Here is the video regarding Proof using Mathematical Induction and the answers to Soakimi's question. Hope this will help Soakimi and anyone else who might be interested!

Malo Viliami hono tali atu e fehu'i 'a Soakimi. Toki ma'u eni e faingamalie keu sio ki he fehu'i. Anway, I think Soakimi will appreciate your help Viliami. I don't have time at the moment to do a video regarding Soakimi's question but I will do one next weekend. I will create a video on part (b) of the question and will cover a few details to help Soakimi or anyone who might not fully understand the principle and the process involved in mathematical induction.

Hi Suliana, just an extra point for clarification. The data were presented in a frequency table and that is why I used the cumulative frequency column. You could have listed the data as shown on the attached diagram and still work out the quartiles. However, if you had more than say 100 data points then listing the data becomes tedious and inefficient. That is why a frequency table and a cumulative frequency column is useful in this analysis. Please note that if the number of data points is EVEN (30 in this case) then there are two middle data points and therefore, take the average of the two. In this case, since the two middle data points are 2.5 and 2.5 then the average is of course 2.5. Sometimes, if you don't have consecutive numbers as data points, the average of the two middle data points is significantly different. The diagram does confirm Latu's solution.

Great work Falesiu, I just realised that we have posted roughly at the same time last night. I was doing my own work and decided to check the forum for I haven't been here for a while. I saw Suliana's post and decided to give her another way of finding a solution. It is good that we both approached the problem from a slightly different perspective. Hopefully, what we have posted here will help Suliana and encourages her and others to post more questions. Have a great day.

Hi Suliana, Latu Falesiu will probably explain in details how to find the Lower Quartile (Q1) and UpperQuartile (Q3) later. However, after looking at the table, I would only need a cumulative frequency column. I will then use the quartile formulas to find Q1, Q2 and Q3. To draw the box and whisker plot. Please see the attached file for details and if you require further explanations then feel free to post your request here.. There maybe other methods to do this but I prefer to solve such questions with the method that I have shown on the file. Note that the frequency table that I have provided does not require the fx column. I will only need the fx column if I want to calculate the mean.

Kataki 'o fakaikiiki mai ho'o fehu'i keu mahino'i.

Another example using a Geometric sequence. See the attached pdf file

Kataki 'oku fu'u general ho'o fiema'u pea 'oku 'ikai keu fu'u mahino'i ho'o 'uhinga. Kaikehe teu fai pe ha fakaofiofi ki ho'o fiema'u. Let's have a look at Solving simultaneous equations first: Example: solve the following equations simultaneously. 3x + 2y = 8 x + y = 3 In this case, teke lava pe 'o solova 'aki 'ae Substitution pe koe Elimination method: 1. Using Substitution Method: Ui e first equation koe Eq1, pea second one koe Eq2. Sai, rearrange Eq2 ke hoko koe: y = -x + 3 Substitute y in Eq2 into y in Eq1. 'o pehe ni: 3x + 2(-x+3) = 8, expand pea simplify then solve for x 3x + -2x + 6 = 8 x + 6 = 8 x = 2. Now find y since x is found to be 2. Teke lava pe 'o ngaue'aki ha taha pe 'oe ongo equation ka teu fili 'ae Eq2 he 'oku matangofua ange ia ke ngaue'aki. x + y = 3 , substitute in x = 2 2 + y = 3 y = 1. solution (2, 1) or x = 2, y = 1 2. Elimination Method: 3x + 2y = 8 x + y = 3 Sai teu liunga 3 'ae Eq2 ke tatau e value 'o x he ongo equation kae lava keu eliminate 'a x, Fakatokanga'i ange teu lava pe 'o eliminate 'a y kapau teu liunga 2 'ae Eq2. Anyway, as a matter of choice, I want to eliminate x 3(Eq2) = 3(x + y = 3) , note that you have to multiply each term by 3 = 3x + 3y = 9 , I will call this new equation Eq3 Now, koe ongo equation 'e pehe ni: 3x + 2y = 8 , Eq1 3x + 3y = 9 , Eq3 Take away Eq1 from Eq3 to avoid having negative numbers but of course you can still take away Eq3 from Eq1 if you are good with directed numbers. 3x + 3y = 9 - 3x + 2y = 8 y = 1, now find x Using Eq2, you can quickly find x x + y = 3 x = 2 Solution (2, 1) or x = 2, y = 1 Note, the answers for both methods are the same. Sai, ta hoko atu ki ho'o fehu'i fekau'aki moe algebraic sequence, are you talking of Arithmetic sequence or Geometric Sequence? 'Oku fiema'u ia kete 'ilo pe koe ha e type of sequence kae lava kete formulate ha ongo equations pea te toki solve simultaneously. Hange ko eni: Let's use an Arithmetic Sequence as an example: Problem: The second term of an Arithmetic sequence is 8 and the ninth term is 29, find the first term, a and the common difference, d We can formulate our two equations as: Using the general formula for Arithmetic Sequence: tn = a + (n-1)d We know, n=2 and n=9, tn = 8 and tn=29, in our scenario. Substitute these values in: 8 = a + (2-1)d, ... Eq1 29 = a + (8-1)d, ... Eq2 We get: 8 = a + d, ... Eq1 29 = a + 7d, ...Eq2 Use the Elimination method to eliminate a in this case (it is easier to eliminate a) Take away Eq1 from Eq2 29 = a + 8d - 8 = a + d 21 = 7d, solve for d d = 3 Since d = 3, we can find a Using 8 = a + d 8 = a + 3 a = 5 so for this sequence, the first term, a = 5, the common difference, d = 3 Just for fun, we can generate the sequence as: 5, 8, 11, 14, 17, 20, 23, 26, 29,...