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1. Re: Maths help needed for questions below.
2. Re: Maths help needed for questions below.
Don't know about the probability stuff but for the linear bit, should you not just set up a simultaneous equations question? As follows roughly:

(1) 4 = 3000m + c
(2) 6 = 5000m + c

I would then do:

(2) 5000m + c = 6
(3) -3000m - c = -4

(4) 2000m + 0c = 2

m = 1/1000 = 1x10^-3

sub m into (1)

4 = 3000(1x10^-3) + c
c = 4 - 3 = 1

so y = (1x10^-3)x + 1

that make sense? Seems too little for 7 marks though I would have thought...
3. Re: Maths help needed for questions below.
Because it's a duplicate of a post you've already made (which has 19 replies!) and it's in the wrong forum.
4. Re: Maths help needed for questions below.
(Original post by gabriel 41)
Hi guys,
Currently, I'm stuck in my business quantitative methods course. I was doing a specimen question for the QCF ABE paper and I've got stuck, so I need you guys to assist me. The questions are dealing with probability and straight line graphs.

The question for probability is "The workforce of a manufacturing company consists of 200 employees, 80 of whom are female. Of these 80 female employees, 60 have a business qualification. Only 34 of the company’s total employees do not have a business qualification."

Calculate the probability that an employee picked at random:
(a) Is male (3 marks)
(b) Has a business qualification (3 marks)
(c) Has a business qualification and is male (4 marks)
(d) Is male or has a business qualification (4 marks)

So for (a) it's pretty simple I just did 1-(80/200) to get the probability of males which is 0.6. Is this right?
So for (b) the approach I took is 1-(34/200) to get 83/100 as the probability of those who have a business qualification. Is this right?
Now the part of (c) was really confusing. I thought of just multiplying the above two probabilities but then saw it's a high 4 mark questions, so decided to take a "harder" approach:
I found that 14 males don't have a business qualification and there are total of 120 males in the organization; hence, 106 males have a business qualification, so the probability of a person chosen at random and is a business qualified male is 106/200 which is simplified to 0.53 as the final answer, so is this right? Because when you multiply the probability of part (a) and part (b) to reach the answer we get 0.498 as the answer which differs from what I did, so which is the right approach? Last question I was doomed to fail because when I added the probability of part (a) and part (b) I got more than 1! Like I did 0.6+0.83 and got 1.43! So I'm pretty sure it's wrong, so how do I go about for part (d)?

Finally, I'm totally lost in the following question and can't make a decent start even! Please help me out:

A linear equation, in the form y = mx + c, where m and c are constants, shows the price y
per unit of a product which a company would be willing to supply any given quantity x. At a
price of £4.00 per unit, the company would be willing to supply 3,000 units of the product.
At the higher price of £6.00 per unit, the company would be willing to supply 5,000 units.
Find the values of m and c, to identify the linear equation in the form y = mx + c. (7 marks).

Thanks
1. a) and b) are right.
c) is simple you're right that onyl 14 males don't have a buisiness degree so the probability is (total-female-male w/o)/total. That make 106/200 or 0.53.
d) is simple it's (total-femal w/o)/total which is 0.9.
2. m= difference in y/difference in x
m=2/2000
=1/1000
so you know that y=4.x=3000, m=1/1000
4=3000/1000+c
1=c
y=x/1000+1000
5. Re: Maths help needed for questions below.
6. Re: Maths help needed for questions below.
(Original post by notnek)
ya but this is a duplicate post because it's always better to have 2 posts where you can confirm the type of answers.
7. Re: Maths help needed for questions below.
(Original post by gabriel 41)
ya but this is a duplicate post because it's always better to have 2 posts where you can confirm the type of answers.
They've been merged. No point posting a thread asking for study help in a forum which is meant to be used for university courses.
8. Re: Maths homework help needed quickly!
(Original post by notnek)
(a) Correct.
(b) Correct.

(c) You can't multiply the probabilities together because an employee having a qualification is not independent of the person's sex, i.e. the probability of having a qualification is different for males and females. Your answer of 0.53 is correct.

(d) Here you can't add the probabilities since the two events are not mutually exclusive (there are people who are both male and have a business qualification). For this you could work out the probability of someone being female who doesn't have a qualification (p) then the probability of someone being male or having a qualification is 1 minus p.

Alternatively, you could add up the number of males and the number of females with a qualification. This is equal to "The number of people who are male or have a business qualification". If this confuses you then I recommend that you draw a Venn diagram to illustrate this info. Actually, you may find that all of these questions become much easier once you have drawn a correct Venn diagram.

Q2)

For this you need to use the information to substitute values into the linear equation y=mx+c. Firstly, when y=4 (it doesn't specify whether y is in pounds or pence so I'll assume it's pounds), x=3000 so you can substitute this in to get:

4=3000m+c

Now use the second peice of info. to generate a second equation in terms of m and c. Then you can solve these two silultaneous equations to find m and c.

