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Cerrena [4.2K]
3 years ago
15

In FGH the measure of

Mathematics
1 answer:
Yakvenalex [24]3 years ago
6 0
Is there supposed to be a picture that goes with this ?
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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
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