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KengaRu [80]
3 years ago
13

1/9, -0.1, -2/12 in order

Mathematics
2 answers:
astra-53 [7]3 years ago
6 0

Answer:

Least to greatest: -2/12 , -0.1 , 1/9

Greatest to least: 1/9, -0.1, -2/12

Step-by-step explanation:

Change all of the numbers so that they are either fractions or decimals. Usually it is easier to change all the numbers to decimal.

Divide:

1/9 = ~0.111 (rounded)

-0.1 = -0.1

-2/12 = - ~0.167 (rounded)

Put the numbers in number order:

-~0.167 , -0.1 , ~0.111

-2/12 , -0.1 , 1/9

~

Basile [38]3 years ago
3 0

Answer:

-2/12, -0.1, 1/9

Step-by-step explanation:

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Does 3 (x-1) = 2x + 9 have one solution, no solution, or infinite solutions?
Allushta [10]

Answer:

one solution

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Step-by-step explanation:

3 (x-1) = 2x + 9

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subtract 2x from each side of the equation:

x - 3 = 9

add 3 to each side:

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Find what equals to x and what equals to y
Nuetrik [128]

Answer:

x=15√2  y=90 degrees.

Step-by-step explanation:

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2 years ago
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AleksAgata [21]

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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
1 year ago
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