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Finger [1]
3 years ago
15

What’s this answer ?

Mathematics
1 answer:
VladimirAG [237]3 years ago
5 0

Answer:

49

Step-by-step explanation:

x² - 14x + c

x² - 2(x)(7) + 7²

c = 7² = 49

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5 0
3 years ago
Read 2 more answers
Let a "binary code" be the set of all binary words, each consisting of 7 bits (i.e., 0 or 1 digits). For example, 0110110 is a c
Dmitry_Shevchenko [17]

Answer:

a) 128 codewords

b) 35 codewords

c) 29 codewords

Step-by-step explanation:

a) Each 7 bits consist of 0 or 1 digits. Therefore the first bit is two choices (0 or 1), the second bit is also two choices (0 or 1), continues this way till the last bit.

So total number of different code words in 7 bits is 2×2×2×2×2×2×2 = 2⁷ = 128

There are 128 different codewords.

b) A code word contains exactly four 1's this means that  it has four 1's and three 0's . Therefore, in 7 bits, we have four of the same kind and three of the same kind. Hence, total number of code words containing exactly four 1's =7!/(4!*3!) = 35 codewords

c) number of code words containing at most two 1's  = codewords containing zero 1's + words containing one 1's + words containing two 1's

Now codewords containing zero 1's = 0000000 so 1 word

Codewords containing one 1's = 1000000,0100000,0010000,0001000,0000100,0000010,0000001. That's seven words

Codewords containing two 1's means word containing two 1's and five 0's. So out of seven, two are of one kind and five are of another kind

Therefore, the total number of such words=7!/(2!*5!)=21

Hence, codewords having at most two 1's = 21+7+1 =29 codewords

8 0
3 years ago
Read 2 more answers
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