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Anna11 [10]
3 years ago
8

What is the answer to all

Mathematics
2 answers:
Burka [1]3 years ago
6 0
11.B. 21+g
12.B.1 because 5/1/5=1
13. 1.35 because 4-2.65=1.35
14. A. x>-1

Bingel [31]3 years ago
4 0
11. 21g
12. ?
13. X=1.35
14. X>-1

(I'm not sure about 12 because I don't have any paper to solve it at the moment. Sorry! Hope this helps anyway!)
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Plz answer this now plz
iragen [17]

Answer:

because M is the centroid of the triangle

=> PM = 2AM

<=> AM = PM/2 = 10 /2 = 5

=> AM = 5

7 0
3 years ago
Please help, will give brainliest, 5 star, and thanks!
Harlamova29_29 [7]
O All real numbers is the range of the function
5 0
3 years ago
Read 2 more answers
Find q. Write your answer in simplest radical form.
yawa3891 [41]

Answer:

q=90 units.

Step-by-step explanation:

5 0
3 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
What is- If x = 5 find the value of (x + 3)2
melamori03 [73]
X=5 we’ve settled that. now all you have to do is pemdas.
P: (Parentheses)
E: Exponents
M: Multiplication
D: Division
A: Addition
S: Subtraction

1.) replace x as 5 so it looks like this:
(5+3)2

2.) solve inside parentheses.
(8)2

3.) 8x2=16

ANSWER: (x+3)2 = 16

and you can check that!

Have a lovely day and mark me brainliest please.
7 0
3 years ago
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