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Gennadij [26K]
3 years ago
11

Prove the following statements: (a) For every integer x, if x is even, then for every integer y, xy is even. (b) For every integ

er x and for every integer y, if x is odd and y is odd then x y is even. (c) For every integer x, if x is odd then x 3 is odd.
Mathematics
1 answer:
Murrr4er [49]3 years ago
3 0

Answer and Step-by-step explanation:

(a) Given that x and y is even, we want to prove that xy is also even.

For x and y to be even, x and y have to be a multiple of 2. Let x = 2k and y = 2p where k and p are real numbers. xy = 2k x 2p = 4kp = 2(2kp). The product is a multiple of 2, this means the number is also even. Hence xy is even when x and y are even.

(b) in reality, if an odd number multiplies and odd number, the result is also an odd number. Therefore, the question is wrong. I assume they wanted to ask for the proof that the product is also odd. If that's the case, then this is the proof:

Given that x and y are odd, we want to prove that xy is odd. For x and y to be odd, they have to be multiples of 2 with 1 added to it. Therefore, suppose x = 2k + 1 and y = 2p + 1 then xy = (2k + 1)(2p + 1) = 4kp + 2k + 2p + 1 = 2(kp + k + p) + 1. Let kp + k + p = q, then we have 2q + 1 which is also odd.

(c) Given that x is odd we want to prove that 3x is also odd. Firstly, we've proven above that xy is odd if x and y are odd. 3 is an odd number and we are told that x is odd. Therefore it follows from the second proof that 3x is also odd.

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a ≈ 5.3

Step-by-step explanation:

Using Pythagoras' theorem

The square on the hypotenuse is equal to the sum of the squares on the other 2 sides, that is

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Delta Airlines' flights from Boston to Seattle are on time 70 % of the time. Suppose 6 flights are randomly selected, and the nu
emmainna [20.7K]

Answer:

P(at least 4)  = 0.74431

Step-by-step explanation:

Here

Probability of the flights on time is = 70% = 70/100= 0.7

q= 0.3

n= 6

As the probability of success and failure is constant and the n is also fixed then the binomail probability distribution can be used to solve this.

P (x= 4) = 6c4 *(0.7)^4 (0.3)^2= 15*0.2401* 0.09= 0.324135

P (x= 5) = 6c5 *(0.7)^5(0.3)^1= 6*0.16807* 0.3=0.302526

P (x= 6) = 6c6 *(0.7)^6 (0.3)^0= 0.117649P

P (x= 0) = 6c0 *(0.7)^0 (0.3)^6= 0.000729

P (x= 1) = 6c1 *(0.7)^1 (0.3)^5=0.010206

P (x= 2) = 6c 2*(0.7)^2 (0.3)^4=0.059535

P (x= 3) = 6c3 *(0.7)^3 (0.3)^3=0.18522

Probability of at least 4 means that probability of greater than 4 inclusive of 4.

This can be obtained in two ways.

First is by subtracting the probabilities of less than 4.

Second is by adding probabilities greater than 4 including 4.

We get the same answer with any of the two methods used.

P(at least 4) = 1- P(x<4)

                     = 1- [ 0.000729+ 0.010206 +0.059535+ 0.18522]

                     = 1- 0.25569

                       = 0.74431

P (at least 4) = P(x=4) +P(X=5) + P(x=6)

                        =0.324135+ 0.302526+ 0.117649

                        =0.74431

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