648. Find the point on y-axis which is equidistant from the points (5, 2), and (-4, 3).

]]>I agree with your answers.

Why this formula? Let's start with the median.

It'll make more sense if the number of data items is 7. Let's say:

2, 4, 6, 8, 10, 12, 14

The median is the middle item in ordered list so the median is 8.

If the list had 8 items, say 2, 4, 6, 8, 10, 12, 14, 16 then there's no middle number. So statisticians have chosen a rule for this: Choose the number half way between the 4th and the 5th numbers ie 0.5(8+10) = 9. They call this the 4.5th number in the list, just to give it a name. It's just a name for it so don't get too bothered by it.

The quartiles are defined to be the numbers one quarter and three quarters of the way along the list.

So for n=7, list = {2, 4, 6, 8, 10, 12, 14} the lower quartile is the 2nd number which is 4, and the upper quartile is 6th number which is 12.

This works nicely when n is a multiple of 4 minus 1. You can easily find actual numbers in the list 1/4, 1/2 and 3/4 of the way along the list.

But what should you do when n isn't a multiple of 4 minus 1 ? To understand the formulas look at how it works when n=7

(7+1)/4 , (7+1)/2 , 3(7+1)/4 give 2nd, 4th and 6th numbers in the list so that's the formula that has been chosen for any value of n.

(n+1)/4, (n+1)/2 and 3(n+1)/4

Of course, in a list where n is not a multiple of 4 minus 1, this leads to apparently silly positions as you have found 1.5th etc. For small lists it isn't really a useful statistic anyway but if n is large it gives a good estimate of the values 1/4, 1/2 and 3/4 of the way along the list.

The situation where I have found it is really useful is when trying to compare two sets of data. Let's say you have the exam results for two classes and you want to consider which class is better. Probably you'd work out the class averages for this. But it's also useful to know how spread out the results are.

Just comparing the range of marks can be misleading because there might be outlying values that distort the overall comparisons. If you calculate the inter quartile range (upper quartile minus lower quartile) you remove the bottom and top quarters of the data and just look at those in the middle 50%. That statistic is a better way to compare the data. But this calculation has to work whatever 'n' is, which is why a formula is needed.

Bob

]]>If you have some centimetre squared paper then you could try plotting the straight line. The answer will be easy to spot then and you'll see a pattern

Bob

]]>4. Volume of cylinder with radius = 4 m.

Height = 8 m.

.

Solution : Volume = 402.1248 cubic meters.

Rounded to 402 cubic meters.

Option (d) is correct.

5. Volume of a cylinder with diameter 21 centimeters and height 15 centimeters.

Here, diameter = 21 centimeters, radius = 10.5 centimeters.

cubic centimeters.

Option (c) is correct.]]>

Anyway, what I wanted to say was: is it possible to ahve some kind of "difficulty level" along with the exercises? I've browsed to the forum and have found almost only things I don't even understand. Not things I can't solve, but things I don't understand at all... I've come here with a rusty high school level of mathematics as to get better, and now I just feel lost. Sorry for having suh a low level, but I'm here to solve that issue!

]]>I made a video explanation here, if Zeeshan 01 would like to have a look.

]]>There are two stages to this problem. I think you have done the first but I'll say anyway.

This is a product of two functions so use the product rule:

You can combine the trig functions into a single trig function using the compound angle formulas. You want a cosine so I'll use

To make the derivative look like the right hand side here:

Hope that helps,

Bob

]]>Welcome to the forum.

Looks like you've got these sorted ; I agree with all your answers.

Bob

]]>Welcome to the forum.

I've been hoping that someone would post an answer to this and I could learn too. No such luck

So I'm having a go myself. Please note: I've never done this before, so what follows may be rubbish. Please comment …. ask for clarification … tell me why it's wrong etc. Maybe between us, we can arrive at the correct answer. And oh yes … your English is good.

So let's work with 3D coordinates with the x-y plane horizontal and z going straight up. Further, let's make the sphere have unit radius (cannot see any harm in that) and centred on the origin, O.

If P is one point of the tetrahedron, then we can specify its position using spherical coordinates theta and phi as shown here: And let P' be the point in the x-y plane below P.

Now, what would be helpful is to have a formula for the volume of a tetrahedron in terms of theta and phi but I cannot find one. Plenty of internet pages giving the formula in vector terms such as

and as a determinant

I could also expand either into a large algebraic formula but it would take ages to enter all the LaTex so you'll have to ask nicely if you want this.

phi can take any random value from 0 to 2pi and theta any from 0 to pi.

So you can construct your function with 8 variables and there it is. Hhhmmmm.

Bob

]]>]]>

Sorry, but I do not understand what you are asking. Please post the whole question.

Bob

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