Step-by-step explanation:
We are asked to simply (2√5 + 3√2)². Using formula: (a + b)² = a² + b² + 2ab. Let's say 2√5 = a, 3√2 = b. So,
→ (a + b)² = a² + b² + 2ab
→ (2√5 + 3√2)² = (2√5)² + (3√2)² + 2(2√5)(3√2)
We are aware about the fact that root means 1/2 and square of root means 2/2 that is 1. Using this we get:
→ (2√5 + 3√2)² = 4(5) + 9(2) + 2(2√5)(3√2)
Solve the brackets, to do so first put the like terms in one box.
→ (2√5 + 3√2)² = 4(5) + 9(2) + 2(2*3)(√5)(√2)
Solve the rest calculations.
→ (2√5 + 3√2)² = 20 + 18 + 2(6)(√10)
→ (2√5 + 3√2)² = 38 + 12√10
Option (a) (38 + 12√10) is the correct option.
Answer:
a) Find the common ratio of this sequence.
Answer: -0.82
b) Find the sum to infinity of this sequence.
Answer: 2.2
Step-by-step explanation:
nth term in geometric series is given by ![4\ th \ term = ar^n-1\\-2.196 = 4r^{4-1} \\-2.196/4 = r^{3} \\r = \sqrt[3]{0.549} \\r = 0.82](https://tex.z-dn.net/?f=4%5C%20th%20%5C%20term%20%3D%20ar%5En-1%5C%5C-2.196%20%3D%204r%5E%7B4-1%7D%20%5C%5C-2.196%2F4%20%3D%20r%5E%7B3%7D%20%5C%5Cr%20%3D%20%5Csqrt%5B3%5D%7B0.549%7D%20%5C%5Cr%20%3D%200.82)
where
a is the first term
r is the common ratio and
n is the nth term
_________________________________
given
a = 4
4th term = -2.196
let
common ratio of this sequence. be r
![4\ th \ term = ar^n-1\\-2.196 = 4r^{4-1} \\-2.196/4 = r^{3} \\r = \sqrt[3]{-0.549} \\r = -0.82](https://tex.z-dn.net/?f=4%5C%20th%20%5C%20term%20%3D%20ar%5En-1%5C%5C-2.196%20%3D%204r%5E%7B4-1%7D%20%5C%5C-2.196%2F4%20%3D%20r%5E%7B3%7D%20%5C%5Cr%20%3D%20%5Csqrt%5B3%5D%7B-0.549%7D%20%5C%5Cr%20%3D%20-0.82)
a) Find the common ratio of this sequence.
answer: -0.82
sum of infinity of geometric sequence is given by = a/(1-r)
thus,
sum to infinity of this sequence = 4/(1-(-0.82) = 4/1.82 = 2.2
Answer:
![Var(X) = E(X^2) -[E(X)]^2 = 4.97 -(1.61)^2 =2.3779](https://tex.z-dn.net/?f=%20Var%28X%29%20%3D%20E%28X%5E2%29%20-%5BE%28X%29%5D%5E2%20%3D%204.97%20-%281.61%29%5E2%20%3D2.3779)
And the deviation would be:

Step-by-step explanation:
For this case we have the following distribution given:
X 0 1 2 3 4 5 6
P(X) 0.3 0.25 0.2 0.12 0.07 0.04 0.02
For this case we need to find first the expected value given by:

And replacing we got:

Now we can find the second moment given by:

And replacing we got:

And the variance would be given by:
![Var(X) = E(X^2) -[E(X)]^2 = 4.97 -(1.61)^2 =2.3779](https://tex.z-dn.net/?f=%20Var%28X%29%20%3D%20E%28X%5E2%29%20-%5BE%28X%29%5D%5E2%20%3D%204.97%20-%281.61%29%5E2%20%3D2.3779)
And the deviation would be:

666 687421568899877665443116