The Taylor series is defined by:

Let a = 0.
Then its just a matter of finding derivatives and determining how many terms is needed for the series.
Derivatives can be found using product rule:

Do this successively to n = 6.

Plug in x=0 and sub into taylor series:

If more terms are needed simply continue the recursive derivative formula and add to taylor series.
Answer:
Step-by-step explanation:
a.
let A= number 5
B=number >3
P(A)=1/6 {5}
P(B)=3/6 {4,5,6}
p(A∩B)=1/6 {5}
P(AUB)=P(A)+P(B)-P(A∩B)=1/6+3/6-1/6=3/6=1/2
b.
let event C=number<5
D=even number
C=[1,2,3,4}
D={2,4,6}
C∩D={2,4}
P(C)=4/6
P(D)=3/6
P(C∩D)=2/6
P(C∪D)=P(C)+P(D)-P(C∩D)=4/6+3/6-2/6=5/6
c.
let event E=number 2
F= an odd number.
E={2}
F={1,3,5}
E∩F={}
P(E)=1/6
P(F)=3/6
P(E∩F)=0
P(E∪F)=P(E)+P(F)-P(E∩F)=1/6+3/6-0=4/6=2/3
Answer:
Very interesting observation
~ Maria
Step-by-step explanation:
Answer:
12
Step-by-step explanation:
1/3 of 12 = 4
1/4 of 12 = 3
4 is 1 more than 3.
The number is 12.
_____
If this does not immediately come to mind, you can solve for the number, x:
x/3 -x/4 = 1 . . . . the difference between 1/3 of x and 1/4 of x is 1
x/12 = 1 . . . . . . . collect terms: (1/3 -1/4 = 4/12 -3/12 = 1/12)
x = 12 . . . . . . . . multiply by 12
Answer:
1
Step-by-step explanation: