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# does the zero mean anything in a set of numbers when calculating mean?

more info well heres the problem 0,10,10,20 and now i have to find the mean well i add it up and now i have 40 but do i divide by 4 or 5? because of the extra zero and is it just really a zero and doesnt count?

## Research, Knowledge and Information :

### How to Find the Mean - Math is Fun - Maths Resources

How to Find the Mean. ... Negative Numbers. How do you handle negative numbers? Adding a negative number is the same as subtracting the number (without the negative).

### How to Calculate the Mean: 4 Steps (with Pictures) - wikiHow

How to Calculate the Mean. ... the 0 in the amount of numbers at the end when you divide your big ... different results than the mean from the same set of numbers.

### probability - Why are mean 0 and standard deviation 1 ...

Why are mean 0 and standard deviation 1 distributions ... number of standard deviations that score is from the mean. The equation is calculating the (score - mean) ...

### If zero is in your number set do you count it when figuring ...

If zero is in your number set do you count it when figuring the mean? ... is the number zero a member of the set of natural numbers? ie does 0 ∈ ℕ The answer is: ...

### How To Analyze Data Using the Average – BetterExplained

You take a set of numbers, ... 1/3 — 1 over anything, really.) The harmonic mean helps us calculate ... How To Analyze Data Using the Average @ BetterExplainedThe ...

### Calculating the Mean and Standard Deviation with Excel ...

Finding the Mean Enter the scores in one of the columns on the ... Calculating the Mean and Standard Deviation with ... If a data set had more than one ...

### Mean Absolute Deviation - MathBitsNotebook(A1 - CCSS Math)

... (average) of all deviations in a set equals zero ... of any value on a number line from zero. ... The mean absolute deviation is the "average" of the ...

Mean absolute deviation is a way to describe variation in a data set. Mean absolute ... Well, how do I ... "Is there a number that can ... is zero, one , two, three ...

### Learn How to Calculate the Mean or Average (Your GPA)

Here's how to calculate the mean or average of a set of numbers. ... Here's how to calculate the mean or average of a set of ... Worked Examples of Calculating the Mean.

## Suggested Questions And Answer :

### does the zero mean anything in a set of numbers when calculating mean?

0,10,10, 20 is 4 numbers tu get the averaej...suma 4 numbers & divide bi 4 =40/4=10 (krats oft kall this the "arithmatik meen")

### show that the function f(x)= sqrt (x^2 +1) satisfies the 2 hypotheses of the Mean Value Theorem

f(x) = sqrt(x^2 + 1) ; [(0, sqrt(8)] Okay, so for the Mean Value Theorem, two things have to be true: f(x) has to be continuous on the interval [0, sqrt(8)] and f(x) has to be differentiable on the interval (0, sqrt(8)). First find where sqrt(x^2 + 1) is continous on. We know that for square roots, the number has to be greater than or equal to zero (definitely no negative numbers). So set the inside greater than or equal to zero and solve for x. You'll get an imaginary number because when you move 1 to the other side, it'll be negative. So, this means that the number inside the square root will always be positive, which makes sense because the x is squared and you're adding 1 to it, not subtracting. There would be no way to get a negative number under the square root in this situation. Therefore, since f(x) is continuos everywhere, (-infinity, infinity), then f(x) is continuous on [0, sqrt(8)]. Now you have to check if it is differentiable on that interval. To check this, you basically do the same but with the derivative of the function. f'(x) = (1/2)(x^2+1)^(-1/2)x2x which equals to f'(x) = x/sqrt(x^2+1). So for the derivative of f, you have a square root on the bottom, but notice that the denominator is exactly the same as the original function. Since we can't have the denominator equal to zero, we set the denominator equal to zero and solve to find the value of x that will make it equal to zero. However, just like in the first one, it will never reach zero because of the x^2 and +1. Now you know that f'(x) is continous everywhere so f(x) is differentiable everywhere. Therefore, since f(x) is differentiable everywhere (-infinity, infinity), then it is differentiable on (0, sqrt(8)). So the function satisfies the two hypotheses of the Mean Value Theorem. You definitely wouldn't have to write this long for a test or homework; its probably one or two lines of explanation at most. But I hope this is understandable enough to apply to other similar questions!

