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Sidana [21]
3 years ago
13

James estimated the total cost of groceries to be $50. The actual cost of the groceries is $65. What is the percent error? Round

your answer to the nearest whole percent. Enter your answer in the box.
_____%
Mathematics
2 answers:
Brrunno [24]3 years ago
8 0

I just took the quiz and it was 23.

Alex777 [14]3 years ago
7 0

To get percent error you take actual-predicted and divide by the actual then multiply by 100 to get the percent.

65-50=15

15/60=0.25

0.25 times 100 =25%

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What is the 8th term of the sequence 1, 3, 9, 27, 81,... .
slega [8]

9514 1404 393

Answer:

  2187

Step-by-step explanation:

This geometric sequence has a first term of 1 and a common ratio of 3. Its general term can be written as ...

  an = a1·r^(n-1)

  an = 3^(n-1)

Then the 8th term is ...

  a8 = 3^(8-1) = 2187

4 0
2 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
3 years ago
What is the mean of the numbers?
velikii [3]

Answer:

1.6666666666666...

Step-by-step explanation:

10/6=5/3=1.6666666...

6 0
3 years ago
6. What is the result of multiplying the digit in the hundredths place by
Setler [38]

Answer:

B.9

Step-by-step explanation:

9*1=9

6 0
3 years ago
Michelle is rolling two six-sided dice, numbered one through six. What is the probability that the sum of her rolls is 8?
Verdich [7]
The probability that the sum of Michelle's rolls is 4 is 0.083
∴ P(A)=0.083
Step-by-step explanation:
Given that Michelle is rolling two six-sided dice, numbered one through six.
To find the probability that the sum of her rolls is 4:

∴ n(s)=36
Let P(A) be the probability that the sum of her rolls is 4
Then the possible rolls with sums of 4 can be written as

n(A)=3
The probability that the sum of her rolls is 4 is given by



=0.083
∴ P(A)=0.083
∴ the probability that the sum of Michelle's rolls is 4 is 0.083

-please give brainliest if correct!
3 0
3 years ago
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