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vlada-n [284]
2 years ago
13

Find the area of the unshaded portion of

Mathematics
1 answer:
Nostrana [21]2 years ago
8 0

Answer: 3055.25

Step-by-step explanation:

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Y=15-4y what is the answer Y=___
Dominik [7]
Y=15 -4y
+4y +4y
5y=15
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5 5
y=3
6 0
3 years ago
Suppose that the length of a side of a cube X is uniformly distributed in the interval 9
Nastasia [14]

Answer:

f(v) = \left \{ {{\frac{1}{3}v^{-\frac{2}{3}}\ 9^3 \le v \le 10^3} \atop {0, elsewhere}} \right.

Step-by-step explanation:

Given

9 < x < 10 --- interval

Required

The probability density of the volume of the cube

The volume of a cube is:

v = x^3

For a uniform distribution, we have:

x \to U(a,b)

and

f(x) = \left \{ {{\frac{1}{b-a}\ a \le x \le b} \atop {0\ elsewhere}} \right.

9 < x < 10 implies that:

(a,b) = (9,10)

So, we have:

f(x) = \left \{ {{\frac{1}{10-9}\ 9 \le x \le 10} \atop {0\ elsewhere}} \right.

Solve

f(x) = \left \{ {{\frac{1}{1}\ 9 \le x \le 10} \atop {0\ elsewhere}} \right.

f(x) = \left \{ {{1\ 9 \le x \le 10} \atop {0\ elsewhere}} \right.

Recall that:

v = x^3

Make x the subject

x = v^\frac{1}{3}

So, the cumulative density is:

F(x) = P(x < v^\frac{1}{3})

f(x) = \left \{ {{1\ 9 \le x \le 10} \atop {0\ elsewhere}} \right. becomes

f(x) = \left \{ {{1\ 9 \le x \le v^\frac{1}{3} - 9} \atop {0\ elsewhere}} \right.

The CDF is:

F(x) = \int\limits^{v^\frac{1}{3}}_9 1\  dx

Integrate

F(x) = [v]\limits^{v^\frac{1}{3}}_9

Expand

F(x) = v^\frac{1}{3} - 9

The density function of the volume F(v) is:

F(v) = F'(x)

Differentiate F(x) to give:

F(x) = v^\frac{1}{3} - 9

F'(x) = \frac{1}{3}v^{\frac{1}{3}-1}

F'(x) = \frac{1}{3}v^{-\frac{2}{3}}

F(v) = \frac{1}{3}v^{-\frac{2}{3}}

So:

f(v) = \left \{ {{\frac{1}{3}v^{-\frac{2}{3}}\ 9^3 \le v \le 10^3} \atop {0, elsewhere}} \right.

8 0
2 years ago
Brian has 40000 to invest in a mix of corporate and municipal bonds. The corporate bond pays 10% simple annual interest and the
jonny [76]

Answer:

Sum of money invested in corporate bonds = 30,000

Step-by-step explanation:

Total sum = 40,000

The rate of interest for corporate bonds = 10 % = 0.1

The rate of interest for municipal bonds = 6 % = 0.06

Total interest = 3600

Let sum of money invested in corporate bonds = x

The sum of money invested in municipal bonds = 40000 - x

x × 0.1 × 1 + (40000 - x) × 0.06 × 1 = 3600

(0.1- 0.06)x + 2400 = 3600

0.04 x = 1200

x = 30,000

Since x =  sum of money invested in corporate bonds

So  sum of money invested in corporate bonds = 30000

4 0
3 years ago
the mean of 10 students is 16 . When a student is added, the mean becomes 18. Find the age of the student added​
MakcuM [25]

Answer:

Step-by-step explanation:

5 0
2 years ago
Student had 500 milliliters of water in a water bottle. She drank 25% of the water before soccer practice. After practice, she d
sesenic [268]
250 milliters of water remains
8 0
2 years ago
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