All categories

3 years ago

Carlos swam to the bottom of a pool that is 12 feet deep. What is the opposite of Carlos elevation relative to the surface.

Mathematics

2 answers:

Kazeer [188]3 years ago

6 0

Answer:

C. 12 feet

Step-by-step explanation:

We have been given that Carlos swam to the bottom of a pool that is 12 feet deep.

Since Carlos swam to the bottom of the pool, so his elevation will be $-12$ feet with respect to surface.

The opposite of Carlos elevation relative to the surface would be opposite of $-12$ feet that is $-(-12)$ feet.

$+12$ feet

Therefore, the opposite of Carlos elevation relative to the surface is 12 feet.

MariettaO [177]3 years ago

5 0

The Anwser Is -12 Because He Swam 12 Feet Deep And It Would Not Be The Other Answers Because He Did Not Swim More Down Into The Pool So It Would Be -12

You might be interested in

What's the answer <br>2(10×10)2

Nookie1986 [14]

2(100)2
200x2
400
Is the answer

7 0

3 years ago

Read 2 more answers

PLZ HELP THIS IS DUE TODAY AND NO LINKS PLZ.

anastassius [24]

Answer:

D is the best answer to the question

8 0

3 years ago

Kendall works as a salesperson and earns a base salary of $82 per week plus a commission of 20% of all her sales. If Kendall had

rjkz [21]

Commission = 20% x 85 = $17

Salary = $82 + $17 = $99 <=== this much she made.

8 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

3 years ago

3(n-t) what is it?

g100num [7]

3(n-t) = 3n- 3t
It is you mean about ?

3 0

3 years ago