In FGH the measure of

A plane is a figure. The intersection of two planes that do not coincide (if it exists) is a

ser-zykov [4K]

Solution:

A plane is a two Dimensional figure.

The Intersection of two planes that do not coincide is always a line.

Consider two planes ,A and B, Following are the possibilities

1. They may be parallel to each other.→→→ No curve obtained

2. The planes A and B can Coincide with each other.→→→Intersection is a plane.

3. The planes A and B are intersecting.→→→→Intersection is a line.

3 0

4 years ago

Read 2 more answers

Help please <3 with number 10

mariarad [96]

30+45= 75= 1 n 15 mins.6:00-1 15= 4:45pm

5 0

3 years ago

What is the linear function? Find the output when the input is n. please help.

melomori [17]

Answer:

y = -x + 6

Example:

-1 +6=5

7 0

3 years ago

I AM IN DESPERATE NED OF QUICK HELP PLEASEEEEEEEEEEEEEEEEEEEEEE!!!!!!!!!!!!!!!!!!!!!!

rosijanka [135]

Answer:

16

Step-by-step explanation:

To solve this equation, we need the formula for perfect squares:

$(a + b)^2 = a^2 + 2ab + b^2$

or

$(a - b)^2 = a^2 - 2ab + b^2$ <em>this is the formula we will use because the signs match the one in the question</em>.

Knowing these, we can set up an equation that assumes that the answer we will end up with is a perfect square.

Work:

$(v - x)^2 = v^2 -8v + x^2$

this is the setup for being able to solve for the unknown $x$ , now we need to solve for $x$ .

$(v - x)^2 = v^2 - 8v + x^2$ (I replaced $2ab$ with $8v$ because I'm setting up this question to be a perfect square).

$(v - x) (v - x) = v^2 - 8v + x^2$

multiply the left side. Remember that your answer will not be $v^2 + x^2$ , but $v^2 - 2vx +x^2$ .

$v^2 - 2vx + x^2 = v^2 + 8v + x^2$

Divide like terms and isolate the variable. In this case, $v^2$ and $x^2$ are on both sides, so dive them and they will be canceled out.

$2vx = 8v$

divide by $2v$ and you will have your value for $x$ .

$x = 4$

Now we plug $x$ into our original equation equation for perfect squares.

$(v - 4)^2 = v^2 - 8v + 16$

Our final answer is 16.

5 0

4 years ago

Suppose we play the following game based on tosses of a fair coin. You pay me $10, and I agree to pay you $n 2 if heads comes up

Artyom0805 [142]

Answer:

In the long run, ou expect to lose $4 per game

Step-by-step explanation:

Suppose we play the following game based on tosses of a fair coin. You pay me $10, and I agree to pay you $n^2 if heads comes up first on the nth toss.

Assuming X be the toss on which the first head appears.

then the geometric distribution of X is:

X $\sim$ geom(p = 1/2)

the probability function P can be computed as:

$P (X = n) = p(1-p)^{n-1}$

where

n = 1,2,3 ...

If I agree to pay you $n^2 if heads comes up first on the nth toss.

this implies that , you need to be paid $\sum \limits ^{n}_{i=1} n^2 P(X=n)$

$\sum \limits ^{n}_{i=1} n^2 P(X=n) = E(X^2)$

$\sum \limits ^{n}_{i=1} n^2 P(X=n) =Var (X) + [E(X)]^2$

$\sum \limits ^{n}_{i=1} n^2 P(X=n) = \dfrac{1-p}{p^2}+(\dfrac{1}{p})^2$ ∵ X $\sim$ geom(p = 1/2)

$\sum \limits ^{n}_{i=1} n^2 P(X=n) = \dfrac{1-p}{p^2}+\dfrac{1}{p^2}$

$\sum \limits ^{n}_{i=1} n^2 P(X=n) = \dfrac{1-p+1}{p^2}$

$\sum \limits ^{n}_{i=1} n^2 P(X=n) = \dfrac{2-p}{p^2}$

$\sum \limits ^{n}_{i=1} n^2 P(X=n) = \dfrac{2-\dfrac{1}{2}}{(\dfrac{1}{2})^2}$

$\sum \limits ^{n}_{i=1} n^2 P(X=n) =\dfrac{ \dfrac{4-1}{2} }{{\dfrac{1}{4}}}$

$\sum \limits ^{n}_{i=1} n^2 P(X=n) =\dfrac{ \dfrac{3}{2} }{{\dfrac{1}{4}}}$

$\sum \limits ^{n}_{i=1} n^2 P(X=n) =\dfrac{ 1.5}{{0.25}}$

$\sum \limits ^{n}_{i=1} n^2 P(X=n) =6$

Given that during the game play, You pay me $10 , the calculated expected loss = $10 - $6

= $4

∴

In the long run, you expect to lose $4 per game

3 0

4 years ago