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astraxan [27]
3 years ago
7

A corporation must appoint a​ president, chief executive officer​ (CEO), chief operating officer​ (COO), and chief financial off

icer​ (CFO). It must also select​ a planning committee with five different members. There are 17 qualified​ candidates, and officers can also serve on the committee. How many different ways can the committee be​ appointed?
Mathematics
1 answer:
Travka [436]3 years ago
6 0

Answer:

There are 524,560 ways that the committee can be​ appointed.

Step-by-step explanation:

There are 9 different positions, a president, 3 officers and five committee members.

So

P - O1 - O2 - O3 - C1 - C2 - C3 - C4 - C5

There are 17 candidates, any of them can be president. So P = 17.

All of the others can be an officers. So O1 = 16, O2 = 15(17 subtracted by the president and O1), O3 = 14.

The officers can be committee members, so, C1 = 16, C2 = 15, C3 = 14, C4 = 13, C5 = 12.

How many different ways can the committee be​ appointed?

This is the sequence C1...C5:

N = 16*15*14*13*12 = 524,560

There are 524,560 ways that the committee can be​ appointed.

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3 years ago
You deposit $400in an account. The account earns $18 simple interest in 9 months. What is the annual interest rate?
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Answer:

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Step-by-step explanation:

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3 years ago
Determine formula of the nth term 2, 6, 12 20 30,42​
nalin [4]

Check the forward differences of the sequence.

If \{a_n\} = \{2,6,12,20,30,42,\ldots\}, then let \{b_n\} be the sequence of first-order differences of \{a_n\}. That is, for n ≥ 1,

b_n = a_{n+1} - a_n

so that \{b_n\} = \{4, 6, 8, 10, 12, \ldots\}.

Let \{c_n\} be the sequence of differences of \{b_n\},

c_n = b_{n+1} - b_n

and we see that this is a constant sequence, \{c_n\} = \{2, 2, 2, 2, \ldots\}. In other words, \{b_n\} is an arithmetic sequence with common difference between terms of 2. That is,

2 = b_{n+1} - b_n \implies b_{n+1} = b_n + 2

and we can solve for b_n in terms of b_1=4:

b_{n+1} = b_n + 2

b_{n+1} = (b_{n-1}+2) + 2 = b_{n-1} + 2\times2

b_{n+1} = (b_{n-2}+2) + 2\times2 = b_{n-2} + 3\times2

and so on down to

b_{n+1} = b_1 + 2n \implies b_{n+1} = 2n + 4 \implies b_n = 2(n-1)+4 = 2(n + 1)

We solve for a_n in the same way.

2(n+1) = a_{n+1} - a_n \implies a_{n+1} = a_n + 2(n + 1)

Then

a_{n+1} = (a_{n-1} + 2n) + 2(n+1) \\ ~~~~~~~= a_{n-1} + 2 ((n+1) + n)

a_{n+1} = (a_{n-2} + 2(n-1)) + 2((n+1)+n) \\ ~~~~~~~ = a_{n-2} + 2 ((n+1) + n + (n-1))

a_{n+1} = (a_{n-3} + 2(n-2)) + 2((n+1)+n+(n-1)) \\ ~~~~~~~= a_{n-3} + 2 ((n+1) + n + (n-1) + (n-2))

and so on down to

a_{n+1} = a_1 + 2 \displaystyle \sum_{k=2}^{n+1} k = 2 + 2 \times \frac{n(n+3)}2

\implies a_{n+1} = n^2 + 3n + 2 \implies \boxed{a_n = n^2 + n}

6 0
2 years ago
Easy question right here!
Bad White [126]
12 glad I could help!
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Read 2 more answers
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Tamiku [17]
George would have 2200
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