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Luda [366]
3 years ago
6

A charity is holding a raffle to raise money. There is one car worth $30,000 and five $100 gift cards being raffled off. Each ti

cket costs $20, and there are a total of 5,000 tickets being sold. Which equation correctly depicts the calculation of the expected value for a ticket?
Mathematics
2 answers:
lesya [120]3 years ago
8 0
30.000+100-5000*20=x(sum of profits)
KengaRu [80]3 years ago
5 0

Answer:

Step-by-step explanation:

Given that a charity is holding a raffle to raise money.

Car is the gift offered and the car is worth 30000

Tickets sold = 5000

Out of 500 tickets one ticket would fetch the car worth 30000 dollars

and five gift cards will bring 100 dollars

Thus we have                             Total

Tickets        1           5       4994      5000

Gain        30000      100        0

Cost               20       20       20

Net gain  29980       80      -20

Prob           1/5000   5/5000  4994/5000

Expected value for a ticket = 29980(1/5000)+80*5(5/5000)-20(4994/5000

=5.996+0.4-19.976

=-13.58

i.e. expected gain by participation=expectedvalue of ticket =-13.58

This negative indicates that the game is at the advantage of the charity conducting raffle and not for the participants

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Which is a perfect square? <br>A 20 <br>B 21 <br>C 24<br>D 25​
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3 years ago
Which of the following in an identity?
sammy [17]
By "which is an identity" they just mean "which trigonometric equation is true?"

What you have to do is take one of these and sort it out to an identity you know is true, or...

*FYI: You can always test identites like this:
Use the short angle of a 3-4-5 triangle, which would have these trig ratios:
sinx = 3/5    cscx = 5/3
cosx = 4/5   secx = 5/4
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Then just plug them in and see if it works. If it doesn't, it can't be an identity!

Let's start with c, just because it seems obvious.
The Pythagorean identity states that sin²x + cos²x = 1, so this same statement with a minus is obviously not true.

Next would be d. csc²x + cot²x = 1 is not true because of a similar Pythagorean identity 1 + cot²x = csc²x. (if you need help remembering these identites, do yourslef a favor and search up the Magic Hexagon.)

Next is b. Here we have (cscx + cotx)² = 1. Let's take the square root of each side...cscx + cotx = 1. Now you should be able to see why this can't work as a Pythagorean Identity. There's always that test we can do for verification...5/3 + 3/4 ≠ 1, nor is (5/3 +  3/4)².

By process of elimination, a must be true. You can test w/ our example ratios:
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5 0
3 years ago
Two part-time instructors are hired by the Department of Statistics and each is assigned at random to teach a single course in p
scZoUnD [109]

Answer:

The probability that they will teach different courses is \frac{2}{3}.

Step-by-step explanation:

Sample space is a set of all possible outcomes of an experiment.

In this case we will write the sample space in the form (x, y).

Here <em>x</em> represents the course taught by the first part-time instructor and <em>y</em> represents the course taught by the second part-time instructor.

Denote every course by their first letter.

The sample space is as follows:

S = {(P, P), (P, I), (P, S), (I, P), (I, I), (I, S), (S, P), (S, I) and (S, S)}

The outcomes where the the instructors will teach different courses are:

s = {(P, I), (P, S), (I, P),(I, S), (S, P) and (S, I)}

The probability of an events <em>E</em> is the ratio of the number of favorable outcomes to the total number of outcomes.

P(E)=\frac{n(E)}{N}

Compute the probability that they will teach different courses as follows:

P(\text{Different courses})=\frac{n(s)}{n(S)}=\frac{6}{9}=\frac{2}{3}

Thus, the probability that they will teach different courses is \frac{2}{3}.

3 0
3 years ago
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