A local church holds an annual raffle to raise money for a new roof. They sell only 500 tickets at $50 each. This year's prizes include: $3,000 in cash, four $100 Amazon gift cards, and two $75 Visa gift cards. You buy one ticket. What is your mathematical expectation for this game

The expectation for an event with outcomes:

{x₁, x₂, ..., xₙ}

Each one with probability:

{p₁, p₂, ..., pₙ}

Is:

Ev = x₁*p₁ + ... + xₙ*pₙ

There are 500 tickets sold.

1 of these, wins $3,000  (this is the event x₁)

4 of these, wins $100     (this is the event x₂)

2 of these, wins $75        (this is the event x₃)

The others do not have a prize.

So the probability of winning the $3000 is equal to the quotient between the number of tickets with that prize (1) and the total number of tickets (500)

p₁ = 1/500

Similarly, the probability of winning $100 will be:

p₂ = 4/500

And for the $75 prize:

p₃ = 2/500

Then the probability of not winning is:

p₄ = 493/500

Then the expected value for a single ticket is:

Ev = $0*493/500 + $75*2/500 + $100*4/500 + $3000*1/500

Ev = $7.1

If you take in account that you pay $50 for the ticket, the actual expectation should be:

E = $7.10 - $50 = -$42.90

i would have to say it is c.

m∠cdf = 43° , m∠gdf = 114°, m∠gde = 23°, m∠edh = 23° , m∠cdh = 157°

step-by-step explanation:

since, it is given that de bisects the angle gdh ⇒ de divides the angle gdh into two equal angles.

⇒ m∠gde = m∠edh

⇒ 8·x   - 1 = 6·x + 5

⇒ x = 3

so, m∠gde = 8·x - 1 = 23°

and m∠edh = 23°

now, it is given ∠cde is an straight angle so, m∠cde = 180°

⇒ m∠cdf + m∠fdg + m∠gde = 180°

⇒ 43° + m∠fdg + 23° = 180°

⇒ m∠fdg = 114°

again, m∠cde = m∠cdh + m∠edh = 180°

m∠cdh + 23° = 180°

⇒ m∠cdh = 157°