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which of the following numbers is biggest (that has the highest numerical value)?

which of the following numbers is biggest (that has the highest numerical value)? 9.5437821799 3274 0.0157493 511.9647158274555

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What is the highest known numerical value? - Math Central


What is the highest known numerical value? ... to know the highest numerical value of some ... a bit of the historical discussion concerning the "biggest number ...
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Chemistry Chapter 8 Flashcards | Quizlet


Chemistry Chapter 8. ... Which of the following sets of 4 quantum numbers correctly describes an ... An atom of which of the following elements has the highest fourth ...
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Names of large numbers - Wikipedia


... names of large numbers have been forced into common usage as a result of hyperinflation. The highest numerical value ... numbers are simply the result of ...
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Numerical digit - Wikipedia


... and each digit has a value. ... The Oksapmin culture of New Guinea uses a system of 27 upper body locations to represent numbers. To preserve numerical ...
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Least to Greatest Calculator, Greatest to Least Calculator ...


Least to Greatest Calculator ... is an online tool to arrange the numbers from least to greatest ... Which of the following student score second highest marks.
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How to: Hold the Largest Possible Number in a Variable ...


A variable holds the largest possible numbers with precision if you declare it ... How to: Hold the Largest Possible Number in a Variable ... Largest Precise Value.
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How to select the highest or lowest value in Excel?


... The Max function and Min function can only find out the highest or lowest value, ... To get the largest 3 or smallest 3 numbers in a range: ... see the following ...
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2 Easy Ways to Find the Mode of a Set of Numbers - wikiHow


... the mode of a set of numbers is the number that ... of finding the biggest "cluster" of identical ... the data points have no numerical value. 3.
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Hexadecimal numbers


Hexadecimal Number System. The hexadecimal ... You will see hexadecimal numbers some of whose digits are letters ... The highest you can count with a given ...
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Suggested Questions And Answer :


which of the following numbers is biggest (that has the highest numerical value)?

?????? hiest numerkal value ?????? 3274
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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.
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how to find slope and y-intercept of y=3x-5

Any line equation which is in the form y=m*x + b is called slope-intercept form. What that gives you is the slope is the number which is multiplied by X (called the coefficient) while the slope intercept is the Y value when X=0. So if we take your first equation: y = 3x - 5 The slope (or m) = 3 The Y intercept = -5 Slope is defined as either "the change in y divided by the change in x" or "rise over run", so that 3, really can be considered as 3/1. Each change in X of 1 will change Y by 3. So slope = 3/1. Graphing any line can come from 2 methods. 1) create a table of values or 2) calculate a single point (x,y) and then apply the slope to find a second coordinate pair (x,y). For the equation y=3x - 5: y = 3x - 5 x y 0 -5 1 -2 Replacing the values for X into the original equation, will come out to the values for Y. So when X=0, y = 3*0 - 5, or simply -5.  When X = 1, Y = 3*1 -5 or simply -2. With these two coordinate pairs of points, you can plot a dot on your graph at each (0,-5) and (1, -2) then draw a straight line which goes through each point and continues straight in each direction, probably ending each end of this line with an arrow to show it continues. I do not have a way to include a picture here of a graph. The second way to graph a line is as follows. You need a starting point that will be on the line. Given the form of y=mx + b, you have a simple point which can be used at the y-intercept. The point is always in the form of (0, b), so in this case it is (0,-5). From that first point, you will apply your slope. The slope is 3 (or technically 3/1 which is a big help). From the initial point (0,-5) you will go UP 3 and RIGHT 1 and that will be the next point that is easy to find. Connect those two points and continue the line in each direction and that will be a graph of your line. Anytime your slope is positive, you will use it by going UP the top number (numerator of the slope) and going RIGHT the bottom number (denominator of the slope). But if your slope is negative (like your second problem is) you will use it by going DOWN the numerator and then RIGHT the denominator. The equation is y= -2/3x + 4 ( / = divided). I need to state the slope and y-intercept. I will not walk through the details on the second equation, but you should have enough information to get the answer from the above example.
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In Excel 2010 what would you type as a formula if amounts were charged at an increment level?