Have a go at that and post your working if you get stuck.
With reference to my earlier question, which was, A linear equation, in the form y = mx + c, where m and c are constants, shows the price y
per unit of a product which a company would be willing to supply any given quantity x. At a
price of £4.00 per unit, the company would be willing to supply 3,000 units of the product.
At the higher price of £6.00 per unit, the company would be willing to supply 5,000 units.
Find the values of m and c, to identify the linear equation in the form y = mx + c. (7 marks).

For this question, is there a graphical or other alternative ways for solving it rather than the routine simultaneous way that I was initially instructed on? (please write in full forms, in your reply)
9. Re: Maths homework help needed quickly!
(Original post by gabriel 41)
With reference to my earlier question, which was, A linear equation, in the form y = mx + c, where m and c are constants, shows the price y
per unit of a product which a company would be willing to supply any given quantity x. At a
price of £4.00 per unit, the company would be willing to supply 3,000 units of the product.
At the higher price of £6.00 per unit, the company would be willing to supply 5,000 units.
Find the values of m and c, to identify the linear equation in the form y = mx + c. (7 marks).

For this question, is there a graphical or other alternative ways for solving it rather than the routine simultaneous way that I was initially instructed on? (please write in full forms, in your reply)
Well, there are two ways that I know of aside from solving simultaneous equations. You can solve for "m" and "c" independently by knowing what these variables mean.

Recall that in the equation y = mx + c, "m" (the coefficient of x) is the slope of the line and c (the constant) is the y-intercept.

Graphically, you can plot the two points A (3000, 4.0) and B (5000, 6.0).
Drawing a line through these two points gives you the function, and so allows you to spot the value of "m" (slope of the line) and the value of "c" (y-intercept of the line).

Another way is to use the values of the coordinates to fine "m" and "c".
The slope is "rise" over "run", or ∆y/∆x. The "rise", or change in price, is 2.0, and "run", or change in quantity supplied, is 2000. Thus, the slope "m" is 1/1000.
After this, the y-intercept can be solved algebraically, not to differently from using simultaneous equations. In the equation y = mx + c, you know "m". You also know that this equation works for all points on the line, say for example A (3000, 4.0). 4 = (1/1000)(3000) + c, thus c = 1.
A way to graphically interpret solving the y-intercept using the slope is that you're decreasing the quantity supplied until it reaches 0, and changing the price proportionally (as dictated by the slope).
10. Re: Maths homework help needed quickly!
(Original post by aznkid66)
Well, there are two ways that I know of aside from solving simultaneous equations. You can solve for "m" and "c" independently by knowing what these variables mean.

Recall that in the equation y = mx + c, "m" (the coefficient of x) is the slope of the line and c (the constant) is the y-intercept.

Graphically, you can plot the two points A (3000, 4.0) and B (5000, 6.0).
Drawing a line through these two points gives you the function, and so allows you to spot the value of "m" (slope of the line) and the value of "c" (y-intercept of the line).

Another way is to use the values of the coordinates to fine "m" and "c".
The slope is "rise" over "run", or ∆y/∆x. The "rise", or change in price, is 2.0, and "run", or change in quantity supplied, is 2000. Thus, the slope "m" is 1/1000.
After this, the y-intercept can be solved algebraically, not to differently from using simultaneous equations. In the equation y = mx + c, you know "m". You also know that this equation works for all points on the line, say for example A (3000, 4.0). 4 = (1/1000)(3000) + c, thus c = 1.
A way to graphically interpret solving the y-intercept using the slope is that you're decreasing the quantity supplied until it reaches 0, and changing the price proportionally (as dictated by the slope).

Thank you for your help; however, if you have time, can you please do the question for me graphically and attach the graphical solution in your post/reply. This is because when I'm doing it graphically I don't seem to get the answer. The scale may be wrong on my axes and I'm getting the gradient as 1000! So please can you do it for me?
11. Re: Maths help needed for questions below.

Give the name of two measures of location and two measures of dispersion that could be useful in describing data. (4 marks)

What is a measure of location? I've heard of measures of dispersion like interquartile range but have never heard a measure of location
12. Re: Maths homework help needed quickly!
(Original post by gabriel 41)
Thank you for your help; however, if you have time, can you please do the question for me graphically and attach the graphical solution in your post/reply. This is because when I'm doing it graphically I don't seem to get the answer. The scale may be wrong on my axes and I'm getting the gradient as 1000! So please can you do it for me?
As you can see, I kinda omitted the part about finding the slope with a graph. You need to understand that the slope is ∆y/∆x, which really isn't a graphical interpretation at all. If you're getting the reciprocal, you might be doing ∆x/∆y. Think of slope as how much 'y' changes after a unit change (of 1) in 'x'.

