Menu Close

Question-189473




Question Number 189473 by Spillover last updated on 17/Mar/23
Answered by a.lgnaoui last updated on 17/Mar/23
x=1   f(1)=((3/4))^3 =((27)/(64))  x=2  f(2)=((3/8))^2 =(9/(64))  x=3  f(3)=((3/(12)))^1 =(1/4)=((16)/(64))  x=4  f(4)=((3/(16)))^0 =1=((64)/(64))  x=5   f(5)=((3/(20)))^(−1) =((20)/3)  x>5  f(x)=0  Median=((27+9+16+64)/(64×2))+((20)/(6×64))=  =     ((23)/(24))=≅0,96  Mode      =((20)/3)−((23)/(24))=((137)/(24))≅5,7
$${x}=\mathrm{1}\:\:\:{f}\left(\mathrm{1}\right)=\left(\frac{\mathrm{3}}{\mathrm{4}}\right)^{\mathrm{3}} =\frac{\mathrm{27}}{\mathrm{64}} \\ $$$${x}=\mathrm{2}\:\:{f}\left(\mathrm{2}\right)=\left(\frac{\mathrm{3}}{\mathrm{8}}\right)^{\mathrm{2}} =\frac{\mathrm{9}}{\mathrm{64}} \\ $$$${x}=\mathrm{3}\:\:{f}\left(\mathrm{3}\right)=\left(\frac{\mathrm{3}}{\mathrm{12}}\right)^{\mathrm{1}} =\frac{\mathrm{1}}{\mathrm{4}}=\frac{\mathrm{16}}{\mathrm{64}} \\ $$$${x}=\mathrm{4}\:\:{f}\left(\mathrm{4}\right)=\left(\frac{\mathrm{3}}{\mathrm{16}}\right)^{\mathrm{0}} =\mathrm{1}=\frac{\mathrm{64}}{\mathrm{64}} \\ $$$${x}=\mathrm{5}\:\:\:{f}\left(\mathrm{5}\right)=\left(\frac{\mathrm{3}}{\mathrm{20}}\right)^{−\mathrm{1}} =\frac{\mathrm{20}}{\mathrm{3}} \\ $$$${x}>\mathrm{5}\:\:{f}\left({x}\right)=\mathrm{0} \\ $$$${Median}=\frac{\mathrm{27}+\mathrm{9}+\mathrm{16}+\mathrm{64}}{\mathrm{64}×\mathrm{2}}+\frac{\mathrm{20}}{\mathrm{6}×\mathrm{64}}= \\ $$$$=\:\:\:\:\:\frac{\mathrm{23}}{\mathrm{24}}=\cong\mathrm{0},\mathrm{96} \\ $$$${Mode}\:\:\:\:\:\:=\frac{\mathrm{20}}{\mathrm{3}}−\frac{\mathrm{23}}{\mathrm{24}}=\frac{\mathrm{137}}{\mathrm{24}}\cong\mathrm{5},\mathrm{7} \\ $$$$ \\ $$$$ \\ $$

Leave a Reply

Your email address will not be published. Required fields are marked *