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Lena [83]
3 years ago
12

If you roll a dice 100 times how many times would you expect to roll a product that is a multiple of 5

Mathematics
1 answer:
Luden [163]3 years ago
3 0

If you're rolling a single cube, then ...

-- There are 6 possible out comes.
-- One of them (5) is a multiple of 5 .
-- The probability of rolling it is (1/6) = <em>16-2/3 %</em> .

If everything is acting perfectly random, then ion 100 rolls,
you'd expect to succeed 16 or 17 times.


If you're rolling a pair of cubes, then ...

-- There are (6 x 6) = 36 possible outcomes.
-- There are seven ways to get a multiple of 5.
(1+4),  (4+1),  (2+3),  (3+2),  (4+6),  (6+4),  (5+5)
-- The probability of rolling one of those is
                                                             (7/36) = <em>19-4/9 %</em> .

If everything is acting perfectly random, then in 100 rolls,
you'd expect to succeed 19 or 20 times.


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Suppose that $p$ and $q$ are positive numbers for which \[\log_9 p = \log_{12} q = \log_{16} (p + q).\] what is the value of $q/
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Given p,q>0 and \log_9p=\log_{12}q=\log_{16}(p+q)=x (say).

Then,

p=9^x\\ q=12^x\\ p+q=16^x

From the above 3 equations,

\frac{q}{p} =(\frac{12}{9} )^x\\ \frac{q}{p} =(\frac{4}{3} )^x\\ \frac{p+q}{p} =(\frac{16}{9} )^x\\ \frac{p+q}{p} =(\frac{4}{3} )^{2x}\\

From the equations, we get

\frac{p+q}{p}=(\frac{q}{p})^2\\ 1+\frac{q}{p}=(\frac{q}{p})^2\\ (\frac{q}{p})^2-\frac{q}{p}-1=0\\ \frac{q}{p}=\frac{1 \pm \sqrt{5}}{2}

Since p,q>0, the negative value is rejected.

\frac{q}{p}=\frac{1 + \sqrt{5}}{2}

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