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.
Answer:
8%
Step-by-step explanation:
Given that:
Reliability of each = 0.92
r = 0.92
1 - (1 - r)(1 - r) = 1 - (0.08)(0.08) = 1 - 0.0064 = 0.9936
Percentage improvement :
1 - (0.9936) / r
1 - (0.9936 / 0.92)
1 - 1.08
= - 0.08
= 0.08 / 100 = 8%
Answer:
96
Step-by-step explanation:
Answer:
1/18 x 9 x 1 = .5 cubic yards. Why would you ever waste time asking this....
Step-by-step explanation:
