All categories

3 years ago

A company makes two models of television.

Mathematics

1 answer:

VLD [36.1K]3 years ago

5 0

Answer:

16/25

Step-by-step explanation:

length of screen B=44*(5/4)=55cm

and its width=32*(5/4)=40cm

so, screen B had dimensions 55cm by 40 cm

screen A area=44*32 cm sqr

screen B area=55*40 cm sqr

so, model A screen area/model B screen area

=(44*32)/(55*40)

=16/25

You might be interested in

.86971 rounded to the 10th

skelet666 [1.2K]

Answer:

Step-by-step explanation:

If the number behind that place is 5 or greater than you round up one but if it is 4 or below than you keep the number the same.

7 0

3 years ago

39.76 divided by 2.8

denpristay [2]

Answer:

14.2

Step-by-step explanation:

6 0

3 years ago

The number of buttons an industrial button machine can manufacture each hour, y, varies directly with the number of active hours

-Dominant- [34]

Answer:

lol

Step-by-step explanation:

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

‼️‼️any help I would appreciate

mylen [45]

Hey there!

In an isosceles triangle the 2 base angles are congruent, therefore angle B would measure 50 degrees just like the measure of angle A.

Thanks for using Brainly!

5 0

3 years ago