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Question Number 189473 by Spillover last updated on 17/Mar/23

Answered by a.lgnaoui last updated on 17/Mar/23

$$x=1f\left(1\right)={\left(\frac{3}{4}\right)}^3=\frac{27}{64}$$$$x=2f\left(2\right)={\left(\frac{3}{8}\right)}^2=\frac{9}{64}$$$$x=3f\left(3\right)={\left(\frac{3}{12}\right)}^1=\frac{1}{4}=\frac{16}{64}$$$$x=4f\left(4\right)={\left(\frac{3}{16}\right)}^0=1=\frac{64}{64}$$$$x=5f\left(5\right)={\left(\frac{3}{20}\right)}^{-1}=\frac{20}{3}$$$$x>5f\left(x\right)=0$$$$Median=\frac{27+9+16+64}{64\times 2}+\frac{20}{6\times 64}=$$$$=\frac{23}{24}=\cong 0,96$$$$Mode=\frac{20}{3}-\frac{23}{24}=\frac{137}{24}\cong 5,7$$$$$$$$$$

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