Draw a graph showing the variation of decay rate with number of active nuclei. 

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According to formula,

$\mathrm{I}=-\lambda \mathrm{N}$

$\therefore \mathrm{I}=(-\lambda) \mathrm{N}+0$

Form of above equation is like equation of a straight line, $y=m x+c .$ Hence graph of $\mathrm{I} \rightarrow \mathrm{N}$ is a straight line with slope $(-\lambda)$.

Here graph is obtained in $4^{\text {th }}$ quadrant because $N$ is positive and $I$ is negative.

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