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Sedbober [7]
3 years ago
14

What is the equation in point-slope form of the line passing through (1, 9) and (−1, 11)?

Mathematics
1 answer:
ololo11 [35]3 years ago
7 0
Point slope form: 
y-y_1=m(x-x_1)

m=slope 
(x1,y1) 

First we need to find the slope using the slope formula:
\frac{y_2-y_1}{x_2-x_1} =slope

Given the two coordinates: 
<span>(1, 9) and (−1, 11)

Plug them into the slope formula: 
</span>\frac{11-9=2}{-1-1=-2} =slope
Simplify: 
\frac{2}{-2} =-1
<span>
Now we know that: 
m=-1  and the (x1,y1) would be </span><span>(1, 9)
</span>
Now we simply plug in the info: 
y-9=-(x-1)

Final answer: y − 9 = −(x − 1)
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Assume the random variable X is normally distributed with meanmu equals 50μ=50and standard deviationsigma equals 7σ=7.Compute th
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Step-by-step explanation:

Previous concepts

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

The Z-score is "a numerical measurement used in statistics of a value's relationship to the mean (average) of a group of values, measured in terms of standard deviations from the mean".  

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Let X the random variable that represent the variable of interest of a population, and for this case we know the distribution for X is given by:

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In a casino game, gamblers are allowed to roll n fair, 6-sided dice. If a 6 shows up on any of them, the gambler gets nothing. I
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Answer:

a) \bf \boxed{W(n) = 3*n*\left( \frac{5}{6} \right)^n\;dollars}

the smallest n that maximizes the expected payoff is n=5.

b) 4

Step-by-step explanation:

a)

The expected amount of $ won for each die would be the average of 1, 2, 3, 4 and 5 which is $3.

Let W(n) the expected money won when rolling n dice.  

n =1

If the gambler rolls only one die, the expected money won would be $3 times the probability of not getting a 6, which is 5/6.

So  

\bf W(1) = 3*1*\frac{5}{6}

n=2

If the gambler rolls 2 dice, the expected money won would be $3 times the probability of not getting a 6 in any of the dice. Since the outcome of the rolling does not depend on the previous rollings, the probability is  

\bf \frac{5}{6}\times\frac{5}{6}=\left( \frac{5}{6} \right)^2

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\bf W(2) = 3*2*\left( \frac{5}{6} \right)^2

n=3

Similarly, since the probability of not getting a 6 in 3 dice equals

\bf \left( \frac{5}{6} \right)^3

\bf W(3) = 3*3*\left( \frac{5}{6} \right)^3

and the formula for the expected money won with n dice would be

\bf \boxed{W(n) = 3*n*\left( \frac{5}{6} \right)^n\;dollars}

In the picture attached there is a plot of the values of the expected money won for n=1 to 20 (See picture)

For n=5 and n=6 we get the maximum profit expected of $6.02816=$6 rounded to the nearest integer.

Hence, the smallest n that maximizes the expected payoff is n=5.

b)

The probability that face k (k=1,2,...or 6) shows up is 1/6,

as this face can be in any of the 10 positions of the arrangement, there are 10 ways that face k can show up.

The probability that face k (k=1,2,...or 6) shows up twice is \bf \left( \frac{1}{6} \right)^2

as this face can be in any of the \bf C(10;2)=\binom{10}{2} (combinations of 10 taken 2 at a time) positions of the arrangement, there are C(10;2) ways that face k can show up twice.

The probability that face k (k=1,2,...or 6) shows up three times is \bf \left( \frac{1}{6} \right)^3

as this face can be in any of the \bf C(10;3)=\binom{10}{2} (combinations of 10 taken 2 at a time) positions of the arrangement, there are C(10;3) ways that face k can show up twice.

So, we infer that the expectation is  

\bf \sum_{k=1}^{10}\binom{10}{k}(1/6)^k=3.6716\approx 4

and the expected number of distinct dice values that show up is 4.

4 0
3 years ago
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