5.1. Find the summarized error Am taking into account a device and round-off error (4_(m)^ast =0 , since the mass measurement was a single one) Delta bar (m)=sqrt ((Delta _(ba))^2+(Delta _(ma))^2)=sqrt (( )^2+( )^2)approx -0.048-0.05 (g) 52. Calculate an aggregate error Delta bar (h) under the formula Delta bar (h)=sqrt ((a_(lambda ))^2+(a_(nc))^2+(a_(nB))^2)=sqrt (( )^2+( )^2+( )^2)approx 1 (cm) Where Delta _(n)=t_(an)sqrt ((sum _(n)^b(bar (h)-bar (h))^2))/(frac (bar (h)(n-1)^2)(n(n-1)))=ldots }sqrt (-)approx ldots (cm)

5.1. To find the summarized error Am, we need to take into account both the device error and the round-off error.<br /><br />Given that the mass measurement was a single one, we have $(4_{m}^{\ast }=0$.<br /><br />The formula for calculating the summarized error Am is:<br />$\Delta \bar {m}=\sqrt {(\Delta _{ba})^{2}+(\Delta _{ma})^{2}}$<br /><br />Substituting the given values, we have:<br />$\Delta \bar {m}=\sqrt {(\quad )^{2}+(\quad )^{2}}\approx -0.048-0.05$ (g)<br /><br />52. To calculate the aggregate error $\Delta \bar {h}$, we use the formula:<br />$\Delta \bar {h}=\sqrt {(a_{\lambda })^{2}+(a_{nc})^{2}+(a_{nB})^{2}}$<br /><br />Substituting the given values, we have:<br />$\Delta \bar {h}=\sqrt {(\quad )^{2}+(\quad )^{2}+(\quad )^{2}}\approx 1\quad (cm)$<br /><br />Where:<br />$\Delta _{n}=t_{an}\sqrt {\frac {\sum _{n}^{b}(\bar {h}-\bar {h})^{2}}}{\frac {\bar {h}(n-1)^{2}}{n(n-1)}}=\ldots }\sqrt {-}\approx \ldots (cm)$