-1.1576e-005 = -1.1576 x 10^005 [i think 005 mean -5] correct

so it will be -1.1576 x 10^-5 => 0.00011576 (i think it is a big value)

Two things wrong here. The first minus means the number is negative. And the decimal point is misplaced.

The way I get the decimal point right is by moving it 5 places left.

and

4e-008 = 4 x 10^008 [i think 008 mean -8]correct

so it will be -4 x 10^-8 => -0.000000004 (i think it is small value)

again two errors I afraid.

A leading minus makes the number negative. This time there is no leading minus so the number is positive.

The -8 on the power of ten means move the point 8 places left.

If you go back to that link again you will find a 'Try It Yourself' section a little way down the page. You can test whether you have done the conversion correctly by entering numbers in the boxes.

I agree with bobbym about whether a number is large or small. If bacteria could talk they would say that the thickness of a human hair is large. If you wanted to say which of the two numbers you gave was smaller then you still need to think a little as one number is negative. If, by smaller, you mean further to the left on the number line, then -0.000011576 is smaller. But, if you ignore negative signs and just consider which has the smaller absolute value then 0.00000004 is smaller.

Bob

]]>needs general help with Math

You might be right.

]]>It was a long time ago and I've forgotten all about this problem. So I have re-worked it from scratch. I still think that post 2 is correct.

And that still gives

The original link in post 1 isn't working for me. Please provide a link to this problem so I can see it in its original form and try my answer.

Bob

]]>1) That is what I got.

2) You are supposed to get 6.2, please recheck your algebra.

These questions have already been asked and answered over here:

http://www.mathisfunforum.com/viewtopic.php?id=22493

Please use that thread.

]]>Welcome to the forum.

The idea for solving this sort of problem is simple enough, but the times in this problem make the arithmetic especially complicated. Yuck! I leave that bit to you and your brother.

Start with a distance time graph to show A and B's journeys. I took me 4 goes before I had a decent graph. This is NOT to scale. It just represents what is going on.

The two towns are shown on the up axis. Times are shown across. I have converted all times into minutes from 9.00am.

The two horizontal lines across the middle represent the near and far side of the river bridge.

If I call the distance between the towns, D, and the speeds V_subscript_a and V_subscript_b, then using distance = speed x time

From this you can get the ratio of the speeds.

Now suppose the bridge has length L and t_sub_a, and t_sub-b are the times to cross:

We also know that

Using these equations you can calculate both times to cross in minutes.

Now call the time at which both arrive at their side of the bridge, t.

For A, calculate the distance he has travelled at this time in terms of his speed and t.

For B, calculate the same distance by finding the time he leaves the bridge (t + t_sub_b), computing his remaining journey time and using his speed.

Equate these distances, substitute in the ratio of the speeds and you are left with an equation in t. Solve for the answer to the question.

Best of luck

Bob

ps. Made a silly mistake with my arithmetic. It's not as bad as I thought and t comes out to be a 'nice' time.

]]>What login?

]]>Welcome to the forum.

I would have used 2 standard deviations too, but, I thought I'd better look up Chebyshev's theorem.

https://en.wikipedia.org/wiki/Chebyshev%27s_inequality

Wiki wrote:

In practical usage, in contrast to the 68-95-99.7% rule, which applies to normal distributions, under Chebyshev's inequality a minimum of just 75% of values must lie within two standard deviations of the mean and 89% within three standard deviations.

question 2: "Determine whether the table describes a probability distribution. " Have you been given a method for doing this ?

LATER EDIT: I had assumed the question wanted you to identify the distribution (eg. binomial) and I couldn't see how you could do that with any certainty. If you don't find a 'fit' that doesn't mean that one doesn't exist and if you find a distribution that is 'close' that doesn't mean it is right.

But, re-reading the question, I wonder if this is just about the fact that the probabilities don't add up to 1. Are you just expected to say it cannot be a probability distribution because of this. It seems a bit simple but maybe that is all this question is about.

Bob

]]>Thanks again!

]]>There's a theorem in geometry that covers all three questions here.

I'll start with the middle question. V is the point where WY and XZ cross.

In triangles XWV and YZV, angle XWY = XZY (using angle props. of a circle) and WVX = ZVY (= 90**) so the triangles are similar. Thus:

In the third question the crossing point is outside the circle, but the theorem still holds.

In triangles TXA and TBY, angle T is common to both, angle XAB = XYB. So the triangles are similar and

Question 1. If you rename point T as point P, then PB.PA = PY.PX

Let X approach Y. Then the equation becomes, in the limit, PB.PA = PY^2

Bob

**By vertically opposite angles this would still be true even if the angle wasn't 90. But you need 90 in order to use Pythagoras theorem to complete the problem.

]]>Try starting with

and differentiate it.

Bob

]]>