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DENIUS [597]
3 years ago
9

After getting trounced by your little brother in a​ children's game, you suspect the die he gave you to roll may be unfair. To​

check, you roll it 48 ​times, recording the number of times each face appears. Do these results cast doubt on the​ die is fairness
Mathematics
1 answer:
iragen [17]3 years ago
6 0

Answer:

The results are

Face     Count         Face          Count

1               6                  4                9

2              12                 5                5

3              8                 6               20

These results are not fair, because they clearly shows that the dice was tricked. We deduct this by observing that he obtained the face 6, 20 times out of 48, which represents 42% of chances to get that result.

Now, a fair dice must have equal probabilities per side, or an approximation to that at least. In this case, we don't have such equality.

So, next time your little brother roll the dice, we probably will get 6.

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Simplify. -4 + 12 + (-9) *<br><br> PLEASE HELP REEEE
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Find a power series for the function, centered at c, and determine the interval of convergence. f(x) = 9 3x + 2 , c = 6
san4es73 [151]

Answer:

\frac{9}{3x + 2} = 1 - \frac{1}{3}(x - \frac{7}{3}) + \frac{1}{9}(x - \frac{7}{3})^2 - \frac{1}{27}(x - \frac{7}{3})^3 ........

The interval of convergence is:(-\frac{2}{3},\frac{16}{3})

Step-by-step explanation:

Given

f(x)= \frac{9}{3x+ 2}

c = 6

The geometric series centered at c is of the form:

\frac{a}{1 - (r - c)} = \sum\limits^{\infty}_{n=0}a(r - c)^n, |r - c| < 1.

Where:

a \to first term

r - c \to common ratio

We have to write

f(x)= \frac{9}{3x+ 2}

In the following form:

\frac{a}{1 - r}

So, we have:

f(x)= \frac{9}{3x+ 2}

Rewrite as:

f(x) = \frac{9}{3x - 18 + 18 +2}

f(x) = \frac{9}{3x - 18 + 20}

Factorize

f(x) = \frac{1}{\frac{1}{9}(3x + 2)}

Open bracket

f(x) = \frac{1}{\frac{1}{3}x + \frac{2}{9}}

Rewrite as:

f(x) = \frac{1}{1- 1 + \frac{1}{3}x + \frac{2}{9}}

Collect like terms

f(x) = \frac{1}{1 + \frac{1}{3}x + \frac{2}{9}- 1}

Take LCM

f(x) = \frac{1}{1 + \frac{1}{3}x + \frac{2-9}{9}}

f(x) = \frac{1}{1 + \frac{1}{3}x - \frac{7}{9}}

So, we have:

f(x) = \frac{1}{1 -(- \frac{1}{3}x + \frac{7}{9})}

By comparison with: \frac{a}{1 - r}

a = 1

r = -\frac{1}{3}x + \frac{7}{9}

r = -\frac{1}{3}(x - \frac{7}{3})

At c = 6, we have:

r = -\frac{1}{3}(x - \frac{7}{3}+6-6)

Take LCM

r = -\frac{1}{3}(x + \frac{-7+18}{3}+6-6)

r = -\frac{1}{3}(x + \frac{11}{3}+6-6)

So, the power series becomes:

\frac{9}{3x + 2} =  \sum\limits^{\infty}_{n=0}ar^n

Substitute 1 for a

\frac{9}{3x + 2} =  \sum\limits^{\infty}_{n=0}1*r^n

\frac{9}{3x + 2} =  \sum\limits^{\infty}_{n=0}r^n

Substitute the expression for r

\frac{9}{3x + 2} =  \sum\limits^{\infty}_{n=0}(-\frac{1}{3}(x - \frac{7}{3}))^n

Expand

\frac{9}{3x + 2} =  \sum\limits^{\infty}_{n=0}[(-\frac{1}{3})^n* (x - \frac{7}{3})^n]

Further expand:

\frac{9}{3x + 2} = 1 - \frac{1}{3}(x - \frac{7}{3}) + \frac{1}{9}(x - \frac{7}{3})^2 - \frac{1}{27}(x - \frac{7}{3})^3 ................

The power series converges when:

\frac{1}{3}|x - \frac{7}{3}| < 1

Multiply both sides by 3

|x - \frac{7}{3}|

Expand the absolute inequality

-3 < x - \frac{7}{3}

Solve for x

\frac{7}{3}  -3 < x

Take LCM

\frac{7-9}{3} < x

-\frac{2}{3} < x

The interval of convergence is:(-\frac{2}{3},\frac{16}{3})

6 0
3 years ago
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