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2 years ago

The sum of 2 consecutive whole numbers is 206. What are the 2 numbers

Mathematics

2 answers:

vlabodo [156]2 years ago

8 0

Answer:

There are no two consecutive whole numbers that sum to 206 (See proof below)

Step-by-step explanation:

let the first number be x

since the numbers are whole and consecutive, the second number must be (x + 1)

we are given that the number sum to 206

hence,

x + (x+1) = 206

x + x + 1 = 206

2x + 1 = 206 (subtract 1 from both sides)

2x = 206 - 1

2x = 205 (divide both sides by 2)

x = 205/2 = 102.5 (NOT A WHOLE NUMBER)

if x = 102.5, then the second number must be

x + 1 = 102.5 + 1 + 103.5 (ALSO NOT A WHOLE NUMBER)

Anni [7]2 years ago

5 0

Answer:

102.5, 103.5

Step-by-step explanation:

a+a+1=206

2a+1=206

2a=205

a=102.5

103.5

I think there is a bug here. They can't be whole numbers.

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Masja [62]

Answer:

first answer is correct

Step-by-step explanation:

Hello, we want to write 2 + 4 + 6 + ... + 20

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for n = 3, 2n=2*3=6

etc

$\displaystyle \sum_{n=1}^{10} {2n} = 2*1 + 2*2 +2*3+2*4+...+2*10=2+4+6+8+...+20$

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2 years ago

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Veseljchak [2.6K]

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3 years ago

Let X and Y be discrete random variables. Let E[X] and var[X] be the expected value and variance, respectively, of a random vari

Ulleksa [173]

Answer:

(a) $E[X+Y]=E[X]+E[Y]$

(b) $Var(X+Y)=Var(X)+Var(Y)$

Step-by-step explanation:

Let X and Y be discrete random variables and E(X) and Var(X) are the Expected Values and Variance of X respectively.

(a)We want to show that E[X + Y ] = E[X] + E[Y ].

When we have two random variables instead of one, we consider their joint distribution function.

For a function f(X,Y) of discrete variables X and Y, we can define

$E[f(X,Y)]=\sum_{x,y}f(x,y)\cdot P(X=x, Y=y).$

Since f(X,Y)=X+Y

$E[X+Y]=\sum_{x,y}(x+y)P(X=x,Y=y)\\=\sum_{x,y}xP(X=x,Y=y)+\sum_{x,y}yP(X=x,Y=y).$

Let us look at the first of these sums.

$\sum_{x,y}xP(X=x,Y=y)\\=\sum_{x}x\sum_{y}P(X=x,Y=y)\\\text{Taking Marginal distribution of x}\\=\sum_{x}xP(X=x)=E[X].$

Similarly,

$\sum_{x,y}yP(X=x,Y=y)\\=\sum_{y}y\sum_{x}P(X=x,Y=y)\\\text{Taking Marginal distribution of y}\\=\sum_{y}yP(Y=y)=E[Y].$

Combining these two gives the formula:

$\sum_{x,y}xP(X=x,Y=y)+\sum_{x,y}yP(X=x,Y=y) =E(X)+E(Y)$

Therefore:

$E[X+Y]=E[X]+E[Y] \text{ as required.}$

(b)We want to show that if X and Y are independent random variables, then:

$Var(X+Y)=Var(X)+Var(Y)$

By definition of Variance, we have that:

$Var(X+Y)=E(X+Y-E[X+Y]^2)$

$=E[(X-\mu_X +Y- \mu_Y)^2]\\=E[(X-\mu_X)^2 +(Y- \mu_Y)^2+2(X-\mu_X)(Y- \mu_Y)]\\$Since we have shown that expectation is linear$\\=E(X-\mu_X)^2 +E(Y- \mu_Y)^2+2E(X-\mu_X)(Y- \mu_Y)]\\=E[(X-E(X)]^2 +E[Y- E(Y)]^2+2Cov (X,Y)$

Since X and Y are independent, Cov(X,Y)=0

$=Var(X)+Var(Y)$

Therefore as required:

$Var(X+Y)=Var(X)+Var(Y)$

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Veseljchak [2.6K]

Answer:

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Step-by-step explanation:

ii)

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iii)

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