mn = 80
m -n = -79
Substitute
m - 80/m = -79
m^2 - 80 = -79m
m^2 + 79m = 80 (NB Notice change of signs)
Quadratic Eq'n
m = { - 79 +/- sqrt[)79)^2 - 4(1)(-80)}] / 2(1)
m = { - 79 +/- sqrt[6241 + 320}]/ 2
m = { -79 +/- sqrt[6561]} / 2
m = { - 79 +/- 81}/2
m = -160 / 2 = -80 or
m = 2/2 = 1
Hence n = 1 + 79 = 80
80
x = 79 + 1 is one of the infinite number of such equations.
80
1
median is 79. Write in order; median is middle number: {72, 89, 67, 81, 75, 79, 81, 80, 62, 64, 83} → {62, 64, 67, 72, 75, 79, 80, 81, 81, 83, 89} To find middle number, count the number of data items, add 1 and divide by 2. If this is a whole number, that is the data item which is the median; if it is a fraction, take the mean average of the data items of the positions either side of the fraction (eg if the fraction is 3.5, take the mean average of the 3rd and 4th data items) There are 11 data items → median is (11+1)/2 = 6th data item → median is 79
80
80 is equal to 80 because it is the same number....... :)ex. (79=79) (100=100)
61, 67, 71, 73, 79, 83, 89, 97, so 79 is nearest
83 is the first prime number after 79.
It is: 79
80
There are an infinite number of sets with mean 80. Here are some: {80, 80, 80}, {80, 80, 80, 80, 80, 80} {79, 80, 81}, {79, 79, 80, 81, 81}, {79, 79, 80, 82} (1, 80, 159}, {-40, 200} To produce a set of n numbers with mean 80, start with any set of n-1 numbers. Suppose their sum is S. Then add the number 80*n-S to the set. You will now have n numbers whose sum is S+80*n-S = 80*n So the mean of this set is 80.
If three of the four numbers are 79, 80, and 65, and the average is 85, then the final number will equal (85 x 4) - 79 - 80 - 65 = 116.
The number of neutrons is 45 because you subtract 80 to 35.
It is 80 and not 79
It can be rounded to 80, 79 or 79.3
71, 73, 79