Statistics

There are 10 decimetres in one metre. Therefore, 1000 metres is equal to 1000 x 10 = 10000 decimetres.

quantitative data is data that is mainly number's hahaha

It was not discovered but invented.

N. R. Narayana Murthy was born on August 20, 1946.

thanx

They are called outliers

a pyramid with a square bottom

It cannot be factored because the discriminant of the quadratic expression is less than zero

7C3 = 7*6*5/(3*2*1) = 35 combinations.

7C3 = 7*6*5/(3*2*1) = 35 combinations.

7C3 = 7*6*5/(3*2*1) = 35 combinations.

7C3 = 7*6*5/(3*2*1) = 35 combinations.

Neither, as a graph can be read from either axis to get a value on the other axis.

However, the x coordinate is usually considered the independent variable as an equation is usually written as y = f(x), so it could be considered the input.

However, from y = f(x), it can often be re-arranged to get x = f-1(y) which would make x the output! eg:

y = 2x + 4

→ 2x + 4 = y

→ 2x = y - 4

→ x = 1/2 y - 2

If they can only be used once each: 720

If they can repeat: 46,656

It's not possible to tell you without the numbers.

You would use combinations instead of permutations so just type in your calculator 96667C5 and you should get 70333953523652148234618

Or, you could use the formula...

r=number chosen

n=number to chose from

n!/((n-r)!(r!))

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If the question was meant to ask - How many 5 number combinations (actually,

permutations) can be made from (the digits) 9,6,6,6,7 ? - The answer would be:

2(5C2) = 2[5!/(3!2!)] = 20

i.e., 96676, 96766, 97666, ...

180

* * * * *

Wrong!

There are 6:

12345, 12346, 12356, 12456, 13456 and 23456. And that is it! No more.

At least two.

We know that in order to be even the last digit must be one of the 4 possible: 0,2,4 or 8. This leaves 6 digits for the 1st number, 5 for the 2nd and 4 for the third.

so 4 x 6 x 5 x 4 = 480 possible 4 digit even numbers

The answer is 15.

(5*6)/2 = 15.

The following are all the possible combinations, where the order is insignificant:

1-2, 1-3, 1-4, 1-5, 1-6, 2-3, 2-4, 2-5, 2-6, 3-4, 3-5, 3-6, 4-5, 4-6, 5-6.

The sum of all those possibilites equals 15.

A third way of finding the answer is to notice that there are 5 ways to combine one class with each of the 5 other classes, 4 ways to combine the second class with the remaining classes, etc. 5 + 4 + 3 + 2 + 1 = 15.

12

You average the #'s and then you divide them to however many #'s there are. There MAY BE a DECIMAL but its okay!

36324

9999 - 999 = 9000

This usually applies to word problems with several variables which are 'connected' in the story. There are also some additional statements about cost or profit.

The constraints are used to write equations. These are graphed and there is usually an enclosed space, Because all the equations form straight lines, hence the name linear programming. Intersections of these lines gives pionts where the max profit or min cost will occur.

Take these points and put them into the cost/profit equation to find the max/min.

The fundamental theorem is that the max/min will occur at these intersection points that is the whole point of graphing and finding the intersections.

Compare data

To determine the number of combinations possible using the numbers 9, 3, 1, and 7, we can use the concept of permutations. A permutation is an arrangement of objects in a specific order. In this case, we want to find the number of permutations of all the numbers.

The formula for permutations of n objects taken r at a time is given by:

P(n, r) = n! / (n - r)!

Where "!" denotes the factorial function.

In this case, we have 4 numbers and we want to arrange all of them, so r = 4.

P(4, 4) = 4! / (4 - 4)!

= 4! / 0!

= 4! / 1

= 4 * 3 * 2 * 1

= 24

Hence, there are 24 different number combinations that can be made using the numbers 9, 3, 1, and 7.