If this is on mymaths.co.uk then the answer to this question is: Integration.
That is how to find the area under the curve.
"integration"
By differentiating the answer and plugging in the x value along the curve, you are finding the exact slope of the curve at that point. In effect, this would be the slope of the tangent line, as a tangent line only intersects another at one point. To find the equation of a tangent line to a curve, use the point slope form (y-y1)=m(x-x1), m being the slope. Use the differential to find the slope and use the point on the curve to plug in for (x1, y1).
The answer is 32 you find the answer by simply multiplying 8*4. Hope this helps your problem!
The main utility of a cumulative frequency curve is to show the distribution of the data points and its skew. It can be used to find the median, the upper and lower quartiles, and the range of the data.
How do you find the area of an unequal quadrangle?
Integration.
"integration"
if its the mymaths one type integration into the box!
integral.
37.6
if this is for www.mymaths.co.uk, integration is all you need. best regards, lokilotus
In statistics you can find the area under a curve to establish what to expect between two input numbers. If there is a lot of area under the curve the graph is tall and there is a higher probability of things occurring there than when the graph is low.
well, it would help to clarify your question, but integrate the fuction and solve on the interval given
(3ab*pi)
The area under N(0,1) from -1 to 1 = 0.6826
Look in any standard normal distribution table; one is given in the related link. Find the area for 2.43 and 1.52; then take the area for 2.43 and subtract the area for 1.52 and that will be the answer. Therefore, .9925 - .9357 = .0568 = area under the normal distribution curve between z equals 1.52 and z equals 2.43.
One word answer: integrate. The area under the acceleration curve, up to time T, is the speed at time T. If you now make a curve of the speed as a function of time, and find the area under that up to time T, that will be the position at time T.