The side of cube is 3m 30cm long, find out its volume

WHY DOES NO ONE ANSWER MY QUESTIONS?? Its kinda annoying that i answer others' questions but no one answers mine. Well if anyone

Alex17521 [72]

Answer:

The test was not statistically significant because if the null hypothesis is true, one could expect to get a test statistic at least as extreme as that observed 21% of the time.

Step-by-step explanation:

This is the right answer,since this result is only observed 21% of the time, so in general it's not significant, so the first 2 are eliminated. The 2 x 0.21 doesn't matter since, the percent is 21% not 42%, so it doesn't even matter. The last question we eliminate is:"The test was not statistically significant because if the null hypothesis is true, one could expect to get a test statistic at least as extreme as that observed 79% of the time" 79% of the time is a pretty good amount to say it's significant, but it only says 21% of the time.So, it leaves us with:The test was not statistically significant because if the null hypothesis is true, one could expect to get a test statistic at least as extreme as that observed 21% of the time.

Hope this helps lol (: is this a psat or somethin?

8 0

2 years ago

Ralph has a collection of dimes and quarters. The ratio of dimes to quarters is 4:5. If the number

Murrr4er [49]

Answer:

180

Step-by-step explanation:

4 times 20 is 80, and 4 times 5 is 100, so add those together and that's the total number of coins

5 0

2 years ago

Find the Fourier series of f on the given interval. f(x) = 1, ?7 < x < 0 1 + x, 0 ? x < 7

Zolol [24]

$f(x)=\begin{cases}1&\text{for }-7$

The Fourier series expansion of $f(x)$ is given by

$\dfrac{a_0}2+\displaystyle\sum_{n\ge1}a_n\cos\frac{n\pi x}7+\sum_{n\ge1}b_n\sin\frac{n\pi x}7$

where we have

$a_0=\displaystyle\frac17\int_{-7}^7f(x)\,\mathrm dx$
$a_0=\displaystyle\frac17\left(\int_{-7}^0\mathrm dx+\int_0^7(1+x)\,\mathrm dx\right)$
$a_0=\dfrac{7+\frac{63}2}7=\dfrac{11}2$

The coefficients of the cosine series are

$a_n=\displaystyle\frac17\int_{-7}^7f(x)\cos\dfrac{n\pi x}7\,\mathrm dx$
$a_n=\displaystyle\frac17\left(\int_{-7}^0\cos\frac{n\pi x}7\,\mathrm dx+\int_0^7(1+x)\cos\frac{n\pi x}7\,\mathrm dx\right)$
$a_n=\dfrac{9\sin n\pi}{n\pi}+\dfrac{7\cos n\pi-7}{n^2\pi^2}$
$a_n=\dfrac{7(-1)^n-7}{n^2\pi^2}$

When $n$ is even, the numerator vanishes, so we consider odd $n$ , i.e. $n=2k-1$ for $k\in\mathbb N$ , leaving us with

$a_n=a_{2k-1}=\dfrac{7(-1)-7}{(2k-1)^2\pi^2}=-\dfrac{14}{(2k-1)^2\pi^2}$

Meanwhile, the coefficients of the sine series are given by

$b_n=\displaystyle\frac17\int_{-7}^7f(x)\sin\dfrac{n\pi x}7\,\mathrm dx$
$b_n=\displaystyle\frac17\left(\int_{-7}^0\sin\dfrac{n\pi x}7\,\mathrm dx+\int_0^7(1+x)\sin\dfrac{n\pi x}7\,\mathrm dx\right)$
$b_n=-\dfrac{7\cos n\pi}{n\pi}+\dfrac{7\sin n\pi}{n^2\pi^2}$
$b_n=\dfrac{7(-1)^{n+1}}{n\pi}$

So the Fourier series expansion for $f(x)$ is

$f(x)\sim\dfrac{11}4-\dfrac{14}{\pi^2}\displaystyle\sum_{n\ge1}\frac1{(2n-1)^2}\cos\frac{(2n-1)\pi x}7+\frac7\pi\sum_{n\ge1}\frac{(-1)^{n+1}}n\sin\frac{n\pi x}7$

3 0

2 years ago

19+7÷2 x 6+1=1 We’re to put brackets to make it =1

dimaraw [331]

It is 41
Please thanks

3 0

2 years ago

A figure can be covered by 28 unit squares,without any gaps or overlaps.what is the area of the figure?

Tom [10]

Remember that the area of a square is just one of its sides squared. $A=s^2$
where
$A$ is the area of the square
$s$ is one of the sides of the square

Unit square is a square whose sides have length 1, so lets use our formula to find the volume of one unit square:
$A=s^2$
$A=1^2$
$A=1$
We now know that each one of the squares has area 1 unit squared. Since there are 28 unit square in our figure, we are going to multiply the area of a unit square by 28 to find the area of our figure:
Area of the figure= $28(1)=28$ units squared

We can conclude that the area of the figure is 28 units squared.

7 0

2 years ago

The side of cube is 3m 30cm long, find out its volume​

The side of cube is 3m 30cm long, find out its volume