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sveticcg [70]
2 years ago
13

For the sequence a n ​ =a n−1 ​ +a n−2 ​ and a 1 ​ =2,a 2 ​ =3 its first term is its second term is its third term is its fourth

term is its fifth term is ​ quence a n ​ =a n−1 ​ +a n−2 ​ and a 1 ​ =2,a 2 ​ =3 its first term is its second term is its third term is its fourth term is its fifth term is ​
Mathematics
1 answer:
taurus [48]2 years ago
5 0

A function assigns the value of each element of one set to the other specific element of another set. The third, fourth, and fifth terms of the sequence are 5, 8, and 13 respectively

<h3>What is a Function?</h3>

A function assigns the value of each element of one set to the other specific element of another set.

Given the nth term of the sequence is given by the formula,

a_n = a_{(n-1)}+a_{(n-2)}

Also, the first term and the second term of the sequence are 2 and 3 respectively, therefore, the other terms of the sequence are,

a_3 = a_{(3-1)}+a_{(3-2)}\\\\a_3 = a_2 + a_1\\\\a_3 = 3 + 2\\\\a_3 = 5

a_4 = a_{(4-1)}+a_{(4-2)}\\\\a_4 = a_3 + a_2\\\\a_4 = 5 + 3\\\\a_4 = 8

a_5 = a_{(5-1)}+a_{(5-2)}\\\\a_5 = a_4 + a_3\\\\a_5 = 8 + 5\\\\a_5 = 13

Hence, the third, fourth, and fifth terms of the sequence are 5, 8, and 13 respectively.

Learn more about Function:

brainly.com/question/5245372

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Evaluate 5x-[21-(3-11)x] when x=-2​
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3 0
3 years ago
X and y are normal random variables with e(x) = 2, v(x) = 5, e(y) = 6, v(y) = 8 and cov(x,y)=2. determine the following: e(3x 2y
andriy [413]

The result for the given normal random variables are as follows;

a. E(3X + 2Y) = 18

b. V(3X + 2Y) = 77

c. P(3X + 2Y < 18) = 0.5

d. P(3X + 2Y < 28) = 0.8729

<h3>What is normal random variables?</h3>

Any normally distributed random variable having mean = 0 and standard deviation = 1 is referred to as a standard normal random variable. The letter Z will always be used to represent it.

Now, according to the question;

The given normal random variables are;

E(X) = 2, V(X) = 5, E(Y) = 6, and V(Y) = 8.

Part a.

Consider E(3X + 2Y)

\begin{aligned}E(3 X+2 Y) &=3 E(X)+2 E(Y) \\&=(3) (2)+(2)(6 )\\&=18\end{aligned}

Part b.

Consider V(3X + 2Y)

\begin{aligned}V(3 X+2 Y) &=3^{2} V(X)+2^{2} V(Y) \\&=(9)(5)+(4)(8) \\&=77\end{aligned}

Part c.

Consider P(3X + 2Y < 18)

A normal random variable is also linear combination of two independent normal random variables.

3 X+2 Y \sim N(18,77)

Thus,

P(3 X+2 Y < 18)=0.5

Part d.

Consider P(3X + 2Y < 28)

Z=\frac{(3 X+2 Y-18)}{\sqrt{77}}

\begin{aligned} P(3X + 2Y < 28)&=P\left(\frac{3 X+2 Y-18}{\sqrt{77}} < \frac{28-18}{\sqrt{77}}\right) \\&=P(Z < 1.14) \\&=0.8729\end{aligned}

Therefore, the values for the given normal random variables are found.

To know more about the normal random variables, here

brainly.com/question/23836881

#SPJ4

The correct question is-

X and Y are independent, normal random variables with E(X) = 2, V(X) = 5, E(Y) = 6, and V(Y) = 8. Determine the following:

a. E(3X + 2Y)

b. V(3X + 2Y)

c. P(3X + 2Y < 18)

d. P(3X + 2Y < 28)

8 0
2 years ago
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