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olga nikolaevna [1]
3 years ago
15

Solve the following congruence equations for X a) 8x = 1(mod 13) b) 8x = 4(mod 13) c) 99x = 5(mod 13)

Mathematics
2 answers:
xxMikexx [17]3 years ago
8 0

Answer:

a) 5+13k  where k is integer

b) 20+13k where k is integer

c)12+13k where k is integer

Step-by-step explanation:

(a)

8x \equiv 1 (mod 13) \text{ means } 8x-1=13k.

8x-1=13k

Subtract 13k on both sides:

8x-13k-1=0

Add 1 on both sides:

8x-13k=1

I'm going to use Euclidean Algorithm.

13=8(1)+5

8=5(1)+3

5=3(1)+2

3=2(1)+1

Now backwards through the equations:

3-2=1

3-(5-3)=1

3-5+3=1

(8-5)-5+(8-5)=1

2(8)-3(5)=1

2(8)-3(13-8)=1

5(8)-3(13)=1

So compare this to:

8x-13k=1

We see that x is 5 while k is 3.

Anyways 5 is a solution or 5+13k is a solution where k is an integer.

b)

8x \equiv 4 (mod 13)

8x-4=13k

Subtract 13k on both sides:

8x-13k-4=0

Add 4 on both sides:

8x-13k=4

We got this from above:

5(8)-3(13)=1

If we multiply both sides by 4 we get:

8(20)-13(12)=4

So x=20 and 20+13k is also a solution where k is an integer.

c)

[tex]99x \equiv 5 (mod 13)[/tex

99x-5=13k

Subtract 13k on both sides:

99x-13k-5=0

Add 5 on both sides:

99x-13k=5

Using Euclidean Algorithm:

99=13(7)+8

13=8(1)+5

Go back through the equations:

13-8=5

13-(99-13(7))=5

8(13)-99=5

99(-1)+8(13)=5

Compare this to 99x-13k=5 and see that x=-1 or -1+13=12 or 12+13k is a solution where k is an integer.

dangina [55]3 years ago
3 0

Answer:

a) x = 5 mod 13.

b)  x = 7 mod 13.

Step-by-step explanation:

a) 8x = 1  mod  13

x = 2,  16 = 3 mod 13

x = 3, 24 = 11 mod 13

x = 4, 32 = 6 mod 13

x = 5 , 40 = 1 mod 13

8x = 40

x = 5 mod 13.

b)   8x = 4 mod 13

x = 7,  56 = 4 mod 13.

7 = 4 mod 13

x = 7 mod 13.

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Items are inspected for flaws by two quality inspectors. Both inspectors inspect every item and the probability that an item has
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a)0.976

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Step-by-step explanation:

Let X_i be an item passed by inspector i

Let Y be the event that there is a fault in an item

The probability that an item has a flaw is 0.1 i.e. P(Y)=0.1

If a flaw is present ,it will be detected by the first inspector with probability 0.92 i.e.P(\bar{X_1}|Y)=0.92

So, P(X_1|Y)=1-0.92=0.08

If a flaw is present ,it will be detected by the second inspector with probability 0.7 i.e.P(\bar{X_2}|Y)=0.7

So,P(X_2|Y)=1-0.7=0.3

If an item does not have a flaw, it will be passed by the first inspector with probability 0.95 i.e. P(X_1|\bar{Y}) = 0.95

So, P(\bar{X_1}|\bar{Y}) = 1-0.95=0.05

If an item does not have a flaw, it will be passed by the second inspector with probability 0.8 i.e. P(X_2|\bar{Y}) = 0.8

So, P(\bar{X_2}|\bar{Y}) = 1-0.8=0.2

a)P(found by atleast one inspector | It has flaw )=1-P(found by none inspector | It has flow )

P(found by atleast one inspector | It has flaw )=1-P(X_1|Y) P(X_2|Y)

P(found by atleast one inspector | It has flaw )=1-0.08 \times 0.3

P(found by atleast one inspector | It has flaw )=0.976

Hence the probability that it will be found by at least one of the two inspectors if it has flaw is 0.976

b)P(Y|X_1)=\frac{P(X_1|Y) P(Y)}{P(X_1|Y) P(Y)+P(X_1|\bar{Y}) P(\bar{Y})}

P(Y|X_1)=\frac{0.08 \times 0.1}{0.08 \times 0.1+0.95 \times 0.9}=0.00926

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P( two inspectors draw different conclusions on the same item)=0.2402

D)

P(Y|(X_1 \cap X_2))=\frac{P(Y \cap X_1 \cap X_2)}{P(X_1 \cap X_2)}\\P(Y|(X_1 \cap X_2))=\frac{P(Y \cap X_1 \cap X_2)}{P(X_1 \cap X_2 \cap Y)+P(X_1 \cap X_2 \cap \bar{Y})}\\P(Y|(X_1 \cap X_2))=0.35

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