Round .1287 to the nearest thousandth
1 answer:
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Answer:
a) n<1 and n>5
b) 0 < n < -4
c) n > 2 and n < -2
Step-by-step explanation:
The signal is given by x[n] = 0 for n < -1 and n > 3
The problem asks us to determine the values of n for which it's guaranteed to be zero.
a) x[n-2]
We know that n -2 must be less than -1 or greater than 3.
Therefore we're going to write down our inequalities and solve for n
Therefore for n<1 and n>5 x [n-2] will be zero
b) x [n+ 3]
Similarly, n + 3 must be less than -1 or greater than 3
Therefore for n< -4 and n>0, in other words, for 0 < n < -4 x[n-2] will be zero
c)x [-n + 1]
Similarly, -n+1 must be less than -1 or greater than 3
Therefore, for n > 2 and n < -2 x[-n+1] will be zero
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;
= [( 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