For example, if you have the equation y=7x, then you can pick any two 'x' values with a difference of 1 (e.g., 0 to 1, 2 to 3, 3 to 4) and observe that the difference in 'y' (e.g., 0 to 7, 14 to 21, 21 to 28). is constant (7).

(Original post by gabriel 41)

Give the name of two measures of location and two measures of dispersion that could be useful in describing data. (4 marks)

What is a measure of location? I've heard of measures of dispersion like interquartile range but have never heard a measure of location

Well, measure of dispersion means how "spread out" the data is. I know two terms that, by their definitions, measure spread (but then again, I may be using "spread" wrongly).
Edit: Interquartile range is one, too.

However, even after finding how "spread out" a set of data is, you still don't know what point(s) the set is spread out from. For example, if you graphed the data, moving all the points up/down and left/right will change the measure of location, but will not change the measure of dispersion. I hope you can, with these hints, deduce what the measure(s) of location are (hint: 3 Ms).
Last edited by aznkid66; 18-06-2012 at 10:13.
13. Re: Maths help needed for questions below.
And one more question to solve / confirm. Here's the question:

Q6 The revenue R (£’000) of a manufacturing process is given by the equation R = 300x – 4x^2
where x is the quantity of manufactured products sold in ’00 units. The production cost
C (£’000) of the same manufacturing process is given by the equation C = 2400 + 40x,
where x is the quantity of manufactured products produced in ’00 units.
(a) Construct a table to calculate the value of C and R, using the values 0, 5, 10, 15, 20,
25, 30 and 35 for x. (4 marks)
(b) Use the values of C and R calculated in (a) to plot a fully labelled graph for C and R
against x. (6 marks)
(c) Use the graph drawn in (b) to estimate the quantity of manufactured products where
the manufacturing process breaks even (i.e. where neither a profit nor loss is made). (2 marks)

So, as you can see in the attached picture (in png format), the answers for (a) I think are right. Please confirm

And another thing - part (b) drawn and attached in the jpg format; however, although I've drawn the graph very accurately, and, I'm surprised, that I only made use of 1 or 2 points on the revenue line from the above table but yet I managed to draw the line but then part (c), my answer being 1700 units, is wrong yet the graph of part (b) shows that's the answer and the graph is drawn as accurately as possible. So what is the right answer? Can anyone help me?
Attached Thumbnails

Last edited by gabriel 41; 18-06-2012 at 19:11.
14. Re: Maths homework help needed quickly!
(Original post by aznkid66)
As you can see, I kinda omitted the part about finding the slope with a graph. You need to understand that the slope is ∆y/∆x, which really isn't a graphical interpretation at all. If you're getting the reciprocal, you might be doing ∆x/∆y. Think of slope as how much 'y' changes after a unit change (of 1) in 'x'.

For example, if you have the equation y=7x, then you can pick any two 'x' values with a difference of 1 (e.g., 0 to 1, 2 to 3, 3 to 4) and observe that the difference in 'y' (e.g., 0 to 7, 14 to 21, 21 to 28). is constant (7).

And one more question to solve / confirm. Here's the question:

Q6 The revenue R (£’000) of a manufacturing process is given by the equation R = 300x – 4x^2
where x is the quantity of manufactured products sold in ’00 units. The production cost
C (£’000) of the same manufacturing process is given by the equation C = 2400 + 40x,
where x is the quantity of manufactured products produced in ’00 units.
(a) Construct a table to calculate the value of C and R, using the values 0, 5, 10, 15, 20,
25, 30 and 35 for x. (4 marks)
(b) Use the values of C and R calculated in (a) to plot a fully labelled graph for C and R
against x. (6 marks)
(c) Use the graph drawn in (b) to estimate the quantity of manufactured products where
the manufacturing process breaks even (i.e. where neither a profit nor loss is made). (2 marks)

So, as you can see in the attached picture (in png format), the answers for (a) I think are right. Please confirm

And another thing - part (b) drawn and attached in the jpg format; however, although I've drawn the graph very accurately, and, I'm surprised, that I only made use of 1 or 2 points on the revenue line from the above table but yet I managed to draw the line but then part (c), my answer being 1700 units, is wrong yet the graph of part (b) shows that's the answer and the graph is drawn as accurately as possible. So what is the right answer? Can anyone help me?

Well, measure of dispersion means how "spread out" the data is. I know two terms that, by their definitions, measure spread (but then again, I may be using "spread" wrongly).
Edit: Interquartile range is one, too.

