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riadik2000 [5.3K]
3 years ago
8

If the probability of getting a particular result in an experiment is 75.3%. What is the probability of not getting that result?

Mathematics
1 answer:
loris [4]3 years ago
5 0
The answer would be C.
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36​% of college students say they use credit cards because of the rewards program. You randomly select 10 college students and a
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5 0
3 years ago
a solid consists of cylinder surmounted by a right circular cone. The height of the cone is 2h. If the volume of the solid is 3
Mademuasel [1]

Answer:

height of cylinder = 4/3 h

Step-by-step explanation:

The solid has a cylinder surmounted with a cone .Therefore, the volume of the solid is the sum of the cone and the cylinder.

volume of the solid = volume of cylinder + volume of cone

volume of the solid = πr²h + 1/3πr²h

let

height of the cylinder = H

recall

the height of the cone = 2h

volume of the solid = πr²h + 1/3πr²h

3(1/3πr²2h) = πr²H + 1/3πr²2h

2πr²h = πr²H + 2/3 πr²h

πr²(2h) = πr²(H + 2/3 h)

divide both sides by πr²

2h = H + 2/3 h

2h - 2/3h = H

H = 6h - 2h/3

H = 4/3 h

height of cylinder = 4/3 h

8 0
3 years ago
Mom’s muffin recipe uses 10 ounces of berries for 2 dozen muffins grandma’s muffin recipe uses 12 ounces of berries for 3 dozen
Rudiy27

Answer:

can u add a photo for me to see

3 0
3 years ago
Work out the circumference of this circle.
Ksenya-84 [330]
Formula for circumference of circle:
C=2πr
here radius, r=4 cm and π=3.142
Put values in formula.
C=2(3.142)(4)
C=25.1 cm (Rounded to 1 decimal place)

Answer: Circumference of the circle is 25.1 cm (Rounded to 1 decimal place)
6 0
3 years ago
Can I get help with finding the Fourier cosine series of F(x) = x - x^2
trapecia [35]
Assuming you want the cosine series expansion over an arbitrary symmetric interval [-L,L], L\neq0, the cosine series is given by

f_C(x)=\dfrac{a_0}2+\displaystyle\sum_{n\ge1}a_n\cos nx

You have

a_0=\displaystyle\frac1L\int_{-L}^Lf(x)\,\mathrm dx
a_0=\dfrac1L\left(\dfrac{x^2}2-\dfrac{x^3}3\right)\bigg|_{x=-L}^{x=L}
a_0=\dfrac1L\left(\left(\dfrac{L^2}2-\dfrac{L^3}3\right)-\left(\dfrac{(-L)^2}2-\dfrac{(-L)^3}3\right)\right)
a_0=-\dfrac{2L^2}3

a_n=\displaystyle\frac1L\int_{-L}^Lf(x)\cos nx\,\mathrm dx

Two successive rounds of integration by parts (I leave the details to you) gives an antiderivative of

\displaystyle\int(x-x^2)\cos nx\,\mathrm dx=\frac{(1-2x)\cos nx}{n^2}-\dfrac{(2+n^2x-n^2x^2)\sin nx}{n^3}

and so

a_n=-\dfrac{4L\cos nL}{n^2}+\dfrac{(4-2n^2L^2)\sin nL}{n^3}

So the cosine series for f(x) periodic over an interval [-L,L] is

f_C(x)=-\dfrac{L^2}3+\displaystyle\sum_{n\ge1}\left(-\dfrac{4L\cos nL}{n^2L}+\dfrac{(4-2n^2L^2)\sin nL}{n^3L}\right)\cos nx
4 0
3 years ago
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