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3 years ago

What do they represent?

Mathematics

1 answer:

andrey2020 [161]3 years ago

8 0

Answer:

Step-by-step explanation:

If you convert the first equation to y=mx+b form, it will be the same as the second equation.

6x - 15y = 15

-15y = -6x + 15

y = $\frac{2}{5}$ x -1

Because the 2 equations are the same now, they are the same line.

You might be interested in

Which expression is equivalent to 6a + 15? A. 6(a + 5) B. 3(2a + 5) C. 3(3a + 12) D. 6(a + 12) PLEASE HELP :)

makvit [3.9K]

The answer is B

A. 6(a+5)
=6a+30 not equivalent to 6a+15
B.3(2a+5)
=6a+15 is equivalent
C.3(3a+12)
=9a+36 not equivalent
D.6(a+12)
=6a+72 not equivalent

So your answer is B

4 0

4 years ago

Read 2 more answers

Nina used plastic rectangles to make 6 rectangular prisms. How many rectangles did she use?

Reil [10]

Answer:

Step-by-step explanation:

6 X 6 = 36

7 0

3 years ago

If stratified random sampling with proportional allocation is used to select a sample of 25 high schools, how many would be sel

gregori [183]

Answer:

02 High Schools would be selected from the stratum with a percent-free-lunch value of 40 less than or equals x.

Step-by-step explanation:

As the sample size needed is 25 and total schools are 100 so this indicate 1 school in each 4 schools is to be selected. This is given as

$n=\frac{n_{total}}{n_{sample}}\\n=\frac{100}{25}\\n=4$

Now as the schools with percent free lunch are 8 so now

$n=\frac{n_{total}}{n_{sample}}\\n=\frac{8}{2}\\n=2$

So only 2 schools will be selected in this regard.

7 0

3 years ago

The Law of Large Numbers states that as the number of trials of an experiment increases, the experimental probability of an even

Lena [83]

The law of large number states that , if number of trials increases in an experiment , in a fair trial where each outcome has same chance of occurring or having equal probabilities,when total number of trials goes higher and higher the probabilities of each single outcome becomes approximately equal.→→In case of Experimental Probability

The Coin Possessed by Jake is a Magic coin.

Now Outcomes received by jake when he tosses the coin certain number of times.He flipped it 100 times, and found that it came up heads 64% of the time. He flipped it another 500 times, and it came up heads 57% of the time. He then flipped it 1000 times, and it came up heads 58% of the time. Then, he flipped it 1500 times, and it came up heads 62% of the time.

Based on the information provided , it appears that coin is not fair . It is Unbiased.So , if we apply law of large numbers here after number of trials will go higher and higher , the chances of coming head will be more than tail i.e theoretical probability of the magic coin coming up heads is> 50%.

To test my hypothesis i have used the information provided by Jake, which shows that coin is not fair. The probability of head has more than tail i.e by 14%, 7%,8% and 12%.

8 0

3 years ago

Let f(x,y,z) = ztan-1(y2) i + z3ln(x2 + 1) j + z k. find the flux of f across the part of the paraboloid x2 + y2 + z = 3 that li

Sophie [7]

Consider the closed region $V$ bounded simultaneously by the paraboloid and plane, jointly denoted $S$ . By the divergence theorem,

$\displaystyle\iint_S\mathbf f(x,y,z)\cdot\mathrm dS=\iiint_V\nabla\cdot\mathbf f(x,y,z)\,\mathrm dV$

And since we have

$\nabla\cdot\mathbf f(x,y,z)=1$

the volume integral will be much easier to compute. Converting to cylindrical coordinates, we have

$\displaystyle\iiint_V\nabla\cdot\mathbf f(x,y,z)\,\mathrm dV=\iiint_V\mathrm dV$
$=\displaystyle\int_{\theta=0}^{\theta=2\pi}\int_{r=0}^{r=1}\int_{z=2}^{z=3-r^2}r\,\mathrm dz\,\mathrm dr\,\mathrm d\theta$
$=\displaystyle2\pi\int_{r=0}^{r=1}r(3-r^2-2)\,\mathrm dr$
$=\dfrac\pi2$

Then the integral over the paraboloid would be the difference of the integral over the total surface and the integral over the disk. Denoting the disk by $D$ , we have

$\displaystyle\iint_{S-D}\mathbf f\cdot\mathrm dS=\frac\pi2-\iint_D\mathbf f\cdot\mathrm dS$

Parameterize $D$ by

$\mathbf s(u,v)=u\cos v\,\mathbf i+u\sin v\,\mathbf j+2\,\mathbf k$
$\implies\mathbf s_u\times\mathbf s_v=u\,\mathbf k$

which would give a unit normal vector of $\mathbf k$ . However, the divergence theorem requires that the closed surface $S$ be oriented with outward-pointing normal vectors, which means we should instead use $\mathbf s_v\times\mathbf s_u=-u\,\mathbf k$ .

Now,

$\displaystyle\iint_D\mathbf f\cdot\mathrm dS=\int_{u=0}^{u=1}\int_{v=0}^{v=2\pi}\mathbf f(x(u,v),y(u,v),z(u,v))\cdot(-u\,\mathbf k)\,\mathrm dv\,\mathrm du$
$=\displaystyle-4\pi\int_{u=0}^{u=1}u\,\mathrm du$
$=-2\pi$

So, the flux over the paraboloid alone is

$\displaystyle\iint_{S-D}\mathbf f\cdot\mathrm dS=\frac\pi2-(-2\pi)=\dfrac{5\pi}2$

6 0

4 years ago