Using probability concepts, it is found that P(S and D) = 0.1275.
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- A probability is the <u>number of desired outcomes divided by the number of desired outcomes</u>.
- In a standard deck, there are 52 cards.
- Of those, 13 are spades, and 13 are diamond.
- The probability of selecting a spade with the first card is 13/52. Then, there is a 13/51 probability of selecting a diamond with the second. The same is valid for diamond then space, which means that the probability is multiplied by 2. Thus, the desired probability is:
Thus, P(S and D) = 0.1275.
A similar problem is given at brainly.com/question/12873219
Answer:
function g is positive over (-∞, ∞)
function g has a y-intercept of (0,4)
function g decreasing over the interval (-∞, 0)
Step-by-step explanation:
Answer:
36°
Step-by-step explanation:
134° is the exterior angle of the triangle.
By the exterior angle theorm
134° = 98° + y
y = 134° - 98° = 36°
Y = $35.25(6) + 40
y = $211.5 + $40= $251.50
the answer is a
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.
