Solve mean(1,2,3,4,5,6,7,8,9,10) | Microsoft Math Solver

m e a n (1, 2, 3, 4, 5, 6, 7, 8, 9, 10)

Evaluate

\frac{1 1}{2} = 5.5

Factor

\frac{1 1}{2} = 5 \frac{1}{2} = 5.5

Quiz

Algebra

5 problems similar to:

m e a n (1, 2, 3, 4, 5, 6, 7, 8, 9, 10)

Similar Problems from Web Search

Need help with Sigma-algebra

https://math.stackexchange.com/questions/540282/need-help-with-sigma-algebra

You're right on target. The (

σ

-)algebra on

Ω

generated by the partition

{A_{1}, A_{2}, A_{3}}

is precisely

{\emptyset, A_{1}, A_{2}, A_{3}, A_{1} \cup A_{2}, A_{1} \cup A_{3}, A_{2} \cup A_{3}, Ω} .

More ...

Subgroup acting on subsets

https://math.stackexchange.com/questions/2664227/subgroup-acting-on-subsets

Label the elements of

Ω

I : = {1, 2}

I I : = {3, 4}

I I I : = {5, 6}

I V : = {7, 8}

. It is clear that

Φ (g h) = Φ (g) Φ (h)

, so we have a homomorphism from

G

into the symmetric group ...

Combinatorics Sequences of length 5 from set

https://math.stackexchange.com/questions/1898838/combinatorics-sequences-of-length-5-from-set

This doesn't work, because the first one doesn't necessarily need to be a letter and the second one doesn't necessarily need to be a number. Assuming that every element can be chosen multiple times, ...

Fermat's 2 Square-Like Results from Minkowski Lattice Proofs

https://math.stackexchange.com/questions/541665/fermats-2-square-like-results-from-minkowski-lattice-proofs

Unfortunately I'm answering my own question, but I was able to prove that if

p \equiv 1 (m o d 8)

p \equiv 3 (m o d 8)

, then

p

is expressible as

x^{2} + 2 y^{2}

for

x, y \in Z

, and if

p \equiv 1 (m o d 3)

...

Density of a set and an approximating sequence

https://math.stackexchange.com/questions/2465030/density-of-a-set-and-an-approximating-sequence

Let

A = Z^{2}

and

C_{n}

be the first

n

points of a spiral the emanates from the origin

(0, 0)

and goes through every integral point on the plane. Think of the spiral as in the following ...

There are only two types of groups of order

6 .

https://math.stackexchange.com/questions/541635/there-are-only-two-types-of-groups-of-order-6

There's an element

a

of order two and an element

b

of order three (Cauchy). If they commute, then

a b

is of order

6

and

G

s cyclic. Otherwise, the elements

1, a, b, b^{2}, a b, b a

are pairwise ...

Copy

Copied to clipboard

1+2+3+4+5+6+7+8+9+10=55

To find the mean of the set 1,2,3,4,5,6,7,8,9,10 first add the members together.

\frac{55}{10}=\frac{11}{2}

The mean (average) of the set 1,2,3,4,5,6,7,8,9,10 is found by dividing the sum of its members by the number of the members, in this case 10.

Topics

Topics

Similar Problems from Web Search

Share