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andreev551 [17]
3 years ago
15

The world famous gambler from Philadelphia, Senor Rick, proposes the following game of chance. You roll a fair die. If you roll

a 1, then Senor Rick pays you $25. If you roll a 2, Senor Rick pays you $5. If you roll a 3, you win nothing. If you roll a 4 or 5, you must pay Senor Rick $10, and if you roll a 6, you must pay Senor Rick $15. Is Senor Rick loco for proposing such a game? Explain.
Mathematics
1 answer:
Orlov [11]3 years ago
7 0

Answer:

You lose $0.833 while Rick gains $0.833, and so Rick is not Loco

Step-by-step explanation:

There are six numbers on a dice. Thus, probability of rolling exactly one number = 1/6

Probability of rolling two is 2/6.

Rolling a 1 pays $25, and probability = 1/6

Rolling a 2 pays $5,and probability = 1/6

Rolling a 3 pays $0, and probability = 1/6

Rolling a 4 or 5 pays -$10, and probability = 2/6

Rolling a 6 pays -$15, and probability = 1/6

Thus,lets calculate the expected value;

EV = (25 × 1/6) + (5 × 1/6) + (0 × 1/6) + (-10 × 2/6) + (-15 × 1/6)

EV = - $0.833

This means you lose $0.833 while Rick gains $0.833, and so Rick is not Loco.

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The graph shows the cost of some taxi journeys<br><br> work out a formula for c in terms of n
Vitek1552 [10]

Answer:

c=\frac{3}{5}(n)+2.5

Step-by-step explanation:

see the attached figure to better understand the problem

Let

c -----> the cost of some taxi journeys

n ----> the number of miles

we know that

The graph is a line

The equation of a line in slope intercept form is

c=m(n)+b

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Observing the graph  we have the points

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The formula to calculate the slope between two points is equal to

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substitute the values

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m=\frac{3}{5}

we have

b=2.5 -----> because the point (0,2.5) is the c-intercept

substitute in the equation of slope-intercept form

c=\frac{3}{5}(n)+2.5

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Alex wants to fence in an area for a dog park. He has plotted three sides of the fenced area at the points E (1, 5), F (3, 5), a
hram777 [196]

Answer:

(1,1)

Step-by-step explanation:

Given: E, F, G, H denote the three coordinates of the area fenced

To find: coordinates of point H

Solution:

According to distance formula,

length of side joining points (x_1,y_1)\,,\,(x_2,y_2) is equal to \sqrt{(x_2-x_1)^2+(y_2-y_1)^2}

So,

EF=\sqrt{(3-1)^2+(5-5)^2}=2\,\,units\\FG=\sqrt{(6-3)^2+(1-5)^2}=\sqrt{9+16}=5\,\,units\\GH=\sqrt{(x-6)^2+(y-1)^2}\\EH=\sqrt{(x-1)^2+(y-5)^2}

Perimeter of a figure is the length of its outline.

EF+FG+GH+EH=16\\2+5+\sqrt{(x-6)^2+(y-1)^2}+\sqrt{(x-1)^2+(y-5)^2}=16\\\sqrt{(x-6)^2+(y-1)^2}+\sqrt{(x-1)^2+(y-5)^2}=16-2-5\\\sqrt{(x-6)^2+(y-1)^2}+\sqrt{(x-1)^2+(y-5)^2}=9

Put (x,y)=(1,1)

\sqrt{(1-6)^2+(1-1)^2}+\sqrt{(1-1)^2+(1-5)^2}=9\\\sqrt{25}+\sqrt{16}=9\\5+4=9\\9=9

This is true.

So, the point (1,1) satisfies the equation \sqrt{(x-6)^2+(y-1)^2}+\sqrt{(x-1)^2+(y-5)^2}=9

So, point H is (1,1).

7 0
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