However, even after finding how "spread out" a set of data is, you still don't know what point(s) the set is spread out from. For example, if you graphed the data, moving all the points up/down and left/right will change the measure of location, but will not change the measure of dispersion. I hope you can, with these hints, deduce what the measure(s) of location are (hint: 3 Ms).
And one more question to solve / confirm. Here's the question:

Q6 The revenue R (£’000) of a manufacturing process is given by the equation R = 300x – 4x 2,
where x is the quantity of manufactured products sold in ’00 units. The production cost
C (£’000) of the same manufacturing process is given by the equation C = 2400 + 40x,
where x is the quantity of manufactured products produced in ’00 units.
(a) Construct a table to calculate the value of C and R, using the values 0, 5, 10, 15, 20,
25, 30 and 35 for x. (4 marks)
(b) Use the values of C and R calculated in (a) to plot a fully labelled graph for C and R
against x. (6 marks)
(c) Use the graph drawn in (b) to estimate the quantity of manufactured products where
the manufacturing process breaks even (i.e. where neither a profit nor loss is made). (2 marks)

So, as you can see in the attached picture (in png format), the answers for (a) I think are right. Please confirm

And another thing - part (b) drawn and attached in the jpg format; however, although I've drawn the graph very accurately, and, I'm surprised, that I only made use of 1 or 2 points on the revenue line from the above table but yet I managed to draw the line but then part (c), my answer being 1700 units, is wrong yet the graph of part (b) shows that's the answer and the graph is drawn as accurately as possible. So what is the right answer? Can anyone help me?
Attached Thumbnails

Last edited by gabriel 41; 18-06-2012 at 19:11.
15. Re: Maths help needed for questions below.
Um, could you re-type the equation for R? "R = 300x - 4x 2" doesn't make much sense ^^; I really don't know what the values should be since I haven't seen the equation. I'll just assume they're correct, and it comes out in the shape of the curve (non-linear is weird, maybe you want to check that) you plotted.

The values for C seem to work with the equation, but not as much with the line you plotted. It's the one that intercepts the y-axis at just above 2000, right? Shouldn't the intercept be just under 2500?

About finding the break-even point, wouldn't that be in intersection R curve and the C line? However, you drew a line connecting the first and last point of R instead of a smooth curve. The intersection of the R curve and the C line should be somewhere behind 10 (according to the diagram you drew). I don't think the intersection between the line you drew for R and the C line has anything to do with the real break-even price, unfortunately...
16. Re: Maths help needed for questions below.
(Original post by aznkid66)
Um, could you re-type the equation for R? "R = 300x - 4x 2" doesn't make much sense ^^; I really don't know what the values should be since I haven't seen the equation. I'll just assume they're correct, and it comes out in the shape of the curve (non-linear is weird, maybe you want to check that) you plotted.

The values for C seem to work with the equation, but not as much with the line you plotted. It's the one that intercepts the y-axis at just above 2000, right? Shouldn't the intercept be just under 2500?

About finding the break-even point, wouldn't that be in intersection R curve and the C line? However, you drew a line connecting the first and last point of R instead of a smooth curve. The intersection of the R curve and the C line should be somewhere behind 10 (according to the diagram you drew). I don't think the intersection between the line you drew for R and the C line has anything to do with the real break-even price, unfortunately...

Sorry! I've corrected it now. Now does anything change? Please explain yourself clearly and tell me exactly where I went wrong
17. Re: Maths help needed for questions below.
Well, where you went wrong was assuming that the curve R(x) = 300x - 4x^2 could be approximated by the line that intersects R(0) and R(35). The break even point should be where the actual R intersects with C, not where an approximation of R intersects with C.

Also, something that I believe is wrong but may not be is that your C seems a bit low. It seems to start at y=2100 instead of y=2400, even though the slope seems fine.
18. Re: Maths help needed for questions below.
(Original post by aznkid66)
Well, where you went wrong was assuming that the curve R(x) = 300x - 4x^2 could be approximated by the line that intersects R(0) and R(35). The break even point should be where the actual R intersects with C, not where an approximation of R intersects with C.

Also, something that I believe is wrong but may not be is that your C seems a bit low. It seems to start at y=2100 instead of y=2400, even though the slope seems fine.
Yes even I wanted to use all points of R but if you look at the points they aren't in one line! They are scattered around and it's impossible to draw one straight line through all the points, so do you mean to say that R is a quadratic U shaped graph ( a parabola)?
19. Re: Maths help needed for questions below.
Yup, just draw a smooth curve. It doesn't have to be exact, as it's only used to help estimate the intersection between R and C.
20. Re: Maths help needed for questions below.
(Original post by aznkid66)
Well, where you went wrong was assuming that the curve R(x) = 300x - 4x^2 could be approximated by the line that intersects R(0) and R(35). The break even point should be where the actual R intersects with C, not where an approximation of R intersects with C.

Also, something that I believe is wrong but may not be is that your C seems a bit low. It seems to start at y=2100 instead of y=2400, even though the slope seems fine.
In fact, the line starts at 2400, you're right but I just extended the line so it seems that it's starting at y=2100, so shouldn't do this or it's okay?