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Minchanka [31]
3 years ago
14

HELP!!!! What are the three ways to create samples spaces of all outcomes?

Mathematics
1 answer:
Kisachek [45]3 years ago
7 0

Answer:

The three most common ways to find a sample space are: To List All the Possible Outcomes. Create a Tree-Diagram. Use a Venn Diagram.  

For example, let's suppose we flip a coin and roll a die.

1. How many outcomes are possible?

2. What is the probability space?

3. Identify the events.

Step-by-step explanation:

sorry

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State the domain and range of the following relation<br> x^2+y^2=16
Vlada [557]

The domain and range is  [-4, 4] and [0, 4]

<h3 /><h3>What is Domain and range?</h3>

The domain of a function is the set of values that we are allowed to plug into our function.

The range of a function is the set of values that the function assumes.

x² + y² = 16

y = √16 - x²

For domain under root should not be negative quantity,

16 - x²≥0

16≥x²

So, x≤4 or x≥4

Thus, the domain is  [-4, 4]

Range:

y is maximum at x=0, y=4

y is minimum at x=4, y=0

Thus, range = [0, 4]

Learn more about domain and range here:

brainly.com/question/12751831

#SPJ1

8 0
2 years ago
For the function defined by f(t)=2-t, 0≤t&lt;1, sketch 3 periods and find:
Oksi-84 [34.3K]
The half-range sine series is the expansion for f(t) with the assumption that f(t) is considered to be an odd function over its full range, -1. So for (a), you're essentially finding the full range expansion of the function

f(t)=\begin{cases}2-t&\text{for }0\le t

with period 2 so that f(t)=f(t+2n) for |t| and integers n.

Now, since f(t) is odd, there is no cosine series (you find the cosine series coefficients would vanish), leaving you with

f(t)=\displaystyle\sum_{n\ge1}b_n\sin\frac{n\pi t}L

where

b_n=\displaystyle\frac2L\int_0^Lf(t)\sin\frac{n\pi t}L\,\mathrm dt

In this case, L=1, so

b_n=\displaystyle2\int_0^1(2-t)\sin n\pi t\,\mathrm dt
b_n=\dfrac4{n\pi}-\dfrac{2\cos n\pi}{n\pi}-\dfrac{2\sin n\pi}{n^2\pi^2}
b_n=\dfrac{4-2(-1)^n}{n\pi}

The half-range sine series expansion for f(t) is then

f(t)\sim\displaystyle\sum_{n\ge1}\frac{4-2(-1)^n}{n\pi}\sin n\pi t

which can be further simplified by considering the even/odd cases of n, but there's no need for that here.

The half-range cosine series is computed similarly, this time assuming f(t) is even/symmetric across its full range. In other words, you are finding the full range series expansion for

f(t)=\begin{cases}2-t&\text{for }0\le t

Now the sine series expansion vanishes, leaving you with

f(t)\sim\dfrac{a_0}2+\displaystyle\sum_{n\ge1}a_n\cos\frac{n\pi t}L

where

a_n=\displaystyle\frac2L\int_0^Lf(t)\cos\frac{n\pi t}L\,\mathrm dt

for n\ge0. Again, L=1. You should find that

a_0=\displaystyle2\int_0^1(2-t)\,\mathrm dt=3

a_n=\displaystyle2\int_0^1(2-t)\cos n\pi t\,\mathrm dt
a_n=\dfrac2{n^2\pi^2}-\dfrac{2\cos n\pi}{n^2\pi^2}+\dfrac{2\sin n\pi}{n\pi}
a_n=\dfrac{2-2(-1)^n}{n^2\pi^2}

Here, splitting into even/odd cases actually reduces this further. Notice that when n is even, the expression above simplifies to

a_{n=2k}=\dfrac{2-2(-1)^{2k}}{(2k)^2\pi^2}=0

while for odd n, you have

a_{n=2k-1}=\dfrac{2-2(-1)^{2k-1}}{(2k-1)^2\pi^2}=\dfrac4{(2k-1)^2\pi^2}

So the half-range cosine series expansion would be

f(t)\sim\dfrac32+\displaystyle\sum_{n\ge1}a_n\cos n\pi t
f(t)\sim\dfrac32+\displaystyle\sum_{k\ge1}a_{2k-1}\cos(2k-1)\pi t
f(t)\sim\dfrac32+\displaystyle\sum_{k\ge1}\frac4{(2k-1)^2\pi^2}\cos(2k-1)\pi t

Attached are plots of the first few terms of each series overlaid onto plots of f(t). In the half-range sine series (right), I use n=10 terms, and in the half-range cosine series (left), I use k=2 or n=2(2)-1=3 terms. (It's a bit more difficult to distinguish f(t) from the latter because the cosine series converges so much faster.)

5 0
4 years ago
What is the y-intercept of the line?​
Vaselesa [24]

Answer:

1

Step-by-step explanation:

The y-intercept is the point at which the line pases the y-axis.

7 0
3 years ago
Hi can someone help me with this its due tommorow!!!
bulgar [2K]

Answer:

1/12

Step-by-step explanation:

Is the answer I'm bro

4 0
3 years ago
Read 2 more answers
the campsite shop sells boxes of funshine cereal the base of each box is 180mm ×60mm rectangle. the shelf where the boxes are di
kvv77 [185]
-the shower block as an area of 144m^2-the length of the shower block is 24m
Area= length x width 
Plug in the values you know. 
144= length*24
Divide both sides by 24. 
length=6
The width of the shower block is 6m. 
I hope this helps!


6 0
3 years ago
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