Question: In Excel 2010 what would you type as a formula if amounts were charged at an increment level? Assume that your data is in cell D3 and you you wish the incremented value to be in cell G3. The syntax for the if-statement in Excel is, IF(logical_test, value_if_true, [value_if_false]) where logical test is: "D3<10000" value_if_true is: D3*1.2 [value_if_false] is: (D3-10000)*1.15+12000 Enter a number in cell D3. Highlight (click on) the cell G3 and enter the following formula, "=IF(D3<10000,D3*1.2,(D3-10000)*1.15+12000)" - excluding the quote marks. Enter a series of numerical values in cells D4 - D8, with some values < 10,000 and some values > 10,000. Highlight cell G3. Copy cell G3 and paste it into the 5 cells beneath it. G4 - G8. Obverse (and check) the results.
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What is the value of SS (sum of squared deviations) for the following sample?

data=2,3,4, 7 num numbers=4, su sum=16, averaej=4, median=3.5, biggest =7 sum av squares=14 varians = (sum av squares) / (num numbers)=3.5 std deviashun=1.87083
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make a place value drawing of a number that has double the number of tens as ones and three times the number of thousands as hundreds write the number.

The Digits we use today are called "Hindu-Arabic Numerals" and look like these: 0   1   2   3   4   5   6   7   8   9 We can use these on their own to count up to 9: 0   1 2 3 4 5 6 7 8 9 ?? But what happens after 9?   Ten Or More ... When we have more than 9 items, we start another column - the "tens" column - and we write down how many "tens" we have, followed by how many "ones" (also called "units").  
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Assuming that the following data represents a poplulation x values with

me assume yu hav data points=1, 2, 3, 3.... then 16 numbers, sum=64, aveaej=4, median=4, biggest=7, smallest=1, raenj=6 sum av squares=40, varians=2.5, std deviashun=1.58114
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6x²+13xy+11x+36y-5y²-7

This is the expansion of (ax+by+c)(dx+ey+f) where a to f represent numerical constants (integers), obeying the following arithmetic rules: ad=6, be=-5, cf=-7 (coefficients of x^2, y^2, constant) The only numbers we have for a, b, c, d, e or f must be in the set {1 1 1 2 3 5 6 7}; no other numbers can be used and the arithmetic rules must always apply. This is because the only factors of 5 are 1 and 5; the only factors of 7 are 1 and 7; but the pairs of factors of 6 are 1 and 6 or 2 and 3. I haven't yet considered the signs plus and minus. We also have the sums of two products: ae+bd=13 (coefficient of xy), af+cd=11 (x), ce+bf=36 (y). The largest of these is 36, and there's only one way of using 4 numbers in the set to get 36 using each one once and that is 7*5+1*1. So from this we can find b, c, e and f. The following table shows all 8 combinations for these coefficients: bf+ce Ref b c e f 1 1 -7 -5 1 2 1 -5 -7 1 3 -7 1 1 -5 4 -5 1 1 -7 5 -1 7 5 -1 6 -1 5 7 -1 7 7 -1 -1 5 8 5 -1 -1 7 The remaining two coefficients are a and d and the possible pairs of factors are 1 and 6, or 2 and 3. Here's a table to show all 8 possible combinations: a and d Ref a d 1 1 6 2 6 1 3 -1 -6 4 -6 -1 5 2 3 6 3 2 7 -2 -3 8 -3 -2 The next bit is tedious. We need to try out all the combinations of b, c, e and f against all those of a and d! Yes, that's 64 in all. What we're looking for is the sum of the products ae+bd=13. I'll spare you the details and just show you the results. Ref 1 from the first table with ref 8 from the second gives the required result and also satisfies af+cd=11. So we have the values of a to f: -3, 1, -7, -2, -5, 1, and the factorisation is: (-3x+y-7)(-2x-5y+1). This can also be written: (3x-y+7)(2x+5y-1), which is ref 5 from the first table with ref 6 from the second. Combinations in red have complementary signs to those shown in black.  
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prove that P1 = P?

