All categories

2 years ago

Soon Yee was putting together her notebook after the first day of

1 answer:

4 0

Answer: Since her art and music sections each only had half the number of sheets of paper as a core subject, together the two sections had the same amount of paper as a core subject. Therefore, it is almost like her notebook had five core subjects, rather than four core subjects and two electives. If she divided the 200 sheets equally among the five core subjects, there would be 200 ÷ 5 = 40 sheets in each section. Now we can see that art would actually have half of this amount, or 20 sheets of paper.

You might be interested in

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

What is the distance between points (-6,8) and (-3,9)

SpyIntel [72]

The distance between points (-6,8) and (-3,9) is 9 units

6 0

3 years ago

What percentage of total hours was spent playing football

Savatey [412]

Answer:

You have to be specific. Do you have a picture that you can send to us so we can help you?

Step-by-step explanation:

3 0

3 years ago

How to write: 3x-9=7y in standard form.

NikAS [45]

Good evening from Canada!

Standard form: ax+by=c
3x-9=7y
3x-7y=9

Hope this helps!

5 0

3 years ago

In the figure below , ABCD is a square . Points are chosen on each pair of adjacent sides of ABCD to form 4 congruent right tria

Lostsunrise [7]

Answer:

5/9 of the area of square ABCD is shaded

Step-by-step explanation:

see the attached figure to better understand the problem

we know that

To find out what fraction of the area of square ABCD is shaded, divide the shaded area by the total area of square ABCD

step 1

Find out the area of square ABCD

The area of a square is

$A=b^{2}$

where

b is the length side of the square

we have

$b=(x+2x)=3x\ units$

$A=(3x)^{2}$

$A=9x^2\ units^{2}$

step 2

Find out the area of the 4 congruent right triangles

$A=4[\frac{1}{2}(x)(2x)]=4x^{2}\ units^2$

step 3

Find out the area of the shaded region

The area of the shaded region is equal to the area of square ABCD minus the area of the 4 congruent right triangles

$A=9x^2-4x^{2}=5x^{2}\ units^{2}$

step 4

Divide the shaded area by the total area of square ABCD

$\frac{5x^{2}}{9x^{2}} =\frac{5}{9}$

therefore

5/9 of the area of square ABCD is shaded

7 0

3 years ago