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3 years ago

14

I need to know the answer ASAP please

1 answer:

cestrela7 [59]3 years ago

6 0

By observing the points you can learn a lot about a function. Concretely $f(x)$ passes through $(1,1)$ but $g(x)$ passes through $(1,-\frac{1}{2})$ that should give you a hint that $g(x)=-\frac{1}{2}x^2$ .

Hope this helps :)

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If g(x)=5x-3 and h(x) = x find (goh) (4)

Alexus [3.1K]

Goh means

G(h(4))
G(4)
5*4-3
17

7 0

3 years ago

Ava started a savings account with 500. After 6 months her savings account balance was 731. Find the rate of change

PIT_PIT [208]

Answer: it would be $38.50 per month

Step-by-step explanation:

231/6=38.50

5 0

3 years ago

(-2,3), (0,1), (2,-4), (3,-1), (2,4) IS IT AN FUNCTION OR NOT

MaRussiya [10]

Answer: No

Explanation: A function is a relation in which each x-coordinate corresponds to exactly one y-coordinate.

In the relation shown here, notice that

the x-coordinate 2 appears twice.

In one case, 2 corresponds to -4 and in

the other case, 2 corresponds to -4.

Since the x-coordinate 2 corresponds to more than one

y-coordinate, no, this relation is not a function.

6 0

2 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

3 years ago

7 1/2 divided by 1 2/3

Rina8888 [55]

Answer:

5

Step-by-step explanation:

6 0

3 years ago

Read 2 more answers