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N76 [4]
3 years ago
11

What is an example for bad debt

Mathematics
2 answers:
Novay_Z [31]3 years ago
7 0
You have Bad credit.
Maslowich3 years ago
5 0
What are the answers given?

one example could be debts with high interest rates.
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Your bathtub has 12 gallons of water in it and you start to drain it. The water drains at a constant rate of 1/2
erastova [34]

Step-by-step explanation:

All ik is that its gonna take 10 minutes to drain.

6 0
3 years ago
Jace gathered the data in the table. He found the approximate line of best fit to be y = –0.7x + 2.36.
MaRussiya [10]
The residual value is -1.14.
Plug 5 into x
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3 years ago
Lelia has 5 1/4 hours of homework to do over the weekend. She does 2/3 of her homework on Saturday so sh can get it out of the w
dedylja [7]
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5 0
3 years ago
If n is a positive integer, how many 5-tuples of integers from 1 through n can be formed in which the elements of the 5-tuple ar
sleet_krkn [62]

This question is incomplete, the complete question is;

If n is a positive integer, how many 5-tuples of integers from 1 through n can be formed in which the elements of the 5-tuple are written in increasing order but are not necessarily distinct.

In other words, how many 5-tuples of integers  ( h, i , j , m ), are there with  n ≥ h ≥ i ≥ j ≥ k ≥ m ≥ 1 ?

Answer:

the number of 5-tuples of integers from 1 through n that can be formed is [ n( n+1 ) ( n+2 ) ( n+3 ) ( n+4 ) ] / 120

Step-by-step explanation:

Given the data in the question;

Any quintuple ( h, i , j , m ), with n ≥ h ≥ i ≥ j ≥ k ≥ m ≥ 1

this can be represented as a string of ( n-1 ) vertical bars and 5 crosses.

So the positions of the crosses will indicate which 5 integers from 1 to n are indicated in the n-tuple'

Hence, the number of such quintuple is the same as the number of strings of ( n-1 ) vertical bars and 5 crosses such as;

\left[\begin{array}{ccccc}5&+&n&-&1\\&&5\\\end{array}\right] = \left[\begin{array}{ccc}n&+&4\\&5&\\\end{array}\right]

= [( n + 4 )! ] / [ 5!( n + 4 - 5 )! ]

= [( n + 4 )!] / [ 5!( n-1 )! ]

= [ n( n+1 ) ( n+2 ) ( n+3 ) ( n+4 ) ] / 120

Therefore, the number of 5-tuples of integers from 1 through n that can be formed is [ n( n+1 ) ( n+2 ) ( n+3 ) ( n+4 ) ] / 120

4 0
3 years ago
HELP!!!!!!!!!!!!!! ILL GIVE BRAINILIST
Scorpion4ik [409]

\frac{1}{3}  \div  \frac{4}{5}  \\ \\  \frac{1}{3}  \times  \frac{5}{4}  \\  \\   \frac{1 \times 4}{3 \times 4}  \times  \frac{5 \times 3}{4 \times 3}  \\  \frac{4}{12}  \times  \frac{15}{12}  \\  \frac{4 \times 5}{12}  \\  \frac{60}{12}  \\ 5  \: \:  \:  \: answer

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2 years ago
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