The first few primes are 1, 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47 (P<50). The sum of consecutive primes that are themselves are primes can be identified: p1=1; p2=1+2=3; p4=1+2+3+5=11; p6=1+2+3+5+7+11=29; p8=1+2+3+5+7+11+13+17=59; p10=1+2+3+...+17+19+23+101, ... My understanding of the questions is that P1={ p1 }; P2={ p1 p2 }; P3={ p1 p2 p4 }; P4={ p1 p2 p4 p6 }; P5={ p1 p2 p4 p6 p8 }; P6={ p1 p2 p4 p6 p8 p10 }. P2 and P4 are defined in this list. In Pn (P sub n), n defines the number of elements, and since n is a finite number (natural number belonging to N), the sets are finite by definition. In the sums p sub k it follows that k will be even (apart from p1) because, after 2, all primes are odd, adding another prime will automatically result in an even number, so a further prime must be added to get an odd number which may be prime. However, not all odd numbers obtained this way will necessarily be prime because not all odd numbers are prime. What I'm not sure of is whether P3={ p1 p2 p4 } should in fact be defined as P4 because the highest pk is p4. If this is indeed the case then the list given earlier should define the sets for P1, P2, P4, P6, P8. P10 instead of P1, P2, P3,... Under this assumption, P4={ p1 p2 p4} while P2 is unaffected. However, the question suggests that P3 and P5 are definable. This takes us back to the original definition. P3 has only three elements under this definition: p1, p2, p4. If p1, p2 and p4 are themselves considered as sets of the primes comprising the sums, then, assuming duplicates are allowed, we have P3={ {1 } { 1 2 } { 1 2 3 5 } }. Five elements from these are, for example, { 1 1 1 2 2 } which also applies to P5.  
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Find solution

The first few primes are 1, 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47 (P<50). The sum of consecutive primes that are themselves are primes can be identified: p1=1; p2=1+2=3; p4=1+2+3+5=11; p6=1+2+3+5+7+11=29; p8=1+2+3+5+7+11+13+17=59; p10=1+2+3+...+17+19+23+101, ... My understanding of the questions is that P1={ p1 }; P2={ p1 p2 }; P3={ p1 p2 p4 }; P4={ p1 p2 p4 p6 }; P5={ p1 p2 p4 p6 p8 }; P6={ p1 p2 p4 p6 p8 p10 }. P2 and P4 are defined in this list. In Pn (P sub n), n defines the number of elements, and since n is a finite number (natural number belonging to N), the sets are finite by definition. In the sums p sub k it follows that k will be even (apart from p1) because, after 2, all primes are odd, adding another prime will automatically result in an even number, so a further prime must be added to get an odd number which may be prime. However, not all odd numbers obtained this way will necessarily be prime because not all odd numbers are prime. What I'm not sure of is whether P3={ p1 p2 p4 } should in fact be defined as P4 because the highest pk is p4. If this is indeed the case then the list given earlier should define the sets for P1, P2, P4, P6, P8. P10 instead of P1, P2, P3,... Under this assumption, P4={ p1 p2 p4} while P2 is unaffected. However, (iv) suggests that P3 and P5 are definable. This takes us back to the original definition. P3 has only three elements under this definition: p1, p2, p4. If p1, p2 and p4 are themselves considered as sets of the primes comprising the sums, then, assuming duplicates are allowed, we have P3={ {1 } { 1 2 } { 1 2 3 5 } }. Five elements from these are, for example, { 1 1 1 2 2 } which also applies to P5. You have presented similar questions recently and I can only hope that this well-meant answer helps you to resolve the problem as you intended.
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