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2 years ago

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