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BartSMP [9]
4 years ago
13

115 + ( -215 ) + 315 + ( -415) + 515 + ... + 3915 + ( -4015 ) =

Mathematics
2 answers:
Likurg_2 [28]4 years ago
5 0
215 because you would add and subtract the numbers as you go them you come out with your answer
quester [9]4 years ago
4 0
215 expression calculator by mapa
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The sales tax rate for shopping malls in New York City is 8.75%.How much sales tax will a customer pay if they buy $396 worth of
yaroslaw [1]
396 x 0.875 = 346.5

396-346.5 = 49.5
The sale tax is $49.5
5 0
4 years ago
This comes with no question
mihalych1998 [28]

Answer:

It has to be 145 degrees.

Step-by-step explanation:

There would be 2 60's and one 95. 360-95-(60*2)=145.

8 0
3 years ago
daphne was given a riddle to solve the sum of two consecutive positive integers is 71 find the two positive integers
Anarel [89]
I believe I already explained this the numbers are 35 and 36 but if you need an explanation someone else posted the same question, but maybe it was you. Anyway, good luck!
8 0
3 years ago
For the function defined by f(t)=2-t, 0≤t<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
Test scores: Scores on a statistics exam had a mean of 75 with a standard deviation of 5. Scores on a calculus exam had a mean o
beks73 [17]

Answer:

CV for statistics exam = 15%

CV for calculus exam = 19%

Since the CV for calculus exam is higher, it has a greater spread relative to the mean than the statistics exam.

Step-by-step explanation:

To find coefficient variation we use the formula:

CV = (SD/mean) * 100

CV for the statistics exam:

where; SD= 5

mean= 75

CV = ( 5/75) *100

= 0.15 or 15%

CV for calculus exam

SD = 11

Mean= 58

CV= (11 /58) * 100

= 0.19 or 19%

8 0
3 years ago
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