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Topics
Pre-Algebra
Mean
Mode
Greatest Common Factor
Least Common Multiple
Order of Operations
Fractions
Mixed Fractions
Prime Factorization
Exponents
Radicals
Algebra
Combine Like Terms
Solve for a Variable
Factor
Expand
Evaluate Fractions
Linear Equations
Quadratic Equations
Inequalities
Systems of Equations
Matrices
Trigonometry
Simplify
Evaluate
Graphs
Solve Equations
Calculus
Derivatives
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mean(3,4)
m
e
a
n
(
3
,
4
)
Evaluate
\frac{7}{2}=3.5
2
7
=
3
.
5
View solution steps
Solution Steps
mean(3,4)
m
e
a
n
(
3
,
4
)
To find the mean of the set 3,4 first add the members together.
To find the mean of the set
3
,
4
first add the members together.
3+4=7
3
+
4
=
7
The mean (average) of the set 3,4 is found by dividing the sum of its members by the number of the members, in this case 2.
The mean (average) of the set
3
,
4
is found by dividing the sum of its members by the number of the members, in this case
2
.
\frac{7}{2}
2
7
Factor
\frac{7}{2} = 3\frac{1}{2} = 3.5
2
7
=
3
2
1
=
3
.
5
Quiz
Algebra
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mean(3,4)
m
e
a
n
(
3
,
4
)
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The best notation for this identity involving pentagonal numbers \omega(n) and the 3x+1 map
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https://math.stackexchange.com/q/1804296
At first some notes: We usually regard the closed formula \begin{align*} \omega(n)=\frac{3n^2+n}{2}\qquad\qquad\qquad\qquad\qquad n\geq 1\tag{1} \end{align*} as simpler than the summation formula ...
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The MGF for X_n is as you already have, (1-p+pe^t)^n. Note that for small t, we have 1-p+pe^t=1+pt+\frac{pt^2}{2}+O(t^3) Thus, we can manipulate nth power by introducing logarithm. (1+pt+\frac{pt^2}{2}+O(t^3))^n=\exp\left(n\log\left(1+pt+\frac{pt^2}{2}+O(t^3)\right)\right) ...
The MGF for
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(
1
−
p
+
p
e
t
)
n
. Note that for small
t
, we have
1
−
p
+
p
e
t
=
1
+
p
t
+
2
p
t
2
+
O
(
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3
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n
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(
1
+
p
t
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2
p
t
2
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|R^2| alone having the distribution as above I have no problem, however, I don't understand why the entire Power_{rx} also have the same distribution. At least, I see they multiple a bunch of ...
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No, it is impossible. Given any list of n numbers X_k without any constraint of its order. Let \begin{cases} \overline{X} &= \frac{1}{n}\sum_{k=1}^n X_k\\ \overline{X^2} &= \frac{1}{n}\sum_{k=1}^n X_k^2 \end{cases} \quad\text{ and }\quad \begin{cases} P &= \max\{ X_k : 1 \le k \le n \}\\ Q &= \min\{ X_k : 1 \le k \le n \} \end{cases} ...
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{
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1
n
X
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3+4=7
To find the mean of the set 3,4 first add the members together.
\frac{7}{2}
The mean (average) of the set 3,4 is found by dividing the sum of its members by the number of the members, in this case 2.
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