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Vesna [10]
2 years ago
14

PLEASE HELP ME!!!! FOR 11 POINTS!!

Mathematics
1 answer:
Alex_Xolod [135]2 years ago
8 0

Answer:

1. .900+.020+.008, 2. .700+.030+.001, 3. 4.5, 4.56, 4.565656, 4.6

Step-by-step explanation:

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PLEASE HELP ME WITH THANKS :)
Maurinko [17]

Answer

12855.04 ft³

Step-by-step explanation:

regular Octagon consists of 8 congruent triangles

B = (13.28 x 11) / 2 x 8=584.32

V = BH = 584.32 x 22 = 12855.04

8 0
3 years ago
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A soup can in the shape of a cylinder is 4 inches tall and has a radius of 2 inches
svlad2 [7]
The volume is around 50.27

the surface area is: 75.4


the perimeter 25.13
7 0
3 years ago
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Jett has to sell carnival tickets worth at least $50. The price of a child ticket is $4, and the price of an adult ticket is $6.
Tpy6a [65]

Answer: 4x + 6y >/ 50

Step-by-step explanation:

The answer is A. 4x + 6y is greater than or equal to 50

7 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
HELP PLEASE I'LL GIVE 30 POINTS!!!​
jasenka [17]

Answer:

x = 45

Step-by-step explanation:

Straight line has 180 degrees; so if you add all the angles up:

2x + x + x = 180

4x = 180

x = 45 degrees

6 0
3 years ago
Read 2 more answers
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