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inysia [295]
3 years ago
13

What is the expected average of rolling a single die

Mathematics
2 answers:
Gekata [30.6K]3 years ago
8 0

Answer:

When you roll a fair die you have an equal chance of getting each of the six numbers 1 to 6. The expected value of your die roll, however, is 3.5.

Step-by-step explanation:

larisa86 [58]3 years ago
5 0

Answer:

Mathwords: Expected Value. A quantity equal to the average result of an experiment after a large number of trials. For example, if a fair 6-sided die is rolled, the expected value of the number rolled is 3.5. This is a correct interpretation even though it is impossible to roll a 3.5 on a 6-sided die.

Step-by-step explanation:

hope it helps

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An amount of $21,000 is borrowed for 10 years at 3.25% interest, compounded annually. If the loan is paid in full at the end of
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The formula for total loan cost is attached.
10 years = 120 payments
MONTHLY Interest = .0325 / 12 = <span> <span> <span> 0.0027083333 </span> </span> </span>

total = <span>0.0027083333 * 21,000 * 120 / (1-(1</span><span><span>.0027083333)-120 </span> </span>
total = 6,825 / (1- <span> <span> <span> 0.7228448406 </span> </span> </span> )
total = 6,825 / <span> <span> <span> 0.2771551594 </span> </span> </span>
total = <span> <span> <span> 24,625.20 </span> </span> </span>


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8. Given the radius and area of the circle, calculate the angle at the centre of the circle.
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The answer is c for this problem
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Four less then five times a number is equal to 11
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Step-by-step explanation:

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Read 2 more answers
Demand The demand function for a product is given by
stira [4]

Answer:

x= -\frac{ln [\frac{5P}{8000-P}]}{0.002}

a) x= -\frac{ln [\frac{5*200}{8000-200}]}{0.002} =1027.062 \approx 1027

b) x= -\frac{ln [\frac{5*800}{8000-800}]}{0.002} =293.893 \approx 294

Step-by-step explanation:

For this case we have the following function:

P= 8000 (1- \frac{5}{5 +e^{-0.002 x}})

We can solve for x like this. First we can reorder the expression like this:

\frac{P}{8000} = 1- \frac{5}{5+e^{-0.002x}}

\frac{5}{5+e^{-0.002x}} = 1 -\frac{P}{8000} = \frac{8000-P}{8000}

\frac{40000}{8000-P} = 5 + e^{-0.002x}

Now we can apply natura log on both sids and we got:

ln[\frac{40000}{8000-P} -5] = ln e^{-0.002x}

ln [\frac{5P}{8000-P}] = -0.002x

And if we solve for x we got:

x= -\frac{ln [\frac{5P}{8000-P}]}{0.002}

Part a

For this case we can replace P = 200 and see what we got for x like this:

x= -\frac{ln [\frac{5*200}{8000-200}]}{0.002} =1027.062 \approx 1027

Part b

For this case we can replace P = 800 and see what we got for x like this:

x= -\frac{ln [\frac{5*800}{8000-800}]}{0.002} =293.893 \approx 294

4 0
3 years ago
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