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2 years ago

Fin find the 31st term in -3, -5, -7, -9

Mathematics

1 answer:

vaieri [72.5K]2 years ago

6 0

Answer:

-63

Find common difference:

-5 -(-3)

-5 + 3

-2

nth term:

a + (n-1)(d) <u>[where a is first term, d is common difference]</u>

-3 + (n-1)(-2)

-3 -2n + 2

-2n -1

So 31st term:

-2(31) -1

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Required information Skip to question NOTE: This is a multi-part question. Once an answer is submitted, you will be unable to re

pogonyaev

The result for the given 14 length bit string is-

(a) The bit strings of length 14 which have exactly three 0s is 2184.

(b) The bit strings of length 14 which have same number of 0s as 1s is 17297280.

<h3>What is a bit strings?</h3>

A bit-string would be a binary digit sequence (bits). The size of the value is the amount of bits contained in the sequence.

A null string is a bit-string with no length.

The concept used here is permutation;

ⁿPₓ = n!/(n-x)!

Where, n is the total samples.

x is the selected samples.

(a) Because there are exactly three 0s, its other digits have always been one, hence the total of permutations is equal to;

n = 14 and x = 3 digits.

¹⁴P₃ = 14!/(14-3)!

¹⁴P₃ = 2184.

Thus, the bit strings of length 14 which have exactly three 0s is 2184.

(b) Because it is a bit string with 14 digits and it can only have digits 0 or 1, we must choose 7 0s and the remaining 7 1s, hence the number possible permutations is equal to;

n = 14 and x = 7 digits.

¹⁴P₇ = 14!/(14-7)!

¹⁴P₇ = 17297280.

Thus, the bit strings of length 14 which have same number of 0s as 1s is 17297280.

(c) The answer is identical to problem a, but the rest of a digits could be either 0 or 1, hence it must be doubled by 2¹¹ because there are 11 digits, each of which can be one of two options;

= 2184×2¹¹

= 4472832

Thus, he bit strings of length 14 which have at least three 1s is 4472832.

To know more about permutation, here

brainly.com/question/12468032

#SPJ4

The correct question is-

How many bit strings of length 14 have:

(a) Exactly three 0s?

(b) The same number of 0s as 1s?

(d) At least three 1s?

3 0

1 year ago

Combine like terms 7y-14y+3=

MaRussiya [10]

-7y+3 is the answer that you are looking for

5 0

3 years ago

Read 2 more answers

How many positive integers under 50 are multiples of neither 4 nor 6

lyudmila [28]

Candidates range from 1 to 50.
50/4=12 positive integers are multiples of 4
50/6=8 positive integers are multiples of 6
50/12=4 positive integers are multiples of 12 (LCM of 4 and 6)
By the inclusion/exclusion principle, the number of multiples of either 4 or 6 is equal to 12+8-4=16.

Therefore, the complement is the number of positive integers that are multiples of neither 4 nor 6 = 50-16=34.