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3 years ago

Find the series shown.

Mathematics

1 answer:

solmaris [256]3 years ago

8 0

Answer:

C $11+13+15+17+...$

Step-by-step explanation:

Consider the series

$\sum\limits_{n=3}^{\infty}(2n+5)$

The nth term of series is $a_n=2n+5$

The bottom index tells you that n starts changing from 3, so

$a_3=2\cdor 3+5=11\\ \\a_4=2\cdot 4+5=13\\ \\a_5=2\cdot 5+5=15\\ \\a_6=2\cdot 6+5=17\\ \\...$

Thus, the sum of all terms is

$11+13+15+17+...$

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Which of the following is the weight of the cargo in a truck if 1/3 of the weight is 516 Ibs?

Anni [7]

Answer:

1548 lb

Step-by-step explanation:

Let the truck's weight be w.

Then (1/3)w = 516 lb

Solve for the truck's weight by multiplying both sides of this equation by 3:

w = 1548 lb (Answer B)

8 0

3 years ago

Suppose the claim size of an auto collision insurance, X, is uniformly distributed on the interval $1,000 to $10,000. What is th

timurjin [86]

Answer:

35.35%

Step-by-step explanation:

If there were no deductibles, the expected claim payment would be:

$E(X) = \frac{10,000 +1,000}{2} \\E(X) =\$5,500$

If the collision insurance claim is under $2,000, then the insurer would not pay anything, but if X > $2,000, then the insurer would pay X - $2,000. The new expected value is:

$E_2(X)=\frac{2,000-1,000}{10,000-1,000} *0+\frac{10,000-2,000}{10,000-1,000}*\frac{(2,000-2,000)+(10,000-2,000)}{2} \\E_2(X)=\frac{8}{9}*\frac{0+8,000}{2}\\ E_2(X)=\$3,555.56$

The percentage reduction on the claim payment is:

$P=(1-\frac{E_2(X)}{E(X)})*100 \\P=(1-\frac{3,555.56}{5,500})*100\\P=35.35\%$

There was a 35.35% reduction.

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.)