$\frac{34^{2} - 16^{2}}{50}$ = _____________.

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Solution

$\frac{34^{2} - 16^{2}}{34 + 16} = \frac{a^{2} - b^{2}}{a + b}$

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Q. Question 4 (i)
The heights of 50 students, measured to the nearest centimeters, have been found to be as follows:

\begin{matrix} 161 & 150 & 154 & 165 & 168 & 161 & 154 & 162 & 150 & 151 162 & 164 & 171 & 165 & 158 & 154 & 156 & 172 & 160 & 170 153 & 159 & 161 & 170 & 162 & 165 & 166 & 168 & 165 & 164 154 & 152 & 153 & 156 & 158 & 162 & 160 & 161 & 173 & 166 161 & 159 & 162 & 167 & 168 & 159 & 158 & 153 & 154 & 159 \end{matrix}

(i) Represent the data given above by a grouped frequency distribution table, taking the class intervals as 160 - 165, 165 - 170, etc.

Question 4 (ii)
The heights of 50 students, measured to the nearest centimeters, have been found to be as follows:
$\begin{matrix} 161 & 150 & 154 & 165 & 168 & 161 & 154 & 162 & 150 & 151 162 & 164 & 171 & 165 & 158 & 154 & 156 & 172 & 160 & 170 153 & 159 & 161 & 170 & 162 & 165 & 166 & 168 & 165 & 164 154 & 152 & 153 & 156 & 158 & 162 & 160 & 161 & 173 & 166 161 & 159 & 162 & 167 & 168 & 159 & 158 & 153 & 154 & 159 \end{matrix}$
$\begin{matrix} Height (in cm) & Number of studentss(frequency) 150 - 155 & 12 155 - 160 & 9 160 - 165 & 14 165 - 170 & 10 170 - 175 & 5 Total & 50 \end{matrix}$
(ii) What can you conclude about their heights from the table?

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342−16250 = _____________.

342−16234+16=a2−b2a+b

$\frac{34^{2} - 16^{2}}{50}$ = _____________.

$\frac{34^{2} - 16^{2}}{34 + 16} = \frac{a^{2} - b^{2}}{a + b}$