# The Sum Of A Number K And 7/5 Is Represented By Point Q. Which Fraction Is One Possible Value Of K

Are all fractions rational numbers? Why?

Rational numbers are those which can be expressed in x/y format where x and y can only be whole numbers or integers. It won't be wrong to say all whole numbers are rational cause they can be expressed as number/1 example:- 3 = 3/1 or 5 = 5/1Where as irrational numbers are those, that can't be expressed as a fraction of whole numbers or integers, example square roots of non perfect squares, like √2 , √3, √5 etcTherefore if a fraction consists an irrational number then the result can't be a rational number. As we can't express square roots of non perfect squares as a fraction of whole numbers or integers.Example √2/7, 5/√7, 9/√5 all are irrational.Therefore we can't say every fraction is a rational number.But every fraction, that contains only whole numbers or integers or square roots of perfect squares will always be a rational numberExample √4/5 is rational as it can be written as 2/5For more information, read this Rational and Irrational Number

How do you find the number of integral solutions of the equation 2x + 3y = 763?

Given that 2x+3y = 763, various combinations of x and y will solve the equation. Here are some of the valuesy=1, x = (763–3)/2 = 760/2 = 380y=3, x = (763–9)/2 = 754/2 = 377y=5, x = (763–15)/2 = 748/2 = 374y=7, x = (763–21)/2 = 742/2 = 371y=9, x = (763–27)/2 = 736/2 = 368y=11, x = (763–33)/2 = 730/2 = 365y=13, x = (763–39)/2 = 724/2 = 362y=15, x = (763–45)/2 = 718/2 = 359y=17, x = (763–51)/2 = 712/2 = 356y=19, x = (763–57)/2 = 706/2 = 353y=21, x = (763–63)/2 = 700/2 = 350y=23, x = (763–69)/2 = 694/2 = 347y=25, x = (763–75)/2 = 688/2 = 344y=27, x = (763–81)/2 = 682/2 = 341y=29, x = (763–87)/2 = 676/2 = 338And so on …..

What is the expected value of winnings?

The sum of the products of the amount and its probability =

100,000 (1/8504860) + 50,000 (1/302500) + 10,000 (1/282735) + 1000 (1/153560) + 100 (1/104560) + 25(1/9540) = .2225.

This is the expected value of the winnings for the cost of $1.60. That means each time you do this, you have an expected loss of -1.60 + .2225 = -$1.377. If you do this 19 times, you're expected to lose 19(-1.377) = -$26.17

Please help me these 5 math questions?!?

Let the consecutive odd numbers be (2n-1) and (2n+1).

(2n-1)/5 + 4(2n+1)/7 = 59

7(2n-1) + 20(2n+1) = 2065

14n - 7 + 40n + 20 = 2065

54n = 2052

n = 38

.'.The consecutive odd numbers are: 75 and 77

Let the consecutive numbers are (n-1),n and (n+1)

.'.(n-1)/4 + n/5 = (n+1)/7 - 5

[5(n-1) + 4n]/20 = [(n+1) - 35]/7

(9n - 5)/20 = (n-34)/7

7(9n-5) = 20(n-34)

63n - 35 = 20n - 680

43n = -645

n = -15

.'.The required numbers are -16, -15 and -14.

The ratio of the numerator to the denominator of a certain fraction is one to four.

Let the number be (p/q). .'.p/q = 1/4, or, q = 4p.

If three is added to the numerator and subtracted from the denominator, the new fraction reduces to one-third.

(p+3)/(q-3) = 1/3

3(p+3) = (q-3)

3p + 9 = 4p - 3

p = 12

.'.q = 48

The original fraction is 12/48.

The ratio of the numerator to the denominator of a certain fraction is three to five.Let the number be p/q. .'.p/q = 3/5 or, p = (3/5)q

If two is added to the numerator and five is subtracted from the denominator, the new fraction reduces to four-fifths.

(p+2)/(q-5) = 4/5

5(p+2) = 4(q-5)

5[(3/5)q + 2] = 4(q-5)

3q + 10 = 4q - 20

q = 30

.'.p = (3/5)*30 = 18

The original fraction is 18/30

The numerator and the denominator of a certain fraction are consecutive odd numbers.

Let the consecutive odd numbers be (2k-1) and (2k+1). .'.The fraction is (2k-1)/(2k+1)

Now, by the problem,

[(2k-1)-9]/(2k+1) = 2/3

(2k-10)/(2k+1) = 2/3

3(2k-10) = 2(2k+1)

6k - 30 = 4k + 2

2k = 32

k = 16

.'.(2k-1) = 31 and (2k+1) = 33

.'.The required fraction is 31/33