There would be about 19 defective products delivered.
First, let's start with the number of defective games.
100000 x 0.0095 = 950
Now, the test will catch 98% of those defects. That means 2% of the defects will get through to the consumers.
0.02 x 950 = 19
You can write
1.5
⋅
10
5
=
15
⋅
10
4
15
⋅
10
4
−
8
⋅
10
4
=
(
15
−
8
)
⋅
10
4
=
7
⋅
10
4
Since the power is positive we move the dec.point 4 to the right:
=
70000
Hope it works
The base is 30.
Hopefully I got this correct, not 100%
Might have to experiment a bit to choose the right answer.
In A, the first term is 456 and the common difference is 10. Each time we have a new term, the next one is the same except that 10 is added.
Suppose n were 1000. Then we'd have 456 + (1000)(10) = 10456
In B, the first term is 5 and the common ratio is 3. From 5 we get 15 by mult. 5 by 3. Similarly, from 135 we get 405 by mult. 135 by 3. This is a geom. series with first term 5 and common ratio 3. a_n = a_0*(3)^(n-1).
So if n were to reach 1000, the 1000th term would be 5*3^999, which is a very large number, certainly more than the 10456 you'd reach in A, above.
Can you now examine C and D in the same manner, and then choose the greatest final value? Safe to continue using n = 1000.