### Difference between mean, average and median?

First, it is usually a good idea to arrange your data in numerical order from lowest to highest.  We can arrange this data set as follows: 2,4,8,9,12,14,14,16,17,19 Now that the data is properly arranged, we can easily calculate the median. The median is the middle number in the data set. For this set, since there are 10 numbers there is no clear middle. In this case we take the two middle values, add them together and divide by 2 12+14 = 26 20/2 = 13 The median value for this data set is 10. mean and average are interchangeable terms. They are calculated by taking the sum of all data points and dividing by the number of data points.  For this data set we have: 2+4+8+9+12+14+14+16+17+19 = 115 115/10 = 11.5 Both mean and median can be useful in different situations. In this situation for example the mean is much lower than the median because data points 2 and 4 are weighed less heavily. When you have data which has a few extremes that may not fairly represent the majority of the data it may be useful to use the median instead of the mean.

### integral from 0 to infinity of (cos x * cos(x^2)) dx

The behaviour of this function f(x)=cos(x)cos(x^2) is interesting. The integral is the area between the curve and the x axis. If the functions cos(x) and -cos(x) are plotted on the same graph, the latter form an envelope for f(x). Between x=(pi)/2 and 3(pi)/2, the curve has 4 maxima and 3 minima; between x=3(pi)/2 and 5(pi)/2 there are 7 maxima and 6 minima; between x=(2n-1)(pi)/2 and (2n+1)(pi)/2 there are 3n+1 maxima and 3n minima (integer n>0), a total of 6n+1. (These are based purely on observation, and need to be supported by sound mathematical deduction.) As n becomes large the envelope appears to fill as the extrema become closer together. As x tends to infinity n also tends to infinity. The envelope apparently has as much area above (positive) as below (negative) the x axis so the total area will be zero as the positive and negative areas cancel out. The question is: do the areas cancel out exactly? As x gets larger, the curve starts to develop irregularities and patterns, but it stays within the envelope, and positive irregularities appear to be balanced by negative irregularities, so the overall symmetry appears to be preserved. f(x)=0 when cos(x)=0 or cos(x^2)=0, which means that x=(2n-1)(pi)/2 or sqrt((2n-1)(pi)/2), where n>0. Between 3(pi)/2 and 5(pi)/2, for example, we have sqrt(3(pi)/2), sqrt(5(pi)/2), ..., sqrt(13(pi)/2), because sqrt(13(pi)/2)<3(pi)/20, and this lies between (2n-1) and (2n+1); so n is defined by 2n-1<(2m+1)^2(pi)/2<2n+1.  For m-1 we have 2(n-z)-1<(2m-1)^2(pi)/2<2(n-z)+1, where z is related to the number of zeroes in the current "batch". For example, take m=3: 2n-1<49(pi)/2<2n+1; 49(pi)/2=76.97 approx., so 2n-1=75, and n=38. Also 2(n-z)-1<25(pi)/2<2(n-z)+1 so, because 25(pi)/2=39.27 approx., 2(n-z)+1=41, n-z=20, and z=18. When m=2, 2n-1=39, n=20; 2(20-z)-1<9(pi)/2<2(20-z)+1; 2(20-z)-1=13, 20-z=7, z=13. The actual number of zeroes, Z, including the end points is 2 more than this: Z=z+2. Now we have an exact way to calculate the number of zeroes in each batch. So Z and n are both related to m. The number of extrema, E=Z-1=z+1. In fact, E=int(2(pi)(m-1)+1), where int(a) means the integer part of a, so as m increases, there are proportionately more extrema over the range (2m-1)(pi)/2 to (2m+1)(pi)/2. The figure of 6n+1 deduced earlier by observation approximates to the mathematical findings, because 2(pi) is approximately equal to 6. But we still need to show, or disprove, that the areas above and below the x axis are equal and therefore cancel out. Unfortunately, if we consider the area under the first maximum (between x=(pi)/2 and sqrt(3(pi)/2)), and the area above the first minimum (between x=sqrt(3(pi)/2) and sqrt(5(pi)/2)), they are not the same, so do not cancel out. More...

