Ask question

Ask question

All categories

2 years ago

7

Please help meeee!!!!!!!!

1 answer:

ivanzaharov [21]2 years ago

6 0

Answer:

$840$

Step-by-step explanation:

This is a good example of permutations. For the first slot on his shelf, Dylan has 7 trophies to choose from. Then 6, 5, and so on. Since he is only fitting 4 trophies on his shelf, we only continue until we have the slots filled. Thus, we have $7\cdot 6 \cdot 5\cdot 4=\boxed{840\:\text{ways}}$ to do so.

*Note: We do not divide by $4!$ because in this case, order matters. A different order of the same four books still produces different arrangements, so the order matters.

You might be interested in

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

Find the ratio of 4 meter to 50 centimetre

ladessa [460]

(4 meters) / (50 centimeters)

= (4 x 100 centimeters) / (50 centimeters)

= (400 centimeters) / (50 centimeters)

= (400) / (50)

= 8 .

4 0

3 years ago

Solve 5(x+9) - 4 = 71 for x

MrRissso [65]

Solving for x you would get X=6

7 0

3 years ago

Read 2 more answers

If 20 minutes from now the time will be 12 noon what time was it 30 minutes ago

EastWind [94]

11:10am

20 minutes before 12 noon is 11:40am
30 minutes before 11:40am is 11:10am

4 0

2 years ago

Maksim231197 [3]

I have no clue I don’t understand this

4 0

1 year ago