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3 years ago

I need help ASAP on number 5

Mathematics

2 answers:

EleoNora [17]3 years ago

7 0

Kim is the answer she drove further

Levart [38]3 years ago

3 0

Kim because she is traveling farther than jay in 12 minutes and he couldn’t beat her there.

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Stacy hits the jackpot one day at the gumball machine. She puts in a quarter and gets 444 gumballs rather than 111. The radius o

inn [45]

Answer: 3617.28 cubic millimeters.

Step-by-step explanation: I used the formula of a sphere since a gumball is a shape of a sphere and plugged the radius into the formula.

I got that the volume of a single gumball is 288π cubic millimeters.

Since there are 4 gumballs, i multiplied the volume by 4.

The total volume of the four gumballs add up to be 1152π .

Lastly, I multiplied 1152 by pi

and got the answer to be 3617.28 cubic millimeters

8 0

4 years ago

What is the perimeter of this figure

lesantik [10]

Answer:

11+11+3+3+3= 31

Step-by-step explanatio

7 0

3 years ago

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PLEASE help me! I’m giving all my points, please help ASAP!!!

Burka [1]

Answer:

I have send u picture

Step-by-step explanation:

I think this will help u

4 0

3 years ago

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A student saves $3 dollars on August 1, $5 on August 2, $7 on August 3, and so on. How much will she save in August 20 ?

Delvig [45]

Here is your answer

Money saved on each successive day is-

$3, &5, $7....

Clearly it forms an AP,

where

a1= 3

common difference, d= 2

n=20

So,

using formula

Tn= a1+(n-1)d

T20= 3+(20-1)2

= 3+ 19×2

= 3+ 38

=41

So, money saved on August 20= $41

Sum of money saved upto August 20 =
n/2 (a1+T20)
= 20/2 (3+41)
= 10× 44
= $440

HOPE IT IS USEFUL

5 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