### Finding a polynomial of a given degree with given zeros: Complex zeros

If the polynomial has an even degree, meaning that the highest power of the variable (for example: x) is an even number (for example: x^4), then all the zeroes could be complex. If odd, there must be at least one real zero. A complex zero is given by the complex expression: a+ib, where a and b are real, and i=sqrt(-1). If the polynomial has real coefficients (all the numbers in at are, or represent, real numbers), as is usually the case, then the imaginary part of the complex zeroes must combine to make a real number. For example, if one complex zero is identified as a+ib, then there will be another zero a-ib such that (x-a-ib)(x-a+ib)=x^2-2ax+a^2+b^2. (Note that, although a and b are real, they are not necessarily rational numbers.) There are no imaginary components in this expansion. Example: x^4+x^3-x-1 is degree 4 polynomial. It has real zeroes 1 and -1 so that it factorises to (x-1)(x+1)(x^2+x+1). The zeroes of x^2+x+1 are found by setting the quadratic to zero and solving using the formula; so x=(-1+sqrt(1-4))/2=(-1+sqrt(-3))/2=(-1+isqrt(3))/2. The two complex roots are -1/2-isqrt(3)/2 and -1/2+isqrt(3)/2. In this case a=-1/2 and b=sqrt(3)/2. Note that a^2+b^2=1. Example: you're told that a degree 5 polynomial has a real zero 1, and two complex zeroes 1+i and -2-i. Find all zeroes and the polynomial. The two missing zeroes will be 1-i and -2+i. The polynomial is (x-1)(x-1-i)(x-1+i)(x+2+i)(x+2-i)=(x-1)(x^2-2x+2)(x^2+4x+5)=x^5+x^4-3x^3-x^2+12x-10. Strictly speaking, the polynomial is a(x^5+x^4-3x^3-x^2+12x-10), where a is a real number, because multiplying a polynomial by a constant doesn't affect its zeroes.

### complex, rational and real roots

The quintic function should have 5 roots.  The changes of sign (through Descartes) tell us the maximum number of positive roots. Since there are two changes of sign there is a maximum of 2 positive roots. To find the number of negative roots we negate the terms with odd powers and check for sign changes: that means that there is at most one negative root, because there's only one sign change between the first and second terms. Complex roots always come in matching or complementary pairs, so that means in this case 2 or 4.  Put x=-1: function is positive; for x=-2 function is negative, so there is a real root between -1 and -2, because the x axis must be intercepted between -1 and -2. This fulfils the maximum for negative roots. That leaves 4 more roots. They could be all complex; there could be two complex and two positive roots. Therefore there are at least two complex roots. To go deeper we can look at calculus and a graph of the function: The gradient of the function is 10x^4-5. When this is zero there is a turning point: x^4=1/2. If we differentiate again we get 40x^3. When x is negative this value is negative so the turning point is a maximum when we take the negative fourth root of 1/2; at the positive fourth root of 1/2 the turning point is a minimum, and these are irrational numbers. The value of the function at these turning points is positive. The fourth root of 1/2 is about 0.84. and once the graph has crossed the x axis between -1 and -2, it stays positive, so all other roots must be complex. The function is 12 when x=0, so we can now see the behaviour: from negative values of x, the function intersects the x axis between -1 and -2 (real root); at x=4th root of 1/2 (-0.84) it reaches a local maximum, intercepts the vertical axis at 12 until it reaches 4th root of 1/2 (0.84) and a local minimum, after which it ascends rapidly at a steep gradient.  [Incidentally, one way to find the real root is to rearrange the equation: x^5=2.5x-6=-6(1-5x/12); x=-(6(1-5x/12))^(1/5)=-6^(1/5)(1-5x/12)^(1/5) We can now use an iterative process to find x. We start with x=0, so x0, the first approximation of the solution, is the fifth root of 6 negated=-1.430969 approx. To find the next solution x1, we put x=x0 and repeat the process to get -1.571266. We keep repeating the process until we get the accuracy we need, or the calculator reaches a fixed value. After just a few iterations, my calculator gave me -1.583590704.]