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3 years ago

Change each of these test scores to:

Mathematics

1 answer:

Veseljchak [2.6K]3 years ago

4 0

Answer:

a) 21/50 -> 42%

b) 7/10 ->70%

c) 3/5 -> 60%

d) 59/100 -> 59%

Step-by-step explanation:

You might be interested in

81 + 7x² +8yt-sx<br> if x = 2 what is the answer

lilavasa [31]

ANSWER:

109 + 8yt - 2s

WORKING:

81 + 7x² + 8yt - sx

81 + 7(2)² + 8yt - s(2)

81 + 7(4) + 8yt - 2s

81 + 28 + 8yt - 2s

109 + 8yt - 2s

6 0

2 years ago

Do the ratios 30/24 and 20/16 form a proportion?

BartSMP [9]

Answer:

Yes.

Step-by-step explanation:

Reduce both fractions.

30/24 = 5/4
20/16= 5/4

6 0

3 years ago

Suppose we are testing the null hypothesis H 0 mu=20 and the alternative H iu =20 , normal population with sigma=6 A random samp

cupoosta [38]

Answer:

The value of the test statistic is $z = -1.5$

Step-by-step explanation:

The formula for the test statistic is:

$z = \frac{X - \mu}{\frac{\sigma}{\sqrt{n}}}$

In which X is the statistic, $\mu$ is the mean, $\sigma$ is the standard deviation and n is the number of observations.

In this problem, we have that:

$\mu = 20, X = 17, \sigma = 6, n = 9$

$z = \frac{X - \mu}{\frac{\sigma}{\sqrt{n}}}$

$z = \frac{17 - 20}{\frac{6}{\sqrt{3}}}$

$z = -1.5$

The value of the test statistic is $z = -1.5$

8 0

4 years ago

A ____________________ is a measurement of the change in variables over a specific period of time.

Montano1993 [528]

Answer:

rate of change

Step-by-step explanation:

The rate of change (ROC) is the speed at which a variable changes over a specific period of time.

5 0

3 years ago

Solve using Fourier series.

Olin [163]

With $2L=\pi$ , the Fourier series expansion of $f(x)$ is

$\displaystyle f(x)\sim\frac{a_0}2+\sum_{n\ge1}a_n\cos\dfrac{n\pi x}L+\sum_{n\ge1}b_n\sin\dfrac{n\pi x}L$
$\displaystyle f(x)\sim\frac{a_0}2+\sum_{n\ge1}a_n\cos2nx+\sum_{n\ge1}b_n\sin2nx$

where the coefficients are obtained by computing

$\displaystyle a_0=\frac1L\int_0^{2L}f(x)\,\mathrm dx$
$\displaystyle a_0=\frac2\pi\int_0^\pi f(x)\,\mathrm dx$

$\displaystyle a_n=\frac1L\int_0^{2L}f(x)\cos\dfrac{n\pi x}L\,\mathrm dx$
$\displaystyle a_n=\frac2\pi\int_0^\pi f(x)\cos2nx\,\mathrm dx$

$\displaystyle b_n=\frac1L\int_0^{2L}f(x)\sin\dfrac{n\pi x}L\,\mathrm dx$
$\displaystyle b_n=\frac2\pi\int_0^\pi f(x)\sin2nx\,\mathrm dx$

You should end up with

$a_0=0$
$a_n=0$
(both due to the fact that $f(x)$ is odd)
$b_n=\dfrac1{3n}\left(2-\cos\dfrac{2n\pi}3-\cos\dfrac{4n\pi}3\right)$

Now the problem is that this expansion does not match the given one. As a matter of fact, since $f(x)$ is odd, there is no cosine series. So I'm starting to think this question is missing some initial details.

One possibility is that you're actually supposed to use the even extension of $f(x)$ , which is to say we're actually considering the function

$\varphi(x)=\begin{cases}\frac\pi3&\text{for }|x|\le\frac\pi3\\0&\text{for }\frac\pi3$

and enforcing a period of $2L=2\pi$ . Now, you should find that

$\varphi(x)\sim\dfrac2{\sqrt3}\left(\cos x-\dfrac{\cos5x}5+\dfrac{\cos7x}7-\dfrac{\cos11x}{11}+\cdots\right)$

The value of the sum can then be verified by choosing $x=0$ , which gives

$\varphi(0)=\dfrac\pi3=\dfrac2{\sqrt3}\left(1-\dfrac15+\dfrac17-\dfrac1{11}+\cdots\right)$
$\implies\dfrac\pi{2\sqrt3}=1-\dfrac15+\dfrac17-\dfrac1{11}+\cdots$

as required.

5 0

3 years ago