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Nataly_w [17]
3 years ago
5

When doing currency exchange why is it better to carry the transaction out to five or six places rather than round to cents. Wha

t am I losing in the transaction? Keep in mind that we are transferring millions of dollars.
Mathematics
2 answers:
k0ka [10]3 years ago
6 0
When rounding earlier, it could be screwing you out of an extra few cents that you might eventually need for tax
soldier1979 [14.2K]3 years ago
4 0
Because exchanging is not always exact so rounding further back will add up per dollar
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Calculus hw, need help asap with steps.
nikdorinn [45]

Answers are in bold

S1 = 1

S2 = 0.5

S3 = 0.6667

S4 = 0.625

S5 = 0.6333

=========================================================

Explanation:

Let f(n) = \frac{(-1)^{n+1}}{n!}

The summation given to us represents the following

\displaystyle \sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n!}=\sum_{n=1}^{\infty} f(n)\\\\\\\displaystyle \sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n!}=f(1) + f(2)+f(3)+\ldots\\\\

There are infinitely many terms to be added.

-------------------

The partial sums only care about adding a finite amount of terms.

The partial sum S_1 is the sum of the first term and nothing else. Technically it's not really a sum because it doesn't have any other thing to add to. So we simply say S_1 = f(1) = 1

I'm skipping the steps to compute f(1) since you already have done so.

-------------------

The second partial sum is when things get a bit more interesting.

We add the first two terms.

S_2 = f(1)+f(2)\\\\S_2 = 1+(-\frac{1}{2})\\\\S_2 = \frac{1}{2}\\\\S_2 = 0.5\\\\\\

The scratch work for computing f(2) is shown in the diagram below.

-------------------

We do the same type of steps for the third partial sum.

S_3 = f(1)+f(2)+f(3)\\\\S_3 = 1+(-\frac{1}{2})+\frac{1}{6}\\\\S_3 = \frac{2}{3}\\\\S_3 \approx 0.6667\\\\\\

The scratch work for computing f(3) is shown in the diagram below.

-------------------

Now add the first four terms to get the fourth partial sum.

S_4 = f(1)+f(2)+f(3)+f(4)\\\\S_4 = 1+(-\frac{1}{2})+\frac{1}{6}-\frac{1}{24}\\\\S_4 = \frac{5}{8}\\\\S_4 \approx 0.625\\\\\\

As before, the scratch work for f(4) is shown below.

I'm sure you can notice by now, but the partial sums are recursive. Each new partial sum builds upon what is already added up so far.

This means something like S_3 = S_2 + f(3) and S_4 = S_3 + f(4)

In general, S_{n+1} = S_{n} + f(n+1) so you don't have to add up all the first n terms. Simply add the last term to the previous partial sum.

-------------------

Let's use that recursive trick to find S_5

S_5 = [f(1)+f(2)+f(3)+f(4)]+f(5)\\\\S_5 = S_4 + f(5)\\\\S_5 = \frac{5}{8} + \frac{1}{120}\\\\S_5 = \frac{19}{30}\\\\S_5 \approx 0.6333

The scratch work for f(5) is shown below.

7 0
2 years ago
What is the answerrrrrrr
sergeinik [125]

Answer:

the answer is 118

Step-by-step explanation:

since one pound is equal to 16 ounces you just convert

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3 years ago
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What is 20% of $ a dollar and 50 Cent
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1.50x.20=.30
It's 30 cents
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There are 3 in. of snow on the ground when it begins to snow 0.5 in./h . Which linear equation represents the total depth of the
dexar [7]

There are 3 in. of snow on the ground when it begins to snow 0.5 in./h.

Initial depth of snow = 3 in.

it begins to snow 0.5 in./h. The constant rate of snow is 0.5. So slope = 0.5

Let x be the number of hours

y be the total depth of the snow

To frame linear equation we use y=mx+b

where m is the slope and b is the y intercept (initial depth of snow)

We know m=0.5  and b=3

Replace it in the equation

y = 0.5x + 3

The linear equation that represents the total depth of the snow(y), in inches, after x hours

is y= 0.5x + 3


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