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Stella [2.4K]
3 years ago
13

If that is = a, find the value of a + 1 over a

1" title="3 - \sqrt{5} " alt="3 - \sqrt{5} " align="absmiddle" class="latex-formula">
​
Mathematics
1 answer:
Doss [256]3 years ago
6 0

Answer:

7+√5/4

Step-by-step explanation:

if that is = a, find the value of a + 1 over a

​Given that a = 3 - √5

a+1/1 = (3-√5)+1/3 - √5

4-√5/3 - √5

Rationalize

4-√5/3 - √5 * 3+√5/3 + √5

= 12 +4√5-3√5-√25/9-5

= 12+√5-5/4

= 7+√5/4

Hence the requred answer is  7+√5/4

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Svetradugi [14.3K]

the answer is a) because its the numerator is bigger than the denominator

8 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
4x +6y =0
PSYCHO15rus [73]
Here, x - 2y = 14
x = 14 + 2y

Now, substitute this value into first equation, 
4x + 6y = 0
4(14 + 2y) + 6y = 0
56 + 8y + 6y = 0
14y = -56
y = -56/14
y = -4

Substitute it into second equation, 
x = 14 + 2(-4)
x = 14 - 8
x = 6

In short, Your Answer would be: (6, -4)

Hope this helps!
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