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valina [46]
3 years ago
5

Monica’s current office is a square. Her office in the company’s new building will be 3 feet wider and 5 feet longer

Mathematics
1 answer:
gregori [183]3 years ago
7 0

Answer:

<h2> The are of the office is 15ft^2</h2>

Step-by-step explanation:

This problem is on the mensuration of flat shapes, this time on the square shape of an office.

given that

length of the office= 5 feet

width of the office= 3 feet

we can continue y solving for the area of the office, seeing that the information given by the question is just enough for the area

area of square= lenght* width

area= 5*3

area= 15 ft^2

The are of the office is 15ft^2

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Question 19 of 25
Ronch [10]

The next step in the series will be Option B

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

A series is the sequence of object in pattern.

From the given figure it can be seen that the red and blue colour semicircle are shifting alternately , and the number series is increasing by 1

Therefore the next step is red on the right side , blue on the left side and no. 4 written on blue

Therefore Option B is the right answer.

To know about Series

brainly.com/question/15415793

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4 0
2 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

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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)

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Use n=3 to find the coefficient of x^n in the expansion of (3x+2)^5
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To find the undefined values, we need to know when the denominator is zero.
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If factors to: (x - 8)(x + 1)

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Answer:

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