Answer:
∑ (-1)ⁿ⁺³ 1 / (n^½)
∑ (-1)³ⁿ 1 / (8 + n)
Step-by-step explanation:
If ∑ an is convergent and ∑│an│is divergent, then the series is conditionally convergent.
Option A: (-1)²ⁿ is always +1. So an =│an│and both series converge (absolutely convergent).
Option B: bn = 1 / (n^⁹/₈) is a p series with p > 1, so both an and │an│converge (absolutely convergent).
Option C: an = 1 / n³ isn't an alternating series. So an =│an│and both series converge (p series with p > 1). This is absolutely convergent.
Option D: bn = 1 / (n^½) is a p series with p = ½, so this is a diverging series. Since lim(n→∞) bn = 0, and bn is decreasing, then an converges. So this is conditionally convergent.
Option E: (-1)³ⁿ = (-1)²ⁿ (-1)ⁿ = (-1)ⁿ, so this is an alternating series. bn = 1 / (8 + n), which diverges. Since lim(n→∞) bn = 0, and bn is decreasing, then an converges. So this is conditionally convergent.
Answer:
I would cost $4.75
Step-by-step explanation:
9.50 / 2
4.75
Hope it´s helpfull :)
The calculation of the expression “four times as large as 124+645” end up in 1141.
<u>Step-by-step explanation:</u>
- The question is asked to Express the calculation “four times as large as 124+645”.
- The given statement can be written in the expression form to perform the calculations.
Here, the phrase 'four times' represents the multiplication of 4.
⇒ “four times as large as 124+645” = 4(124) + 645
So, the first step is to multiply 4 with 124.
⇒ 4 × 124
⇒ 496
The expression is now modified as “496+645"
We know that, the final result is the addition of the two numbers 496 and 645 which is calculated as
⇒ 496+645
⇒ 1141
Therefore, the calculation of the expression “four times as large as 124+645” end up in 1141.
Using the z-distribution and the formula for the margin of error, it is found that:
a) A sample size of 54 is needed.
b) A sample size of 752 is needed.
In a sample with a number n of people surveyed with a probability of a success of , and a confidence level of , we have the following confidence interval of proportions.
In which z is the z-score that has a p-value of .
The margin of error is of:
90% confidence level, hence, z is the value of Z that has a p-value of , so .
Item a:
The estimate is .
The sample size is <u>n for which M = 0.03</u>, hence:
Rounding up, a sample size of 54 is needed.
Item b:
No prior estimate, hence
Rounding up, a sample of 752 should be taken.
A similar problem is given at brainly.com/question/25694087