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Ronch [10]
3 years ago
5

Prove that

Mathematics
1 answer:
pentagon [3]3 years ago
6 0

LStep-by-step explanation:

(a) To proof this statement choose m = 4 and n = -1, from here we have 2(4) + 7(-1) = 1 as required. Hence th statement is true.

(b) To proof this statement choose m = 1 and n = -1, from here we have 15(1) + 12(-1) = 3 as required. Hence the statement is true.

(c) To proof this statement, suppose m and n are integers, since integers are closed under addition and multiplication then 2m, 4n and 2m + 4n are all integers. Obviously 2m and 4n are even integers, therefore their sum is also an even integer. This contradicts the equation 2m + 4m = 7, hence there do not exist integers m and n that satisfy the statement.

(d) To proof this statement, suppose m and n are integers, then 12m, 15n and 12m + 15n are all integers, since integers are closed under addition and multiplication. Factoring out 3 from the sum, we have: 3(4m + 5n). Therefore 3(4m + 5n) = 1 is a contradiction, because it will give 4m + 5n = 1/3 and addition of two integers can never give a fraction.

(e) To proof this statement, suppose m and n are integers, then 15m + 16n = t can be rewritten as 3(5m) + 8(2n) = t. Since integers are closed under multiplication, then 5m and 2n are integers. Let 5m = r and 2n = s, then we have: 3r + 8s = t as required.

(f) To proof this statement, suppose that m and n are both negative integers, we see that 12m + 15n = 1 is clearly impossible because the sum would be less than 0. Therefore, if there exist integers m and n such that 12m + 15n = 1, then they must be positive.

(g) To proof this statement, suppose m is an integer and has the form 4k + 1 for some integer k, then m + 2 = 4k + 1 + 2 = 4k + 3 = 4k + 4 - 1 = 4(k + 1) - 1. Since k is an integer then k + 1 is an integer because integers are closed under addition. Therefore, put j = k + 1. Hence we have 4j - 1.

(h) suppose that m is odd, then there is an n such that m = 2n + 1, therefore m² = (2n + 1)² = 4n² + 4n + 1 = 4(n² + n) + 1. Here, we need a lemma that says: suppose k is an integer, k(k + 1) is an even integer. This can be proven intuitively, when k is odd, the term in the bracket becomes even, and the whole expression is even. If k is even, then automatically the whole expression becomes even. Hence, our lemma is proven intuitively. This implies that n² + n is an even integer and can be replaced by 2k for some k that is an integer. Therefore, we have 4(2k) + 1 = 8k + 1 as required.

(i) suppose that m and n are odd integers such that mn = 4k - 1 for some k that is an integer. If we assume that neither m nor is in the for 4j - 1, then:

mn = (4a + 1)(4b + 1) for some a and b that are integers. Expanding that we have mn = 4(4ab + a + b) + 1 which is not in the form 4k - 1 for some k that is an integer as assumed. Therefore, if m and n are odd integers such that mn is in the form 4k - 1 for some k that is an integer, then m and n are in the form 4j - 1 for some j that is an integer.

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The statement is False,

The function and its inverse are reflected about the line y=x. This means that the x and y values are replaced.

So, if a function has a point (2,3) the same point on its inverse will moved be (3,2). This happens for all the points.

Therefore, the statement is false. 
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An urn contains three white balls and two red balls. The balls are drawn from the urn, oneat a time without replacement, until a
Tpy6a [65]

Answer:

p ( X = 1 ) = 0.6 , p ( X = 2 ) = 0.3 , p ( X = 3 ) = 0.1

Verified

E ( X ) = 1.5

Step-by-step explanation:

Solution:-

- An urn contains the following colored balls:

                       Color                  Number of balls

                       White                            3

                       Red                               2

- A ball is drawn from urn without replacement until a white ball is drawn for the first time.

- We will construct cases to determine the distribution of the random-variable X: The number of trials it takes to get the first white ball.

- We have three following case:

1) White ball is drawn on the first attempt ( X =  1 ). The probability of drawing a white ball in the first trial would be:

              p ( X = 1 ) = ( Number of white balls ) / ( Total number of ball )

              p ( X = 1 ) = ( 3 ) / ( 5 )

2) A red ball is drawn on the first draw and a white ball is drawn on the second trial ( X = 2 ). The probability of drawing a red ball first would be:

      p ( Red on first trial ) = ( Number of red balls ) / ( Total number of balls )

      p ( Red on first trial ) = ( 2 ) / ( 5 )

- Then draw a white ball from a total of 4 balls left in the urn ( remember without replacement ).

   p ( White on second trial ) = ( Number of white balls ) / ( number of balls left )

   p ( White on second trial ) = ( 3 ) / ( 4 )

- Then to draw red on first trial and white ball on second trial we can express:

                p ( X = 2 ) =  p ( Red on first trial ) *  p ( White on second trial )

                p ( X = 2 ) =  ( 2 / 5 ) * ( 3 / 4 )

                p ( X = 2 ) =  ( 3 / 10 )  

3) A red ball is drawn on the first draw and second draw and then a white ball is drawn on the third trial ( X = 3 ). The probability of drawing a red ball first would be ( 2 / 5 ). Then we are left with 4 balls in the urn, we again draw a red ball:

   p ( Red on second trial ) = ( Number of red balls ) / ( number of balls left )

   p ( Red on second trial ) = ( 1 ) / ( 4 )    

 

- Then draw a white ball from a total of 3 balls left in the urn ( remember without replacement ).                  

   p ( White on 3rd trial ) = ( Number of white balls ) / ( number of balls left )

   p ( White on 3rd trial ) = ( 3 ) / ( 3 ) = 1

- Then to draw red on first two trials and white ball on third trial we can express:

                p ( X = 3 ) =  p ( Red on 1st trial )*p ( Red on 2nd trial )*p ( White on 3rd trial )

                p ( X = 3 ) =  ( 2 / 5 ) * ( 1 / 4 ) * 1

                p ( X = 3 ) =  ( 1 / 10 )  

- The probability distribution of X is as follows:

    X          1                  2                      3

p ( X )      0.6               0.3                  0.1

- To verify the above the distribution. We will sum all the probabilities for all outcomes ( X = 1 , 2 , 3 ) must be equal to 1.

          ∑ p ( Xi ) = 0.6 + 0.3 + 0.1

                         = 1 ( proven it is indeed a pmf )

- The expected value E ( X ) of the distribution i.e the expected number of trials until we draw a white ball for the first time:

               E ( X ) = ∑ [ p ( Xi ) * Xi  ]

               E ( X ) = ( 1 ) * ( 0.6 ) + ( 2 ) * ( 0.3 ) + ( 3 ) * ( 0.1 )

               E ( X ) = 0.6 + 0.6 + 0.3

               E ( X ) = 1.5 trials until first white ball is drawn.

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