ANTHONY SURVEYED 320 STUDENTS AND FOUND OUT THAT 120 OF THEM HAVE A PET BASED ON THIS INFORMATION HOW MANY

Jordan has 55 she earns 67 by doing chores how many money does Jordan have now

elixir [45]

Answer:

$122

Step-by-step explanation:

amount she had + amount she earns = amount she has now

55 + 67 =

122

3 0

3 years ago

Read 2 more answers

Which of the above numbers is the value of expression equal to?

babymother [125]

Answer:

D.

$4.33 \times {10}^{6}$

4 0

3 years ago

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You are choosing between two plans at a health club. Plan A offers an initial membership fee of $75. And you pay $15 per month.

Dennis_Churaev [7]

Answer:

5 months

Step-by-step explanation:

Let the number of months equalising both plans fee = x

Plan A total fee = 75 + 15x , Plan B total fee = 20 + 26x

Equalise both: 75+15x = 20+26x ; 75-20 = 26x-15x ; 55 = 11x ; x = 55 / 11

x = 5

Proof : 75 + 15 (5) = 20 + 26 (5) ; LHS 150 = RHS 150

6 0

3 years ago

The area of the photo will be half the area of the entire ad. What is the value of x

Ksenya-84 [330]

Answer:

Need more information

Step-by-step explanation:

8 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