Answer:
It means 
also converges.
Step-by-step explanation:
The actual Series is::
The method we are going to use is comparison method:
According to comparison method, we have:
If series one converges, the second converges and if second diverges series, one diverges
Now Simplify the given series:
Taking"n^2"common from numerator and "n^6"from denominator.
![=\frac{n^2[7-\frac{4}{n}+\frac{3}{n^2}]}{n^6[\frac{12}{n^6}+2]} \\\\=\frac{[7-\frac{4}{n}+\frac{3}{n^2}]}{n^4[\frac{12}{n^6}+2]}](https://tex.z-dn.net/?f=%3D%5Cfrac%7Bn%5E2%5B7-%5Cfrac%7B4%7D%7Bn%7D%2B%5Cfrac%7B3%7D%7Bn%5E2%7D%5D%7D%7Bn%5E6%5B%5Cfrac%7B12%7D%7Bn%5E6%7D%2B2%5D%7D%20%5C%5C%5C%5C%3D%5Cfrac%7B%5B7-%5Cfrac%7B4%7D%7Bn%7D%2B%5Cfrac%7B3%7D%7Bn%5E2%7D%5D%7D%7Bn%5E4%5B%5Cfrac%7B12%7D%7Bn%5E6%7D%2B2%5D%7D)
![\sum_{n=1}^{inf}a_n=\sum_{n=1}^{inf}\frac{[7-\frac{4}{n}+\frac{3}{n^2}]}{[\frac{12}{n^6}+2]}\ \ \ \ \ \ \ \ \sum_{n=1}^{inf}b_n=\sum_{n=1}^{inf} \frac{1}{n^4}](https://tex.z-dn.net/?f=%5Csum_%7Bn%3D1%7D%5E%7Binf%7Da_n%3D%5Csum_%7Bn%3D1%7D%5E%7Binf%7D%5Cfrac%7B%5B7-%5Cfrac%7B4%7D%7Bn%7D%2B%5Cfrac%7B3%7D%7Bn%5E2%7D%5D%7D%7B%5B%5Cfrac%7B12%7D%7Bn%5E6%7D%2B2%5D%7D%5C%20%5C%20%5C%20%5C%20%5C%20%5C%20%5C%20%5C%20%5Csum_%7Bn%3D1%7D%5E%7Binf%7Db_n%3D%5Csum_%7Bn%3D1%7D%5E%7Binf%7D%20%5Cfrac%7B1%7D%7Bn%5E4%7D)
Now:
![\sum_{n=1}^{inf}a_n=\sum_{n=1}^{inf}\frac{[7-\frac{4}{n}+\frac{3}{n^2}]}{[\frac{12}{n^6}+2]}\\ \\\lim_{n \to \infty} a_n = \lim_{n \to \infty} \frac{[7-\frac{4}{n}+\frac{3}{n^2}]}{[\frac{12}{n^6}+2]}\\=\frac{7-\frac{4}{inf}+\frac{3}{inf}}{\frac{12}{inf}+2}\\\\=\frac{7}{2}](https://tex.z-dn.net/?f=%5Csum_%7Bn%3D1%7D%5E%7Binf%7Da_n%3D%5Csum_%7Bn%3D1%7D%5E%7Binf%7D%5Cfrac%7B%5B7-%5Cfrac%7B4%7D%7Bn%7D%2B%5Cfrac%7B3%7D%7Bn%5E2%7D%5D%7D%7B%5B%5Cfrac%7B12%7D%7Bn%5E6%7D%2B2%5D%7D%5C%5C%20%5C%5C%5Clim_%7Bn%20%5Cto%20%5Cinfty%7D%20a_n%20%3D%20%5Clim_%7Bn%20%5Cto%20%5Cinfty%7D%20%20%5Cfrac%7B%5B7-%5Cfrac%7B4%7D%7Bn%7D%2B%5Cfrac%7B3%7D%7Bn%5E2%7D%5D%7D%7B%5B%5Cfrac%7B12%7D%7Bn%5E6%7D%2B2%5D%7D%5C%5C%3D%5Cfrac%7B7-%5Cfrac%7B4%7D%7Binf%7D%2B%5Cfrac%7B3%7D%7Binf%7D%7D%7B%5Cfrac%7B12%7D%7Binf%7D%2B2%7D%5C%5C%5C%5C%3D%5Cfrac%7B7%7D%7B2%7D)
So a_n is finite, so it converges.
Similarly b_n converges according to p-test.
P-test:
General form:
if p>1 then series converges. In oue case we have:
p=4 >1, so b_n also converges.
According to comparison test if both series converges, the final series also converges.
It means also converges.
Answer:
x=-6 y=-1
Step-by-step explanation:
// Solve equation [2] for the variable y
[2] y = -x - 7
// Plug this in for variable y in equation [1]
[1] 3x - 2•(-x -7) = -16
[1] 5x = -30
// Solve equation [1] for the variable x
[1] 5x = - 30
[1] x = - 6
// By now we know this much :
x = -6
y = -x-7
// Use the x value to solve for y
y = -(-6)-7 = -1
Solution :
{x,y} = {-6,-1}
Answer: p = −15
Step-by-step explanation:
Add 11p to both sides.
−10p+11p=−11p−15+11p
p=−15
Answer:
The z-score for SAT exam of junior is much small than his ACT score. This means he performed well in his ACT exam and performed poor in his SAT exam.
Step-by-step explanation:
Mean SAT scores = 1026
Standard Deviation = 209
Mean ACT score = 20.8
Standard Deviation = 4.8
We are given SAT and ACT scores of a student and we have to compare them. We cannot compare them directly so we have to Normalize them i.e. convert them into such a form that we can compare the numbers in a meaningful manner. The best way out is to convert both the values into their equivalent z-scores and then do the comparison. Comparison of equivalent z-scores will tell us which score is higher and which is lower.
The formula to calculate the z-score is:
Here, μ is the mean and σ is the standard deviation. x is the value we want to convert to z score.
z-score for junior scoring 860 in SAT exam will be:
z-score for junior scoring 16 in ACT exam will be:
The z-score for SAT exam of junior is much small than his ACT score. This means he performed well in his ACT exam and performed poor in his SAT exam.
Answer:
a= 78
Step-by-step explanation:
To solve subtract:
180-102= 78
Hope this helps!
