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DochEvi [55]
3 years ago
11

If you ask the first 50 females who enter a restaurant to complete a survey about their experience, who is the sample?

Mathematics
2 answers:
balandron [24]3 years ago
8 0

Answer:

the first girl is the answer

Anettt [7]3 years ago
6 0

Answer:

the first girl

Step-by-step explanation:

because the first girl to take the survey is a sample but the rest that follows are just extras(can i get brainliest pls)

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Round 14494 to the nearest thousand feet
klasskru [66]
14,000 because 494 would round down since it is not above 500.

7 0
3 years ago
Does this graph represent a function? <br>A=Yes <br>B=No​
RSB [31]

Answer: no

Step-by-step explanation:

it doesn't pass the vertical line test

8 0
3 years ago
Read 2 more answers
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
3 years ago
Find the value of x number on top is 6 and other is 9
faltersainse [42]

Answer:

x = - 12, 3

Step-by-step explanation:

By tangent secant theorem:

x(x + 9) =  {6}^{2}  \\  \\  {x}^{2}  + 9x = 36 \\  \\  {x}^{2}  + 9x - 36 = 0 \\  \\  {x}^{2}  + 12x - 3x - 36 = 0 \\  \\ x(x + 12) - 3(x + 12) = 0 \\  \\ (x + 12)(x - 3) = 0 \\  \\ x + 12 = 0, \:  \: x - 3 = 0 \\  \\ x =  - 12, \:  \: x = 3 \\  \\

5 0
2 years ago
can someone please answer the first four for me? Also can someone please explain it to me, because I don't fully understand​
pentagon [3]

1                           2

a^{2} +b^{2} =c^{2} \\9^{2} + 13^{2} =c^{2}\\81+169=c^{2} \\250=c^{2} \\125=c        a^{2} +b^{2} =c^{2} \\16^{2} +18^{2} =c^{2} \\256+324=c^{2} \\580=c^{2} \\290=c

3                          4

a^{2} +b^{2} =c^{2} \\19^{2} +14^{2} =c^{2} \\361+196=c^{2} \\557=c^{2} \\278.5=c      a^{2}+ b^{2}=c^{2} \\10^{2} +11^{2}=c^{2} \\100+121=c^{2} \\221=c^{2} \\110.5=c

6 0
3 years ago
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