$14$:$19$

$.$II $4$G

The very first entries have been made $($those of Mayetu Neo$),$ and the student got an average of $56\mathrm{.}07\%$ and the final outcome is a pass. The example below shows how to determine the average mark using Mayetu Neo as an example:

Average $=(\frac{\text{T}\text{u}\text{t}1\text{M}\text{a}\text{r}\text{k}}{\text{T}\text{u}\text{t}1\text{t}\text{o}\text{t}\text{a}\text{l}\text{m}\text{a}\text{r}\text{k}})$ Tut $1$ weighting $+(\frac{\text{T}\text{u}\text{t}2\text{M}\text{a}\text{r}\text{k}}{\text{T}\text{u}\text{t}2\text{t}\text{o}\text{t}\text{a}\text{l}\text{m}\text{a}\text{r}\text{k}})$ Tut $2$ weighting

$+(\frac{\text{T}\text{u}\text{t}3\text{M}\text{a}\text{r}\text{k}}{\text{T}\text{u}\text{t}3\text{t}\text{o}\text{t}\text{a}\text{l}\text{m}\text{a}\text{r}\text{k}})$ Tut $3$ weighting $+(\frac{\text{T}\text{u}\text{t}4\text{M}\text{a}\text{r}\text{k}}{\text{T}\text{u}\text{t}\text{t}\text{o}\text{t}\text{a}\text{l}\text{m}\text{a}\text{r}\text{k}})$ Tut $4$ weighting

$=(\frac{34}{30})5+(\frac{25}{45})20+(\frac{31}{50})35+(\frac{22}{50})40=56\mathrm{.}07\%$

$2$

CPT$2601$ Assessment $\frac{1}{2024}$

$(1\mathrm{.}1)$ Copy the tables as they are into an excel spread sheet. Make sure that the information lies in the cells as they appear in the image: so the student names should lie in cells C$5$:C$9,$ the marks for Tut $1$ should lie in Cells D$5$:D$9$ etc.

$(1\mathrm{.}2)$ Determine the averages for all the students using absolute references were necessary in cells G$5$:G$9.$ Provide two images:

$($a$)$ Image of table with the Excel code in full view

$($b$)$ Image with the actual values

$[22$ marks$]$

Question $2$

Using the Vlookup function to find Mayetu Neo's mark for Tut $1.$

$($a$)$ Image of table with the Excel code in full view

$($b$)$ Image with the actual value

$[14$ marks$]$