First, convert the fraction on the left from a mixed fraction to an improper fraction then multiply the two fractions. See full explanation below:Explanation:First, convert the mixed fraction\displaystyle{1}\frac{{1}}{{4}} ...

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\displaystyle\frac{{7}}{{3}} Explanation:First change\displaystyle{5}\frac{{1}}{{4}}into an improper fraction. \displaystyle{5}\frac{{1}}{{4}}=\frac{{21}}{{4}} so\displaystyle\ \text{ }\ {5}\frac{{1}}{{2}}\times\frac{{4}}{{9}}\ \text{ }\ \to\ \text{ }\ \frac{{21}}{{4}}\times\frac{{4}}{{9}} ...
See below.Explanation:You multiply the\displaystyle{N}^{{r}} s and\displaystyle{D}^{{r}} s. \displaystyle\frac{{7}}{{4}}\times\frac{{3}}{{5}}\displaystyle\Rightarrow\frac{{{7}\times{3}}}{{{4}\times{5}}} ...
You find the product by using the rule for multiplying fractions. See the full explanation below.Explanation:The rule for multiplying fractions is to multiply the numerators and denominators ...
https://math.stackexchange.com/questions/610627/probability-of-choosing-coins-from-a-bag-why-doesnt-binomial-coefficient-work
Pr(ND) = \frac{1}{4\choose1} is implying that the probability of choosing a nickel then a dime is \frac{1}{4}. For the first one, with the choosing of 2 pennies, it's 4 choose 2 becuase there are ...
During a night, each chameleon changes its colour to one of the other four colours with equal probability.

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https://math.stackexchange.com/questions/1296373/during-a-night-each-chameleon-changes-its-colour-to-one-of-the-other-four-colou
The number of permutations without fixed points in S_5 is given by: 5!\left(\frac{1}{2!}-\frac{1}{3!}+\frac{1}{4!}-\frac{1}{5!}\right)=\color{red}{44}\tag{1} by the inclusion-exclusion ...
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\left< \begin{array} { l l } { 2 } & { 3 } \\ { 5 } & { 4 } \end{array} \right> \left< \begin{array} { l l l } { 2 } & { 0 } & { 3 } \\ { -1 } & { 1 } & { 5 } \end{array} \right>

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