I was taught that the conventional way, when the last integer is an exact 5, is to round even down and round odd up.
Examples: 100.1245 the four is even so this rounds down to 100.124
100.1235 the three is odd so this rounds up to 100.124
The premise is that there there are an equal number of odd digits as even digits, so the rounding balances out, over the long run. I see that my method conflicts with CPhill. You might check with your teacher and see how s/he wants you to do it.
the diagram shows a sector OPQ of a circle with centre O. the radius of the circle is 18m and the angle POQ is 2π / 3 radians.
a) find the length of the arc PQ, giving the answer as a multiple of π
b) the tangents to the cirlce at the point P and Q meet at the point T. and the angles TPO AND TQO are both right angles
i) Angle PTQ = ∝ radians. find ∝ in terms of pi
ii)Find the area of the shaded region bounded by the arc PQ and the tangents TP and TQ
a) arc length PQ = 18 * (2 pi)/3 = 36 pi / 3 = 12 pi
b) i) Angle PTQ will be supplemental to angle POQ = pi - (2pi)/3 = pi/3 = 60°
ii ) The area of the circular sector formed by arc PQ = (1/2)r^2 (2pi/3) = (1/2)(18)^2 (2pi)/3 = 108 pi m^2 (1)
[PTQO] will from a kite ...OQ = 18 and triangle TPO will form a 30-60-90 right triangle with QO = 18 opposite the 30° angle (angle QTO) and QT opposite the 60° angle (angle QOT)....so QT = 18√3
So the area of this kite = product of the legs of triangle OQT = QO * QT = 18 * 18√3 = 324√3 (2)
So...the area bounded by arc PQ and the tangents TP and TQ = (2) - (1) = 324√3 - 108 pi ≈ 221.9 m^2
A toddler is stacking wooden blocks. He has 4 blocks that are all different colors (red, blue, green, and yellow). How many different towers of blocks can he make?
4 ways to choose the 1st block * 3 ways to choose the 2nd * 2 ways to choose the 3rd * 1 way to choose the last = 4 * 3 * 2 * 1 = 24
A committee of 3 people is to be selected from a group of 10 people. The order in which people are selected doesn’t matter. How many different committees can be formed?
We have 10 people and we wish to choose any 3 of them....so...10C3 = 120 different committees
An ice cream shop has 20 flavors. If Louis wants a double scoop cone, how many different flavor combinations can he get?
We can choose 2 flavors from 20 and order these in 2 ways...so
20C2 * 2 = 190 * 2 = 380 combos