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3 years ago

8

a club is holding a raffle. Ten thousand tickets have been sold for $2 each. There will be a 1st prize of $3000, 3 second prizes

of $1000 each, 5 third prizes of $500 each, and 20 consolation prizes of $100 each. Find the expected winnings of a single ticket

1 answer:

seropon [69]3 years ago

4 0

Answer:

1000(1/20000) +

300(2/20000) +

10(20/20000) +

=

1800/20000 = .09

Step-by-step explanation:

Each ticket purchased is expected to win 9 cents

Each ticket purchased cost 75 cents

If you interpret expected winnings per ticket to include the cost then each ticket is expected to lose 66 cents.

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Find the gcf of <br> 24, 14, 21

kobusy [5.1K]

Answer:

1

Step-by-step explanation:

All the numbers provided can have a factor of 1.

eg.)

1*24=24

1*14=14

1*21=21

This is because there is only one factor they all commonly share, 1.

4 0

3 years ago

Read 2 more answers

Question 10 (5 points)

11Alexandr11 [23.1K]

I think the answer is A

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3 years ago

Read 2 more answers

Ship collisions in the Houston Ship Channel are rare. Suppose the number of collisions are Poisson distributed, with a mean of 1

alexandr1967 [171]

Answer:

$a) \simeq 0.3012$ $b) \simeq 0.0494$ c) $\simeq 0.2438$

Step-by-step explanation:

Rate of collision,

1.2 collisions every 4 months

or, $\frac{1.2}{4}$

= 0.3 collisions per month

So, the Poisson distribution for the random variable no. of collisions per month (X) is given by,

P(X =x) = $\frac{e^{-\lambda}\times {\lambda}^{x}}}{x!}
$

for x ∈ N ∪ {0}

= 0 otherwise --------------------------------------(1)

here, $\lambda = 0.3$ collision / month

No collision over a 4 month period means no collision per month or X =0

Putting X = 0 in (1) we get,

P(X = 0) = $\frac{e^{-0.3}\times {\0.3}^{0}}{0!}
$

$\simeq 0.7408182207$ ------------------------------------(2)

Now, since we are calculating this for 4 months,

so, P(No collision in 4 month period)

= $0.7408182207^{4}$

$\simeq 0.3012$ -----------------------------------------------------------(3)

2 collision in 2 month period means 1 collision per month or X =1

Putting X =1 in (1) we get,

P(X =1) = $\frac{e^{-0.3}\times {\0.3}^{1}}{1!}
$

$\simeq 0.2222454662$ ------------------------------------(4)

Now, since we are calculating this for 2 months, so ,

P(2 collisions in 2 month period)

= $0.2222454662^{2}$

$\simeq 0.0494$ -----------------------------------------(5)

1 collision in 6 months period means

$\frac{1}{6}$ collision per month

Now, P(1 collision in 6 months period)

= P( X = 1/6] (which is to be estimated)

= $\frac {P(X=0)\times 5 + P(X =1)\times 1}{6}$

= \frac {0.7408182207 \times 5 + 0.2222454662 \times 1}{6}[/tex]

$\simeq 0.6543894283$ -------------------------------------------(6)

So,

P(1 collision in 6 month period)

= $0.6543894283^{6}$

$\simeq 0.0785267444$ ------------------------------------------------(7)

So,

P(No collision in 6 months period)

= $(P(X =0)^{6}$

$\simeq 0.1652988882$ ---------------------------------(8)

so,

P(1 or fewer collision in 6 months period)

= (8) + (7 ) = 0.0785267444 +0.1652988882

$\simeq 0.2438$ ---------------------------------------------(9)

7 0

3 years ago

2(m + 1) = 16 <br> full answer

Serhud [2]

Answer:

2m+2=16, then do the isolate m but doing the inverse operation of 2 which is now -2 and add that to 16 would be left with 2m=14 and 14/2 is 7 so m=7

Step-by-step explanation:

3 0

2 years ago

Which of the following is the correct factorization of the polynomial below? 8x3 + 64y3 A. (4x + 4y) (2x+8y) B. (2x + 4y)(4x2 -

stepladder [879]

Answer:

B. (2x+4y)(4x2-8xy+16y2)

Step-by-step explanation:

hope this helps!

6 0

3 years ago