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3 years ago

Solve A=1\2h(b+c) for b

Mathematics

1 answer:

mr_godi [17]3 years ago

6 0

B=2A/h - c
i hope that is correct cuz there are no values

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What is the missing number in this pattern?<br> 0, 2, 6, ----, 30, 62

mario62 [17]

Answer:

Step-by-step explanation:

so the pattern is they add 2 then they add 4 then they add 8 then the add 16 basically they are double the number being added each number so the missing number is 14

5 0

3 years ago

Read 2 more answers

A cruise left port A and traveled towards port B 225 km away. After 1.5 hours of travel, the cruise was stopped for a half an ho

Softa [21]

Answer:

Step-by-step explanation:

v = the original speed of the cruise

d = 225 km the distance passed

t = d/v = 225/v the total time of the travel

v*1.5 + (v + 10)(d/v - 1.5 - 0.5) = d (using the formula distance = speed*time)

1.5v + (v + 10)(225/v - 2) = 225

Look at the images to see the solved equation

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

3 years ago

If there are 26 students in each third-grade class and 3 third-grade classes in the school, how many students will attend an ass

ddd [48]

Answer:

Step-by-step explanation:

26 • 3 = 78 - 4 = 74

8 0

3 years ago

Find the perimeter and area

stepladder [879]

Answer:

32 times square root of 3. is the perimeter, area is 48 times square root of 3

Step-by-step explanation:

4 0

3 years ago