da672c…: “Rather than a bitter circle”

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\pgfmathsetmacro {\nextphi }{\curphi + \tdplotsuperfudge *\viewphistep } \par \begin {scope}[opacity=1] \par \par \tdplotcheckdiff {\nextphi }{360}{\origviewphistep }{#2}{} \tdplotcheckdiff {\nextphi }{0}{\origviewphistep }{#2}{} \par \tdplotcheckdiff {\nextphi }{90}{\origviewphistep }{#3}{} \tdplotcheckdiff {\nextphi }{450}{\origviewphistep }{#3}{} \end {scope} \par \foreach \curtheta in{\viewthetastart ,\viewthetainc ,...,\viewthetaend } { \par \pgfmathsetmacro {\curlongitude }{90 -\curphi } \pgfmathsetmacro {\curlatitude }{90 -\curtheta } \par \ifthenelse {\equal {\leftright }{-1.0}}{\pgfmathsetmacro {\curphi }{\curphi -\origviewphistep } }{} \par \pgfmathsetmacro {\tdplottheta }{mod(\curtheta ,360)} \pgfmathsetmacro {\tdplotphi }{mod(\curphi ,360)} \par \pgfmathparse {\tdplotphi <0} \ifthenelse {\equal {\pgfmathresult }{1}}{ \pgfmathsetmacro {\tdplotphi }{\tdplotphi + 360} }{}\par \pgfmathparse {\tdplottheta >\tdplotuppertheta } \pgfmathsetmacro {\logictest }{1 -\pgfmathresult } \par \pgfmathparse {\tdplottheta <\tdplotlowertheta } \pgfmathsetmacro {\logictest }{\logictest * (1 -\pgfmathresult )} \par \pgfmathsetmacro {\tdplottheta }{\tdplottheta + \viewthetastep } \pgfmathparse {\tdplottheta >\tdplotuppertheta } \pgfmathsetmacro {\logictest }{\logictest * (1 -\pgfmathresult )} \par \pgfmathparse {\tdplottheta <\tdplotlowertheta } \pgfmathsetmacro {\logictest }{\logictest * (1 -\pgfmathresult )} \par \pgfmathparse {\tdplotphi >\tdplotupperphi } \pgfmathsetmacro {\logictest }{\logictest * (1 -\pgfmathresult )} \par \pgfmathparse {\tdplotphi <\tdplotlowerphi } \pgfmathsetmacro {\logictest }{\logictest * (1 -\pgfmathresult )} \par \pgfmathsetmacro {\tdplotphi }{\tdplotphi + \viewphistep } \par \pgfmathparse {\tdplotphi <0} \ifthenelse {\equal {\pgfmathresult }{1}}{ \pgfmathsetmacro {\tdplotphi }{\tdplotphi + 360} }{}\par \pgfmathparse {\tdplotphi >\tdplotupperphi } \pgfmathsetmacro {\logictest }{\logictest * (1 -\pgfmathresult )} \par \pgfmathparse {\tdplotphi <\tdplotlowerphi } \pgfmathsetmacro {\logictest }{\logictest * (1 -\pgfmathresult )} \par \par \pgfmathsetmacro {\tdplottheta }{\curtheta } \pgfmathsetmacro {\tdplotphi }{\curphi } \par \ifthenelse {\equal {#6}{parametricfill}}{\ifthenelse {\equal {\logictest }{1.0}}{\pgfmathsetmacro {\radius }{#1} \pgfmathsetmacro {\tdplotr }{\radius *360} \par \pgfmathlessthan {\radius }{0} \pgfmathsetmacro {\phaseshift }{180 * \pgfmathresult } \par \pgfmathsetmacro {\colorarg }{#5} \pgfmathsetmacro {\colorarg }{\colorarg + \phaseshift } \pgfmathsetmacro {\colorarg }{mod(\colorarg ,360)} \par \pgfmathlessthan {\colorarg }{0} \pgfmathsetmacro {\colorarg }{\colorarg + 360*\pgfmathresult } \par \pgfmathdivide {\colorarg }{360} \definecolor {tdplotfillcolor}{hsb}{\pgfmathresult ,1,1} \color {tdplotfillcolor} }{}}{\pgfsetfillcolor {#5} } \pgfsetstrokecolor {#4} \par \ifthenelse {\equal {\leftright }{-1.0}}{\pgfmathsetmacro {\curphi }{\curphi + \origviewphistep } }{} \par \ifthenelse {\equal {\logictest 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(\tdplotstopx ,\tdplotstopy ) {$2\pi $}; \par \pgfmathsetmacro {\tdplotstopy }{\tdploty + (360+\tdplothuestep )/2*\tdplotyscale } \node [tdplot_screen_coords,anchor=west,xshift=5pt] at (\tdplotstopx ,\tdplotstopy ) {$\pi $}; } \newcommand {\tdplotgetpolarcoords }[3]{\pgfmathsetmacro {\vxcalc }{#1} \pgfmathsetmacro {\vycalc }{#2} \pgfmathsetmacro {\vzcalc }{#3} \pgfmathsetmacro {\vcalc }{ sqrt((\vxcalc )^2 + (\vycalc )^2 + (\vzcalc )^2) } \par \pgfmathsetmacro {\vxycalc }{ sqrt((\vxcalc )^2 + (\vycalc )^2) } \par \pgfmathsetmacro {\tdplotrestheta }{asin(\vxycalc /\vcalc )} \pgfmathparse {\vzcalc <0} \ifthenelse {\equal {\pgfmathresult }{1}}{\pgfmathsetmacro {\tdplotrestheta }{180 -\tdplotrestheta } } {} \ifthenelse {\equal {\vxcalc }{0.0}}{\pgfmathparse {\vycalc <0} \ifthenelse {\equal {\pgfmathresult }{1}}{\pgfmathsetmacro {\tdplotresphi }{270} } {\pgfmathparse {\vycalc >0} \ifthenelse {\equal {\pgfmathresult }{1}}{\pgfmathsetmacro {\tdplotresphi }{90} } {\pgfmathsetmacro 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A student is conducting a survey for their statistics class. They decide to poll 20% of the 1200 full-time students on campus and 40% of the 4000 part-time students. What percent of the total student population did the student survey? First give the exact answer. $\answer {\frac {184000}{5200}}$

Now, use a calculator and round your answer to 1 decimal point. $\answer {35.4}\%$