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3 years ago

Which shapes make up this composite figure?

Mathematics

1 answer:

Alona [7]3 years ago

3 0

Answer: a sphere and a cylinder

Step-by-step explanation:

The composite figure has two semi-circles on each end (which is basically a whole sphere). In the middle is a cylinder. So, d is the correct option.

A cone has a triangular form, so it can't be the other answers.

You might be interested in

Find the value of each variable. Write an<br> equation then solve showing ALL the work<br> A2x+10°

AveGali [126]

Answer:

x = 25

Step-by-step explanation:

The three straight lines in the sides of the triangle means that ALL 3 SIDES ARE EQUAL.

This is an equilateral triangle. So all 3 angles are equal.

We know sum of 3 angles in a triangle is 180. Since 3 of the angles are equal, each angle is:

180/3 = 60

The top angle is given as "2x + 10", thus we can say "2x + 10" is equal to 60 degrees. Now we equate and solve for x:

$2x+10=60\\2x=60-10\\2x=50\\x=\frac{50}{2}\\x=25$

The value of x is 25

4 0

3 years ago

PLEASE HELP I’M SO CONFUSED

bekas [8.4K]

Answer:

a. $45 an hour

b. 5.25 hours

Step-by-step explanation:

a. The plumber charges $75 per job, so it is irrelevant to the number of hours they work. 45(h) is $45 multiplied by the number of hours they work.

First, plug the numbers into the equation, the cost being C(h):

311.25 = 45(h) + 75

Next, subtract $75 as it is a set price that is always payed:

236.25 = 45(h)

Then, divide both sides by 45 to find how many hours were worked:

h = 5.25

Hours = $5.25

5 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

Could someone pls help me and explain this if you know it!!

yanalaym [24]

Answer:

25% on question one

Step-by-step explanation:

For example, on question one, 5 students drink more than half a dozen sodas a week. If there is a total of 20 students you'd divide 5 by 20 and that should be the answer.

8 0

3 years ago

If 3t-7=5t, then 6t#

Inessa05 [86]

Here, 3t - 7 = 5t
5t - 3t = -7
2t = -7
t = -7/2

Then, 6t = 6(-7/2) = 3(-7) = -21

In short, Your Answer would be -21

Hope this helps!

3 0

3 years ago