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3 years ago

4,11,18 what is the 109th value of this sequence ?

Mathematics

1 answer:

kompoz [17]3 years ago

5 0

Answer:

109th value is 760

Step-by-step explanation:

the nth term rule is 7n-3

7×109=763

763-3=760

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30008876543*98765435678

Sladkaya [172]

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3 years ago

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1.) A man invested $600 at 8% per annum for 5 years. Calculate: a.) The simple interest payable b.) The total amount of money th

Bezzdna [24]

Answer:

Following are the solution to this question:

Step-by-step explanation:

Using formula:

$\text{Simple Interest} = \frac{P \times R \times T}{100} \\\\\text{ Total amount} = \text{principle}+ \text{Simple Interest}$

In question (1):

$Principle= \$ 600\\Rate = \ 8 \%\\Time= \ 5$

In point a:

$\to \frac{P \times R \times T}{100} \\\\\to \frac{(600 \times 8 \times 5)}{100}\\\\ \to 240$

In point b:

$\to amount = principle \ +\ simple \ interest$

$= 600 + 240 \\\\= 840$

In question (2):

$Principle= \$ 12,450\\Rate = \ 7.25 \%\\Time= \ 6$

In point a:

$\to \frac{P \times R \times T}{100} \\\\\to \frac{(12,450 \times 7.25 \times 6)}{100}\\\\ \to 5,415.75$

In point b:

$\to amount = principle \ +\ simple \ interest$

$= 12,450 + 5,415.75 \\\\= 17,865.75$

The 3 question is incomplete, that's why it can't be solved.

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3 years ago

Judy paid $10.48 for a number of

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Answer:

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Step-by-step explanation:

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2 years ago

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Alenkasestr [34]

Answer:

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Step-by-step explanation:

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3 years ago

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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:

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

7 0

3 years ago