[tex]\huge\textbf{Hey there!}[/tex]
[tex]\mathbf{7\dfrac{1}{6} = 5\dfrac{2}{5} + p}[/tex]
[tex]\rightarrow\mathbf{7 \dfrac{1}{6} = 5\dfrac{2}{5} + p}[/tex]
[tex]\mathbf{\rightarrow p + 5\dfrac{2}{5}= 7\dfrac{1}{6}}[/tex]
[tex]\mathbf{\rightarrow {p + \dfrac{27}{5}=\dfrac{43}{6}}}[/tex]
[tex]\text{SUBTRACT }\rm{\dfrac{27}{5}}\text{ to BOTH SIDES}[/tex]
[tex]\mathbf{p + \dfrac{27}{5} - \dfrac{27}{5} = \dfrac{43}{6}- \dfrac{27}{5}}[/tex]
[tex]\text{CANCEL out: }\rm{\dfrac{27}{5} - \dfrac{27}{5}}\text{ because it gives you 0}[/tex]
[tex]\text{KEEP: }\rm{\dfrac{43}{6} - \dfrac{27}{5}}\text{ because it help solve for the p-value}[/tex]
[tex]\text{NEW EQUATION: }\rm{p = \dfrac{43}{6} - \dfrac{27}{5}}[/tex]
[tex]\large\textsf{SIMPLIFY IT!}[/tex]
[tex]\mathbf{p = \dfrac{53}{30} \approx p = 1\dfrac{23}{30}}[/tex]
[tex]\huge\boxed{\textsf{Therefore, your answer is: \boxed{\mathsf{p = 1\dfrac{23}{30}}}}}\huge\checkmark[/tex]
[tex]\huge\textbf{Good luck on your assignment \&}\\\huge\textbf{enjoy your day!}[/tex]
~[tex]\frak{Amphitrite1040:)}[/tex]
