First align the decimal points and the numbers, then add the extra 0's if needed. Lastly, add and the total answer is 14.225.
so I need to do a list of numbers like this:
98, 100, 102, 104, 106, 108, 110.
104 is the median number
add 98+100+102+104+106+108+110=728
divide by seven 728 / 7 = 104
so the difference between the median 104 and the avarage 104 is 0 they are
104=104
hope this helps
If the acceptable percent error is 2.5%, then the amount it can be over or under 16 oz is 0.4oz.
16 + 0.4
16 - 0.4
16.4 is the greatest, 15.6 is the least
Answer:
Given:
length of the wire = 0.20 meters
magnetic field strength = 0.45 newtons/amperes meter
speed = 10.0 meters per second
emf = B * l * v
B = flux density ; l = length of the wire ; v = velocity of the conductor
emf = 0.45 newtons / ampere meter * 0.20 meters * 10.0 meters/seconds
emf = 0.90 volts
The emf produced is 0.90 volts.
Step-by-step explanation:
C(x) = 200 - 7x + 0.345x^2
Domain is the set of x-values (i.e. units produced) that are feasible. This is all the positive integer values + 0, in case that you only consider that can produce whole units.
Range is the set of possible results for c(x), i.e. possible costs.
You can derive this from the fact that c(x) is a parabole and you can draw it, for which you can find the vertex of the parabola, the roots, the y-intercept, the shape (it open upwards given that the cofficient of x^2 is positive). Also limit the costs to be positive.
You can substitute some values for x to help you, for example:
x y
0 200
1 200 -7 +0.345 = 193.345
2 200 - 14 + .345 (4) = 187.38
3 200 - 21 + .345(9) = 182.105
4 200 - 28 + .345(16) = 177.52
5 200 - 35 + 0.345(25) = 173.625
6 200 - 42 + 0.345(36) = 170.42
10 200 - 70 + 0.345(100) =164.5
11 200 - 77 + 0.345(121) = 164.745
The functions does not have real roots, then the costs never decrease to 0.
The function starts at c(x) = 200, decreases until the vertex, (x =10, c=164.5) and starts to increase.
Then the range goes to 164.5 to infinity, limited to the solutcion for x = positive integers.
