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3 years ago

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Mirna earned $120 baby-sitting during the spring break. She needs to save $90 for the German Club trip. She wants to spend the r

emainder of the money shopping. Use s for the amount of money Mirna can spend. Enter your inequality in the most simplified form.

1 answer:

Verizon [17]3 years ago

3 0

Answer:

$30

Step-by-step explanation:

s + 90 = 120

s = 120 - 90

s = 30

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A line passes through the points (–4,4) and (4,3). What is its equation in point-slope form?

larisa [96]

M=-1/8 or -0.125 hope this helped

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2 years ago

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How large a sample should be selected to provide a 95% confidence interval with a margin of error of 6? Assume that the populati

marshall27 [118]

Using the z-distribution, it is found that a sample of 171 should be selected.

<h3>What is a z-distribution confidence interval?</h3>

The confidence interval is:

$\overline{x} \pm z\frac{\sigma}{\sqrt{n}}$

The margin of error is:

$M = z\frac{\sigma}{\sqrt{n}}$

In which:

$\overline{x}$ is the sample mean.

z is the critical value.

n is the sample size.

$\sigma$ is the standard deviation for the population.

For this problem, the parameters are:

$z = 1.96, \sigma = 40, M = 6$

Hence we solve for n to find the needed sample size.

$M = z\frac{\sigma}{\sqrt{n}}$

$6 = 1.96\frac{40}{\sqrt{n}}$

$6\sqrt{n} = 40 \times 1.96$

$\sqrt{n} = \frac{40 \times 1.96}{6}$

$(\sqrt{n})^2 = \left(\frac{40 \times 1.96}{6}\right)^2$

n = 170.7.

Rounding up, a sample of 171 should be selected.

More can be learned about the z-distribution at brainly.com/question/25890103

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1 year ago

Which of the following could be the graph of this equation?

Anna35 [415]

Answer: OPTION C.

Step-by-step explanation:

Given a function f(x), the range of the inverse of f(x) will be the domain of the function f(x) and the range of the domain of f(x) will be the range of the inverse function.

For example, if the point (2,1) belongs to f(x), then the point (1,2) belongs to the inverse of f(x).

Observe that in the graph of the function f(x) the point (-3,1) belongs to the function, then the point (1,-3) must belong to the inverse function.

Therefore, you need to search the option that shown the graph wich contains the point (1,-3).

Observe that the Domain f(x) is (-∞,0) then the range of the inverse function must be (-∞,0).

This is the graph of the option C.

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3 years ago

If $190 is invested at an interest rate of 11% per year and is compounded continuously, how much will the investment be worth in

dangina [55]

Answer:

$295.01

Step-by-step explanation:

Using the compounding continuously formula, we are looking for A. We have that P = 190; r = .11 (always use the decimal form of the rate!); and t = 4. Filling in we have:

$A=190e^{(.11)(4)}$

Simplifying that multiplication:

$A=190e^{.44}$ .

On your calculator, raise e to the .44 power to get

A = 190(1.552707219) and

A = $295.01

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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

3 years ago