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kherson [118]
3 years ago
10

You are given that p(event

Mathematics
1 answer:
likoan [24]3 years ago
7 0
You do not.

We have not been told if the events are mutually exclusive or not.
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This graph shows a proportional relationship. What is the constant of proportionality? Enter your answer as a ratio in simplifie
Juli2301 [7.4K]

Answer:

k=\frac{5}{4}

Step-by-step explanation:

we know that

A relationship between two variables, x, and y, represent a proportional variation if it can be expressed in the form k=\frac{y}{x} or y=kx

In a proportional relationship the constant of proportionality k is equal to the slope m of the line and the line passes through the origin

In this problem we have that

The line pass through the points

(0,0)\ and\ (\frac{2}{5},\frac{1}{2})  

<em>Find the value of the constant of proportionality k</em>

For x=2/5, y=1/2  

substitute

k=\frac{y}{x}

k=\frac{1}{2}:\frac{2}{5}=\frac{5}{4}

4 0
3 years ago
Is (0,0) a solution of the graphed inequality?
sweet [91]
No because it is not within the shaded area
5 0
3 years ago
Read 2 more answers
"Three times a number increased by eleven is seventeen" translates to which of the following equations?
ololo11 [35]
3x + 11 = 17 i think
7 0
3 years ago
Read 2 more answers
Calculus Problem
Roman55 [17]

The two parabolas intersect for

8-x^2 = x^2 \implies 2x^2 = 8 \implies x^2 = 4 \implies x=\pm2

and so the base of each solid is the set

B = \left\{(x,y) \,:\, -2\le x\le2 \text{ and } x^2 \le y \le 8-x^2\right\}

The side length of each cross section that coincides with B is equal to the vertical distance between the two parabolas, |x^2-(8-x^2)| = 2|x^2-4|. But since -2 ≤ x ≤ 2, this reduces to 2(x^2-4).

a. Square cross sections will contribute a volume of

\left(2(x^2-4)\right)^2 \, \Delta x = 4(x^2-4)^2 \, \Delta x

where ∆x is the thickness of the section. Then the volume would be

\displaystyle \int_{-2}^2 4(x^2-4)^2 \, dx = 8 \int_0^2 (x^2-4)^2 \, dx \\\\ = 8 \int_0^2 (x^4-8x^2+16) \, dx \\\\ = 8 \left(\frac{2^5}5 - \frac{8\times2^3}3 + 16\times2\right) = \boxed{\frac{2048}{15}}

where we take advantage of symmetry in the first line.

b. For a semicircle, the side length we found earlier corresponds to diameter. Each semicircular cross section will contribute a volume of

\dfrac\pi8 \left(2(x^2-4)\right)^2 \, \Delta x = \dfrac\pi2 (x^2-4)^2 \, \Delta x

We end up with the same integral as before except for the leading constant:

\displaystyle \int_{-2}^2 \frac\pi2 (x^2-4)^2 \, dx = \pi \int_0^2 (x^2-4)^2 \, dx

Using the result of part (a), the volume is

\displaystyle \frac\pi8 \times 8 \int_0^2 (x^2-4)^2 \, dx = \boxed{\frac{256\pi}{15}}}

c. An equilateral triangle with side length s has area √3/4 s², hence the volume of a given section is

\dfrac{\sqrt3}4 \left(2(x^2-4)\right)^2 \, \Delta x = \sqrt3 (x^2-4)^2 \, \Delta x

and using the result of part (a) again, the volume is

\displaystyle \int_{-2}^2 \sqrt 3(x^2-4)^2 \, dx = \frac{\sqrt3}4 \times 8 \int_0^2 (x^2-4)^2 \, dx = \boxed{\frac{512}{5\sqrt3}}

7 0
2 years ago
Excel - How to graph different APRs using one formula for an excel beginner? Image Attached
Levart [38]

Answer:

Please see the attachments. For reasons I cannot fathom, the Brainly censor seems to think there are unapproved words contained in this text, so I cannot post it as text. Among other things, that means you probably will not be able to copy and paste the formulas into your spreadsheet—you will have to type them.

I apologize if you are offended by any of the words contained in this answer. I hope you will take them in the intended context—a math problem about the use of Excel.

3 0
3 years ago
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