Find the area. 3 ft. 9 ft. 4ft. 5 ft. Cid +

10(4-3i)= whats the answer

LenaWriter [7]

Answer: Hi!

To solve this equation, we first distribute to the terms inside of the parentheses.

10*4 = 40

10*-3i = -30i

Our equation now looks like this:

40 - 30i

There is nothing left to simplify, so you're done!

Hope this helps!

5 0

3 years ago

The circles centered at points A, C, and D have radii of length AC.

natita [175]

it is the second one

Step-by-step explanation:

3 0

3 years ago

Suppose you own a home worth $185,325 and your property tax is 1.5%. How much are you liable to pay at the end of the year?

jarptica [38.1K]

Answer:

$ 2779.88

Step-by-step explanation:

185 325 x 1.5 % =

185 325 x .015 = 2779.88

8 0

2 years ago

Finding angles June 23,2022

nikitadnepr [17]

Answer:

61 degrees

Step-by-step explanation:

Considering angle <GJI and <IKD, these two angles are corresponding angles. Recall that corresponding angles are equal hence,

m<GJI = m<IKD = 119 degrees

Again

m<GJI and m<AJG are on a straight line at a point. Recall that the sum of angles on a straight line is 180 degrees. Hence

m<GJI + m<AJG = 180

119 + m<AJG = 180

m<AJG = 180 - 119

= 61 degrees

8 0

3 years ago

Let X have the uniform distribution U(0, 2) and let the conditional distribution of Y , given that X = x, be U(0, x). Find the j

Elina [12.6K]

Answer:

$f(x,y) = \frac{1}{x} \frac{1}{2}= \frac{1}{2x} , 0\leq x \leq 2 , 0\leq y \leq x$

$E(Y|x) = \int_{x=y}^2 y \frac{1}{x} dx= y ln x \Big|_{x=y}^2 =y ln 2 -y ln y = y(1-lny) \$

Step-by-step explanation:

We have two random variables X and Y. $X \sim Unif(0,2)$ and given that X=x, Y has uniform distribution (0,x)

From the definition of the uniform distribution we have the densities for each random variable given by:

$f_X (x) =\frac{1}{2} , 0\leq x\leq 2$

$f_{Y|X} (y|x) = \frac{1}{x}, 0\leq y \leq x$

And on this case we can find the joint density with the following formula:

$f(x,y) = f_{Y|X}(y|x) f_X (x)$

And multiplying the densities we got this:

$f(x,y) = \frac{1}{x} \frac{1}{2}= \frac{1}{2x} , 0\leq x \leq 2 , 0\leq y \leq x$

Now with the joint density we can find the expected value E(Y|x) with the following formula:

$E(Y|x) = \int y f_{Y|X}(y|x)dx$

And replacing we got:

$E(Y|x) = \int_{x=y}^2 y \frac{1}{x} dx= y ln x \Big|_{x=y}^2 =y ln 2 -y ln y = y(1-lny) \$

5 0

4 years ago