Math Is Fun Forum

Hello!

2. a. It is not, because they ask for one equation in 2c. Bit of a giveaway (:

2. b. $2 per hour for 3.25 hours. See 2c.

2. c. The cost in dollars is double the time in hours.

For question 1, I personally hate eyeballing graphs and having to say "well it looks to be around about such and such sort of maybe". Points are much more convenient. But one way of solving it is by using the formulas y = mx + b and m = (y2-y1)/(x2-x1) and the endpoints of the lines.
For example, the last line looks to me to extend from (2,3) to (7,-2). The slope (m) is (-2-3)/(7-2) = -1. That is the change in y for each increase of +1 of x. To find the y-intercept (b) use this relation to find y when x=0. The last condition is y = -x + 5 after x=2.

7a. An erf expression that is about 62.3918845810973043106238234927825020672554920384931% I would expect a bit of error in a typical answer because you would not normally find the exact conversion for 6/19 of a s.d.

Similarly for 8a. because you would not normally convert exactly -59/511 of a s.d. The answer is about 54.595969158773076487246800747118403970727804830356%

Hello!

Why is that?

Hi,
I do not know where you read that, but it sounds like a simplification for the sake of deriving some result. The chance of a trillion heads in a row has a very real probability. It Is 1/(2^10^12). This is such a ridiculously small number that I don't have access to an engine that can evaluate it. But it is not zero.

The way probability works is if the probability of an event is 10^-50, it will take an average of 10^50 events before it occurs

Hi Fruityloop,

If you believe in this formula (which is definitely the best we have), you will have to explain some discrepancies between its results and those mentioned at the shared birthdays page.

If I sum all of the probabilities from 1 to 11 (for the chance of at least one match), I get 49.4592%, which is close enough to the expected 50%.
If I sum all of the probabilities from 1 to 28 for 57 people, I get 80.9279%, which is not even close to the expected 99%.
If I sum all of the probabilities from 1 to 15 for 30 people, I get 67.7786%, which differs somewhat from the expected 70.6%.

Also, If I extend to include 0 pairs, I get the answers I would expect to from that page.
I think the formula is correct, I am just leaving something out in the sums

Length is 44/3 and width is 28/3.

If the width is supposed to be THREE more than half the length you get 14 and 10. Besides that I can't see a simple adjustment to make integers

Hi,

This last interpretation is not quite correct either. Although it gets very close to the right answer for 23 people, if I extend to 57 people I get probabilities that add up to over 200%.

It is very easy to do simulations of this kind in Excel. I write 1 in cell A1 and =(FACT(23)*COMBIN(365,23-A1)*COMBIN(23-A1,A1))/(365^23*2*A1) in cell B1. Replace 23 with any number of people desired. But this last expression must be corrected

Hi bobbym,

I am trying to replace two with one to confirm the results given in the shared birthday link, so that we may now have a way to find the probability of 3 or 4 or 20 matches for more people, but am not getting any correspondence at all. It should be a very simple adjustment, but it is proving quite annoying lol

Hi,

I agree with you about the perimeter, but the book is correct about the angle. There is a rule that the area is equal to half the product of the diagonals (2400mm^2 = 24cm^2 in this case). There is another rule that the area is equal to the side length squared times the sine of either angle.

Which is close to 73.74 degrees.

To your second post, it is part of the definition of a rhombus that the side lengths are equal. Regarding the angles, unless the rhombus is a square (where all angles are equal), there will be two different angles because the opposite angles are corresponding. One pair of angles will be acute, the other obtuse, and since they all sum to 360 each type of angle must sum to 180

For problem three, I think the algebra is extremely tedious to do the old-fashioned way. I have the computed answer with no working if that is all you wanted (I'm not sure it is).

Hi, I would just use the cosine rule for the first two.

For problem one, say c=14 and b=16. For problem two, say c=16 and b=14.

Bob is right that there are two solutions for problem one. For problem two there are two solutions algebraically, but because one of them is negative there is actually only one answer.