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son4ous [18]
3 years ago
12

Help answering this simple Series Question? I don't know what I did wrong!

Mathematics
1 answer:
Tatiana [17]3 years ago
5 0
Looks like you just evaluated the summand for the given value of n, whereas the question is asking you to find the value of the sum for the first n terms.

Let S_k=\displaystyle\sum_{n=1}^k\frac3{(-2)^n}. Then S_k is the kth partial sum.

S_1 happens to be the first term in the series, which is why that box is marked correct:

S_1=\displaystyle\sum_{n=1}^1\frac3{(-2)^n}=\frac3{(-2)^1}=-1.5

But the next partial sum is not correct:

S_2=\displaystyle\sum_{n=1}^2\frac3{(-2)^n}=\frac3{(-2)^1}+\frac3{(-2)^2}=-0.75

and this is not the same notion as the second term (which indeed is 0.75) in the series.
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A true-false quiz with 10 questions was given to a statistics class. Following is the probability distribution for the score of
disa [49]

Answer:

E(X)=\sum_{i=1}^n X_i P(X_i)  

And if we use the values obtained we got:  

E(X)=5*0.05 +6*0.15 +7*0.33 +8*0.28+ 9*0.12 +10*0.07=7.48  

For this case this value means that the expected score is about 7.48

Step-by-step explanation:

For this case we assume the following probability distribution:

X         5       6         7       8        9        10

P(X)   0.05   0.15  0.33  0.28   0.12   0.07

First we need to find the expected value (first moment) and the second moment in order to find the variance and then the standard deviation.

In order to calculate the expected value we can use the following formula:  

E(X)=\sum_{i=1}^n X_i P(X_i)  

And if we use the values obtained we got:  

E(X)=5*0.05 +6*0.15 +7*0.33 +8*0.28+ 9*0.12 +10*0.07=7.48  

For this case this value means that the expected score is about 7.48

In order to find the standard deviation we need to find first the second moment, given by :  

E(X^2)=\sum_{i=1}^n X^2_i P(X_i)  

And using the formula we got:  

E(X^2)=(5^2 *0.05)+(6^2 *0.15)+(7^2 *0.33)+(8^2 *0.28)+ (9^2 *0.12 +(10^2 *0.07))=57.46  

Then we can find the variance with the following formula:  

Var(X)=E(X^2)-[E(X)]^2 =57.46-(7.48)^2 =1.5096  

And then the standard deviation would be given by:  

Sd(X)=\sqrt{Var(X)}=\sqrt{1.5096}=1.229  

5 0
4 years ago
Read 2 more answers
A bin contains 10 cards labeled A to J and 20 cards labeled 1 to 20. The cards are shuffled and one card is chosen at random. Wh
Fudgin [204]
So you ave a 3 out of 10 chance to get a letter in eh word DIG then you have a 1 in 2 chance of drawing an odd number add those together.

13 out of 30 chance

Simplify if necessary, hope this helps
7 0
3 years ago
Find each sum or difference
alexdok [17]
The P is 16x^2 -12 and Perimeter is the distance around the edge of a shape. So you divide your equation by 4 and the answer you get is 4x^2 - 3
5 0
3 years ago
Read 2 more answers
In a bag of counters, there are 3 times as many red counters as
Anni [7]

Answer:

The ratio is 6:2:1.

Step-by-step explanation:

Let's denote:

Blue counter b

Red counter r

Yellow counter y.

There is given

r=3b (three times more red as blue) [1]

and

b=2y (twice more blue as yellow.) [2]

Now, replacing [2] to [1] gives

3*2y=r

6y=r.

Hence, the ratio is 6:2:1, because r=6y and b=2y.

7 0
3 years ago
น. <br><img src="https://tex.z-dn.net/?f=x%20%3D%201%20-%20%20%5Csqrt%7B2%7D%20find%20%20%5C%3A%28x%20-%20%20%5Cfrac%7B1%7D%7Bx%
Nastasia [14]

Answer:

Step-by-step explanation:

\frac{1}{x}=\frac{1}{1-\sqrt{2}}\\\\=\frac{1*(1+\sqrt{2})}{(1-\sqrt{2})(1+\sqrt{2})}\\\\=\frac{1+\sqrt{2}}{1^{2}-(\sqrt{2})}\\\\=\frac{1+\sqrt{2}}{1-2}\\\\=\frac{1+\sqrt{2}}{-1}\\\\= -[1+\sqrt{2}]\\\\x-\frac{1}{x}=1-\sqrt{2}-(-[1+\sqrt{2}])\\\\=1-\sqrt{2}+1+\sqrt{2}\\\\=2

8 0
3 years ago
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