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3 years ago

suppose Richard walks one kilometer every 10 minutes how many meters are there can you walk in an hour at this new rate

1 answer:

3 0

One kilometer is 1000 meters. One hour is 60 minutes, or six segments of 10 minutes each. If you can walk 1000 meters in 10 minutes, you can walk 6000 in an hour (that is REALLY fast, by the way)

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Im dying bro help The vertices of a quadrilateral are (2, 3), (3, 8), (7, 5), and (6, 2). Draw the quadrilateral in the coordina

Solnce55 [7]

Okay to the first number in the ordered pair is the x and the second is the y. So (x,y) when you put a dot on the coordinate plane such as (2,3), you have to go the the horizontal line which is x and go over right 2 spaces(depending on how many numbers is one space) then go up 3. Do that same thing for each of the order pairs. Eventually you will have 4 dots. Connect those dots then you will have the quadrilateral. I hope this helps you!

7 0

2 years ago

The equation K = mv2 represents the energy an object has based on its motion. The kinetic energy, K, is based on the mass of the

docker41 [41]

<span>The answer is She should have multiplied by 2 instead of dividing by 2.
First one.</span>

6 0

2 years ago

Read 2 more answers

A dozen box of bagels costs $4.98. Because of inflation the price increases by 4% a year.

Olenka [21]

A=p (1+r)^t
A future value
P present value 4.98
R rate 0.04
T time

After 3 years
A=4.98×(1+0.04)^(3)
A=5.60

7.50=4.98 (1.04)^t
Solve for t
T=log(7.50÷4.98)÷log(1.04)
T=10 years

Hope it helps:-)

3 0

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:

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

7 0

2 years ago

Given: AB with point C between A and B. AB = 10x + 20, AC = 2x + 13, and CB = 10x - 9.

Feliz [49]

Answer:46

Step-by-step explanation:

7 0

2 years ago