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iragen [17]
3 years ago
9

Suppose you open a new game at the county fair. When patrons win, you pay them $3.00; when patrons lose, they pay you $1.00. If

the probability of a patron winning is p = .20, then how much can you expect to win (or lose) in the long run? Hint: You need to compute the expected value of the mean.A) win 0.20 cents per playB) win 0.60 cents per playC) lose 0.80 cents per playD) lose 2.20 dollars per play
Mathematics
1 answer:
tatyana61 [14]3 years ago
3 0

Answer:

E(X) = 1 *0.8 - 3*0.2 = 0.2

so at the long run we can conclude that the best option is :

A) win 0.20 cents per play

Step-by-step explanation:

Previous concepts

The expected value of a random variable X is the n-th moment about zero of a probability density function f(x) if X is continuous, or the weighted average for a discrete probability distribution, if X is discrete.

The variance of a random variable X represent the spread of the possible values of the variable. The variance of X is written as Var(X).  

Solution to the problem

Let X the random variable who represent the ampunt of money win/loss at the game defined.

The probability of loss $3.00 for this game is 0.2 and the probability of win is 1-0.2=0.8 and you will recieve $1.00 if you win. The expected value is given by:

E(X) = \sum_{i=1}^n X_i P(X_i)

And for this case if we replace we got:

E(X) = 1 *0.8 - 3*0.2 = 0.2

so at the long run we can conclude that the best option is :

A) win 0.20 cents per play

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6) Two coastguard stations P and Q are 17km apart, with due East of P. A ship S is observed in distress on a bearing 048° fromP
adelina 88 [10]

Answer:

The ship S is at 10.05 km to coastguard P, and 12.70 km to coastguard Q.

Step-by-step explanation:

Let the distance of the ship to coastguard P be represented by x, and its distance to coastguard Q be represented by y.

But,

<P = 048°

<Q = 360^{o} - 324^{o}

     = 036^{o}

Sum of angles in a triangle = 180^{o}

<P + <Q + <S = 180^{o}

048° + 036^{o} + <S = 180^{o}

84^{o} + <S = 180^{o}

<S  = 180^{o} -  84^{o}

    = 96^{o}

<S = 96^{o}

Applying the Sine rule,

\frac{y}{Sin P} = \frac{x}{Sin Q} = \frac{z}{Sin S}

\frac{y}{Sin P} = \frac{z}{Sin S}

\frac{y}{Sin 48^{o} } = \frac{17}{Sin 96^{o} }

\frac{y}{0.74314} = \frac{17}{0.99452}

⇒ y = \frac{12.63338}{0.99452}

       = 12.703

y = 12.70 km

\frac{x}{Sin Q} = \frac{z}{Sin S}

\frac{x}{Sin 36^{o} } = \frac{17}{Sin 96^{o} }

\frac{x}{0.58779} = \frac{17}{0.99452}

⇒ x = \frac{9.992430}{0.99452}

      = 10.0475

x = 10.05 km

Thus,

the ship S is at a distance of 10.05 km to coastguard P, and 12.70 km to coastguard Q.

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Answer:

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Step-by-step explanation:


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A salesperson is paid a flat rate plus a fixed percentage of her sales. Last month, she sold $16,000 worth of goods and was paid
Feliz [49]

Answer:

<em>She will be paid $1,350</em>

Step-by-step explanation:

<u>Linear Modeling</u>

Some events can be modeled as linear functions. If we are in a situation where a linear model is suitable, then we need two sample points to make the model and predict unknown behaviors.

The linear function can be expressed in the slope-intercept format:

y=mx+b, where m and b are constants.

The payments for a salesperson will be linearly modeled. There are two known points: When the sales were $16,000, the payment was $1,600. This makes the point (16,000;1,600).

We also know when the sales were $12,000, the payment was $1,400. The point is (12,000;1,400)

Let's use the points to find the values of m and b.

Using (16,000;1,600):

1,600=m*16,000+b

Using (12,000;1,400):

1,400=m*12,000+b

Subtracting both equations:

200=16,000m-12,000m

200=4,000m

Solving:

m=200/4,000=0.05

Using the first equation and the value of m:

1,600=0.05*16,000+b

1,600=800+b

Solving:

b=800

The equation is now complete:

y=0.05x+800

She sold $11,000 this month, so the payment is:

y=0.05\cdot 11,000+800

Y=550+800=1,350

She will be paid $1,350

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