I think it is the last one or the first one if I'm wrong I'm sorry
The sum of any arithmetic sequence is the average of the first and last terms times the number of terms.
Any term in an arithmetic sequence is:
a(n)=a+d(n-1), where a=initial term, d=common difference, n=term number
So the first term is a, and the last term is a+d(n-1) so the sum can be expressed as:
s(n)(a+a+d(n-1))(n/2)
s(n)=(2a+dn-d)(n/2)
s(n)=(2an+dn^2-dn)/2
However we need to know how many terms are in the sequence.
a(n)=a+d(n-1), and we know a=3 and d=2 and a(n)=21 so
21=3+2(n-1)
18=2(n-1)
9=n-1
10=n so there are 10 terms in the sequence.
s(n)=(2an+dn^2-dn)/2, becomes, a=3, d=2, n=10
s(10)=(2*3*10+2*10^2-2*10)/2
s(10)=(60+200-20)/2
s(10)=240/2
s(10)=120
(76.4)(0.32)(R - 112) + (0.35)(20R + 435) = 54
24.448(R - 112) + 7R + 152.25 = 54
24.448R - 2738.176 + 7R + 152.25 = 54
31.448R - 2585.926 = 54
31.448R = 54 + 2585.926
31.448R = 2639.926
R = 2639.926/31.448
R = 83.9457
Answer:
Step-by-step explanation:
Given that you choose at random a real number X from the interval [2, 10].
a) Since this is a contnuous interval with all number in between equally likely
E = probability for choosing a real number is U(2,10)
pdf of E is 
b) P(X>5) = 
For
P(X<5 or x>7) = 1-P(5<x<7)
= 
