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Yakvenalex [24]
3 years ago
8

Michael borrowed $3,000 at 7% simple interest for two years. How much interest is owed at the end of the two years on this loan?

​
Mathematics
1 answer:
babunello [35]3 years ago
3 0

Answer:

h

Step-by-step explanation:

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29/5 divided by nine and a half
trapecia [35]

29/5 is 5.8   5.8 dividen by 9 1/2 is 0.61052631578

Hope this helps! Have a great day!!!

3 0
3 years ago
In relation defined by the equation y=2x+1 for all x>0, y is a function because of x because
nekit [7.7K]

that would be the letter d :)

each value of x has a unique value of y

4 0
3 years ago
*BRAIN TEASER*
Mademuasel [1]

Answer:

idk but go dtep by step

Step-by-step explanation:

6 0
3 years ago
Read 2 more answers
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
What are the solution and the answer?
dedylja [7]

Answer:

I think its 40 if i did it right Ill show you how i did it

Step-by-step explanation:

First I added 12+12 and the is 24.

The I added 8+8 and that was 16.

Lastly I added 16+24= 40

7 0
2 years ago
Read 2 more answers
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