*URGENT 20 POINTS*

Anyone help me please

Whitepunk [10]

Assuming you are solving for x...

5 0

3 years ago

Jeff gave 1/4 of a sum of money to his wife. Then he divided the remainder equally among his 4 children.

Semenov [28]

Answer:

A) The fraction of sum of money did each child receive is $\frac{3 x}{16}$

B) The sum of money did Jeff have $ 3200

Step-by-step explanation:

Given as :

Let The sum of money did Jeff have = $ x

The fraction of money did Jeff's wife get = $\frac{1}{4}$ of $ x

The remaining money Jeff will have = $ x - $\frac{1}{4}$ of $ x

I.e The remaining money Jeff will have = $\frac{4 x - x}{4}$

= $\frac{3 x}{4}$

A ) The remaining amount of money is divided equally among 4 children

So, The fraction of sum of money did each child receive = $\frac{\frac{3 x}{4}}{4}$

I.e The fraction of sum of money did each child receive = $\frac{3 x}{16}$

B ) If each child will receive $ 600

∴, $\frac{3 x}{16}$ = $ 600

Or, 3 x = $ 600 × 16

Or, 3 x = $ 9600

∴ x = $\frac{9600}{3}$

I.e x = $ 3200

So, The sum of money did Jeff have $ 3200

Hence ,

A) The fraction of sum of money did each child receive is $\frac{3 x}{16}$

B) The sum of money did Jeff have $ 3200 Answer

8 0

3 years ago

3. Find two possible lengths for CD if C, D, and E

Kaylis [27]

Given :

C, D, and E are col-linear, CE = 15.8 centimetres, and DE= 3.5 centimetres.

To Find :

Two possible lengths for CD.

Solution :

Their are two cases :

1)

When D is in between C and E .

. . .

C D E

Here, CD = CE - DE

CD = 15.8 - 3.5 cm

CD = 12.3 cm

2)

When E is in between D and C.

. . .

D E C

Here, CD = CE + DE

CD = 15.8 + 3.5 cm

CD = 19.3 cm

Hence, this is the required solution.

3 0

3 years ago

Please 100 points and brainliest if u do them all plsssssss I rlly need help asap

AysviL [449]

Answer:

<u>Question 1</u>

F

T

The volume of figure A can be found by multiplying (5 · 7)(2).

<u>Question 2</u>

T

F

The total volume of the figure is 442 mm³

<u>Question 3</u>

T

F

T

The volume of the triangular prism is 108 in³

Step-by-step explanation:

<u>Formula used</u>

Volume of a prism = base area × height

Area of a rectangle = width × length

Area of a triangle = 1/2 × base × height

Volume of a cube = s³ (where s is the side length)

---------------------------------------------------------------------------------------------------

<h3><u>Question 1</u></h3>

⇒ Volume of Figure A = (5 · 7)(2)

= 70 cm³

⇒ Volume of Figure B = (13 · 7)(2)

= 182 cm³

⇒ Total Volume = Volume of Figure A + Volume of Figure B

= 70 + 182

= 252 cm³

The first statement is false.

<u>Rewritten statement</u>:

The volume of figure A can be found by multiplying (5 · 7)(2).

---------------------------------------------------------------------------------------------------

<h3><u>Question 2</u></h3>

⇒ Volume of rectangular prism = (15 · 15)(2)

= 450 mm³

⇒ Volume of central cube = 2³

= (2 · 2 · 2)

= 8 mm³

⇒ Total Volume = Volume of rectangular prism - Volume of cube

= 450 - 8

= 442 mm³

The third statement is false.

<u>Rewritten statement</u>:

The total volume of the figure is 442 mm³

---------------------------------------------------------------------------------------------------

<h3><u>Question 3</u></h3>

⇒ Volume of rectangular prism = (12 · 3)(8)

= 288 in³

⇒ Volume of triangular prism = (1/2 · 12 · 6)(3)

= 108 in³

⇒ Total Volume = Volume of rectangular prism + Volume of triangular prism

= 288 + 108

= 396 in³

The second statement is false.

<u>Rewritten statement</u>:

The volume of the triangular prism is 108 in³

7 0

2 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