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3 years ago

Simplify m18 over m61

Mathematics

1 answer:

Bond [772]3 years ago

6 0

${m}^{ - 53}$

1) Use quotient rule:

$\frac{ {x}^{a} }{ {x}^{b} } = {x}^{a - b}$

2) Simply

$8 - 61 \: \: to \: - 53 = {m}^{ - 53}$

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The integers $r$ and $k$ are randomly selected, where $-3 < r < 6$ and $1 < k < 8$. what is the probability that the

e-lub [12.9K]

Denote by $R,K$ the random variables representing the integer values $r,k$ , respectively. Then $R\sim\mathrm{Unif}(-2,5)$ and $K\sim\mathrm{Unif}(2,7)$ , where $\mathrm{Unif}(a,b)$ denotes the discrete uniform distribution over the interval $[a,b]$ . So $R$ and $K$ have probability mass functions

$p_R(r)=\begin{cases}\dfrac18&\text{for }r\in\{-2,-1,\ldots,5\}\\\\0&\text{otherwise}\end{cases}$

$p_K(k)=\begin{cases}\dfrac16&\text{for }k\in\{2,3,\ldots7\}\\\\0&\text{otherwise}\end{cases}$

We want to find $P\left(\dfrac RK\equiv0\pmod n\right)$ , where $n$ is any integer.

We have six possible choices for $K$ :

(i) if $K=2$ , then $\dfrac RK$ is an integer when $R=\pm2,0,4$ ;

(ii) if $K=3$ , then $\dfrac RK$ is an integer when $R=0,3$ ;

(iii) if $K=4$ , then $\dfrac RK$ is an integer when $R=0,4$ ;

(iv) if $K=5$ , then $\dfrac RK$ is an integer when $R=0,5$ ;

(v) if $K=6$ or $K=7$ , then $\dfrac RK$ is an integer only when $R=0$ in both cases.

If the selection of $R,K$ are made independently, then the joint distribution is the product of the marginal distribution, i.e.

$p_{R,K}(r,k)=p_R(r)\cdot p_K(k)=\begin{cases}\dfrac1{48}&\text{for }(r,k)\in[-2,5]\times[2,7]\\\\0&\text{otherwise}\end{cases}$

That is, there are 48 possible events in the sample space. We counted 12 possible outcomes in which $\dfrac RK$ is an integer, so the probability of this happening is $\dfrac{12}{48}=\dfrac14$ .