d) a, 3D7-2" — п+2. 16. Find the solution to each of these recurrence relations with the given initial conditions. Use an iterative ap- proach such as that used in Example 10. a) a, = -a„-1, aŋ = 5 b) a, = a,-1 + 3, a, = 1 c) a, = a„-1 - n, a, = 4 d) a, = 2a,-1 - 3, a, = -1 e) a, = (n+ 1)a,-1,4 = 2 f) a, = 2na„-1, aŋ = 3 g) a, = -a„-1 + n – 1, a, = 7 17. Find the solution to each of these recurrence relations and 23. %3D %3D 24. Links initial conditions. Use an iterative approach such as that used in Example 10. a) a, = 3a,-1, a, = 2
Correlation
Correlation defines a relationship between two independent variables. It tells the degree to which variables move in relation to each other. When two sets of data are related to each other, there is a correlation between them.
Linear Correlation
A correlation is used to determine the relationships between numerical and categorical variables. In other words, it is an indicator of how things are connected to one another. The correlation analysis is the study of how variables are related.
Regression Analysis
Regression analysis is a statistical method in which it estimates the relationship between a dependent variable and one or more independent variable. In simple terms dependent variable is called as outcome variable and independent variable is called as predictors. Regression analysis is one of the methods to find the trends in data. The independent variable used in Regression analysis is named Predictor variable. It offers data of an associated dependent variable regarding a particular outcome.
I was wondering if someone could explan to me what I am doing wrong in these problems I looked up the answers and they where drastically differnet than what I was getting.
Assignment three CS 250
Cyrus Barkhurst
CS 250
3/6/20
2.1 number 16
a)
.An = -an-`1, a0 = 5
a1= -5
A2 = -(-5)
A3= -(-(-5))
A(n)= 5(-1^n)
b)
.An= a(n+1)+3 a(0) = 1 the brackets in the text mean that there is a subscript
a1 = 1+3 = 4
a2 = (1+3)+3
a3 = ((1+3)+3)+3
a(n)= 1+3n
- c)
A(n) = a(n-1)-n, a(0) = 4
a1 = (4-0)-1
A2 = ((4-0)-1)-2
A3 = (((4-0)-1)-2)-3
a(n) = 4- n(n+1)/2
- d)
A(n) = 2a(n-1)- 3, a0=-1
A1 = 2(2)-3
A2=2(2(2))-3)-3
A3= 2((2(2))-3)-3)-3
a(n)= -2^n+2+3
- e)
a(n)= (n+1)a(n-1) a(0)=2
a1= (2+1)
a2= (2+1)+1
a3= (2+1)+1)+1
a(n)= 2(n+1)!
f)
a(n)= 2na(n-1), a(0) = 3
a1=3(1)+1
A2 = 3(3(1)+1)+1
A3 =3(3(3(1)+1)+1)+1
a(n) = 3.2^n(n)!
g).
a(n)= -a(n-1)+n-1 a(0)= 7
A(1) = -7+1-1
A(2)=-(-7+1-1)+2-1
A(3)=-(-(-7+1-1)+2-1)+3-1)
A(n) = ((n+1)/2)-8
Thanks
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