B.)155
a1+d(n-1)
a1+d(6-1)=50
a1+5d=50
a1+d(11-1)=85
a1+10d=85
a1=50-5d
(50-5d)+10d=85
50-5d+10d=85
5d+50=85
5d=35
d=7
a1+7(6-1)=50
a1+35=50
a1=15
15+7(n-1)
15+7(21-1)
15+7(20)
15+140=155
200-170=30
The percentage of increase:
(30/170)*100=17.6471%
Answer:
0.0833333333...
Step-by-step explanation:
The 3 goes on forever, just for your information.
Hope this helps!
-3×2= -6
-6-27= -33
so if we peak option B.
(-3x +91)(x+31)
solving it
-3x (x+31)+91(x+31)
-3x^2 -93x +91x +2821
we will take (-93x+91x)= -2x
so it's going to be
-3x^2 -2x +2821
this is a quadratic equation hence
2+/-√-2^2-(4×-3×2821)divided by 2×-3
2+/-√4+33852divide by -6
2+/-√33856 divide by -6
2+/- 184 divided by -6
2+/- 30.67
when you add you'll get
32.67
round it off and you get 33
which is the same as what we got at the top
so expression B is the same as
-3×2-27 which is 33
This question is incomplete, the complete question is;
An aircraft seam requires 27 rivets. The seam will have to be reworked if any of these rivets is defective. Suppose rivets are defective independently of one another, each with the same probability. (Round your answers to four decimal places.)
If 25% of all seams need reworking, what is the probability that a rivet is defective?
Answer:
the probability that a rivet is defective is 0.0106
Step-by-step explanation:
Given the data in the question,
let event X represent seam fails and event B represent rivet fails.
now, 25% of all seams need reworking;
i.e P(X) = 25% = 0.25
the probability that a rivet is defective will be;
P( A ) = 1 - P( B' )²⁷
so
0.25 = 1 - P( B' )²⁷
P( B' )²⁷ = 1 - 0.25
P( B' )²⁷ = 0.75
P( B' ) =
P( B' ) = 0.9894
so
P( B ) = 1 - 0.9894
P( B ) = 0.0106
Therefore, the probability that a rivet is defective is 0.0106