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xz_007 [3.2K]
3 years ago
9

Suppose that a and b are integers, a ≡ 4 ( mod 13 ) and b ≡ 9 ( mod 13 ) . Find the integer c with 0 ≤ c ≤ 12 such that: c ≡ 9 a

( mod 13 ) c ≡ 11 b ( mod 13 ) c ≡ a + b ( mod 13 ) c ≡ a 2 + b 2 ( mod 13 ) c ≡ a 2 − b 2 ( mod 13 )
Mathematics
1 answer:
g100num [7]3 years ago
3 0

Answer:

A) For c ≡ 9 a ( mod 13 ) ; C is 10

B) For c ≡ 11 b ( mod 13 ) ; C is 8

C) For c ≡ a + b ( mod 13 ); C is 0

D) For c ≡ a² + b² ( mod 13 ); C is 6

E) For c ≡ a² − b² ( mod 13 ) ; C is 0

Step-by-step explanation:

This is a modular arithmetic problem where a ≡ 4 ( mod 13 ) and b ≡ 9 ( mod 13 ).

And 0 ≤ c ≤ 12.

A) c ≡ 9 a ( mod 13 )

Substituting the value of a to obtain;

c ≡ 9 x4 ( mod 13 ) = 36 mod 13

To find 36 mod 13 using the Modulo Method, we first divide the Dividend (36) by the Divisor (13).

Second, we multiply the whole part of the Quotient in the previous step by the Divisor (13).

Then finally, we subtract the answer in the second step from the Dividend (36) to get the answer. Here is the math to illustrate how to get 36 mod 13 using Modulo Method:

36 / 13 = 2.769231

2 x 13 = 26

36 - 26 = 10

Thus, the answer to "What is 36 mod 13?" is 10

So C = 10

B) c ≡ 11 b ( mod 13 ) = 11 x 9 ( mod 13 ) = 99 ( mod 13 )

Using the same method as above,

99 ( mod 13 );

99 / 13 = 7.6155

7 x 13 = 91

99 - 91 = 8

So, C = 8

C)c ≡ a + b ( mod 13 ) = 4 + 9 (mod 13) = 13 (mod 13)

Thus;

13 / 13 = 1

1 x 13 = 13

13 - 13 = 0

So, C = 0

D)c ≡ a² + b² ( mod 13 ) = 4² + 9²( mod 13 ) = 16 + 81 ( mod 13 ) = 97 ( mod 13 )

Thus;

97 / 13 = 7.46154

7 x 13 = 91

97 - 91 = 6

So, C = 6

E)c ≡ a² − b² ( mod 13 )= 4² - 9²( mod 13 ) = 16 - 81 ( mod 13 ) = - 65 ( mod 13 )

Thus;

-65 / 13 = - 5

-5 x 13 = - 65

-65 - (-65) = 0

So, C = 0

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Following are the solution to the given choices:

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Using chebyshev's theorem :

\to P(|(x-\mu)|\leq k \sigma)\geq 1- \frac{1}{k^2}\\\\here \\ \to \mu=28.75\\\\ \to \sigma=4.44

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\to |(x-\mu)|= |20.17-28.75|=8.58 and \sigma=4.44 \\\\so, \\k= \frac{ |(x-\mu)| }{\sigma}

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\to (1- \frac{1}{k^2}) \times 100 \% =84\% \\\\\to (1- \frac{1}{k^2})=0.84 \\\\\to \frac{1}{k^2} =0.16 \\\\\to \frac{1}{k}=0.4\\\\ \to k=2.5 \\\\\to |(x-\mu)| \leq k \sigma  \\\\= |(x-\mu)|\leq 2.5 \times 4.44  \\\\  = |(x-\mu)|\leq 1.11 \\\\ =  (28.75\pm 11.1) \\\\\to \text{fares lies between}(17.65,39.85)

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\to 99.7 \% \\ lie \ between=28.75 \pm z(.03)\times \sigma \\\\ =28.75\pm 2.97 \times 4.44\\\\=(28.75\pm 13.1)\\\\=(15.65,41.85)\\

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using standard normal variate

x=20.17\\\\  z=-2 \\\\x=37.13\\\\ z=2\\\\\to P(20.17

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