There are 1420 students enrolled at Senn. If 60% of these students walk to school, how many students are walking to school?

Mathematics

2 answers:

den301095 [7]3 years ago

6 0

Answer:

852 walk to school

Step-by-step explanation:

i got this because i divided the 1420 into 10 which became a 142, multiply by 6 and you get 852

also you could

it is really simple also, all you have to do is multiply the number of student, S times the precent of students who walk, so s times w, which is 60 precent turend into .6 precent so 1420 x .6

pychu [463]3 years ago

3 0

Answer:

852 students walking to school

Step-by-step explanation:

Lets start with a simple example so you can get it

if 60% of 100 students walk to school that means

60 students walk to school... right??

so lets simplify it more that means you have multiplied 100 (students) by 60%

or $\frac{60}{100}$ (note: you can say that 60% is equal to $\frac{60}{100}$ )

so the problem told 60% of 1420 students walk to school that means we should

do this: $1420*\frac{60}{100} = 852$

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Find the volume of the region between the planes x plus y plus 2 z equals 2 and 4 x plus 4 y plus z equals 8 in the first octant

Alex787 [66]

Find the intercepts for both planes.

Plane 1, x + y + 2z = 2:

$y=z=0\implies x=2\implies (2,0,0)$

$x=z=0\implies y=2\implies(0,2,0)$

$x=y=0\implies 2z=2\implies z=1\implies(0,0,1)$

Plane 2, 4x + 4y + z = 8:

$y=z=0\implies4x=8\implies x=2\implies(2,0,0)$

$x=z=0\implies4y=8\impliesy=2\implies(0,2,0)$

$x=y=0\implies z=8\implies(0,0,8)$

Both planes share the same x- and y-intercepts, but the second plane's z-intercept is higher, so Plane 2 acts as the roof of the bounded region.

Meanwhile, in the (x, y)-plane where z = 0, we see the bounded region projects down to the triangle in the first quadrant with legs x = 0, y = 0, and x + y = 2, or y = 2 - x.

So the volume of the region is

$\displaystyle\int_0^2\int_0^{2-x}\int_{\frac{2-x-y}2}^{8-4x-4y}\mathrm dz\,\mathrm dy\,\mathrm dx=\displaystyle\int_0^2\int_0^{2-x}\left(8-4x-4y-\frac{2-x-y}2\right)\,\mathrm dy\,\mathrm dx$

$=\displaystyle\int_0^2\int_0^{2-x}\left(7-\frac72(x+y)\right)\,\mathrm dy\,\mathrm dx=\int_0^2\left(7(2-x)-\frac72x(2-x)-\frac74(2-x)^2\right)\,\mathrm dx$