x\sqrt{3x}\right)-x^2\sqrt{5}\left(2x^2\sqrt{15}+x\sqrt{3x}\right)

2x^4\sqrt{150}+x^3\sqrt{30x}-2\sqrt{75}x^4-x^3\sqrt{15x}

2x^4\cdot5\sqrt{6}+x^3\sqrt{30x}-2\cdot5\sqrt{3}x^4-x^3\sqrt{15x}

10x^4\sqrt{6}+x^3\sqrt{30x}-10\sqrt{3}x^4-x^3\sqrt{15x}

10x^4\sqrt{6}+x^3\sqrt{30x}-10x^4\sqrt{3}-x^3\sqrt{15x}

Hence final answer is 10x^4\sqrt{6}+x^3\sqrt{30x}-10x^4\sqrt{3}-x^3\sqrt{15x}


5 0
2 years ago
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Answer:

D) 140

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140% of 10 is 14 and 14 added to ten is 24

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(9/ 4)/ 18 = 9/ (4 /18) A.True B.False
kotegsom [21]
False

<span>(9/ 4)/ 18
= 9/4 * 1/18
= 9/72
= 1/8 </span>
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3 years ago
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