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shtirl [24]
3 years ago
10

you go to a school in a college town. You know there are 2000 students enrolled in school. You polled 100 and 60 are students wh

at is the population of nonstudents?
Mathematics
1 answer:
antoniya [11.8K]3 years ago
5 0
Your answer would be1200. The average of 60/100, is 60/100 * 20 is 1200/2000. 

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Cd's at sound house are marked 1/3 off. what percent is this
Goshia [24]

1/3=.33333333, which is 33%

5 0
3 years ago
PLEASE HELP! A positive number is 1/3 of another number. The sum of the numbers is five less than the square of the larger numbe
antoniya [11.8K]

Answer: The numbers are 1 and 3.

Step-by-step explanation:

Let x = smaller number , y= larger number.

As per given,

x=\dfrac{y}{3}   ...(i)

x+y=y^2-5       ...(ii)

Put value of x from (i) in (ii)

\dfrac{y}{3}+y=y^2-5\\\\\Rightarrow\ \dfrac43y=y^2-5\\\\\Rightarrow\ 3y^2-4y-15=0\\\\\Rightarrow\ 3y^2-9y+5y-15=0\\\\\Rightarrow\ 3y(y-3)+5(y-3)=0\\\\\Rightarrow\ (y-3)(3y+5)=0\\\\\Rightarrow\ y=3 \ \ or\ y=\dfrac{-5}{3}

Since numbers are positive , so y=3 is correct.

And x will be 1  [from (i)]

Hence, the numbers are 1 and 3.

7 0
2 years ago
PLS I NEED HELP NOW PRONTO!!!!!
weeeeeb [17]

Answer:

The mean for Stem is 2.5

The mean for Leaf is 81.25

8 0
2 years ago
Write the equation in standard form<br><br> y = -10x + 6
Mumz [18]
Step (1) Flip the equation

−10x+6=y


Step (2) Add -6 to both sides

−10x+6+−6=y+−6

−10x=y−6

Step (3) Divide both sides by -10

−10x−10=y−6−10

ANSWER
x =  \frac{ - 1}{10} y +  \frac{3}{5}
5 0
3 years ago
Let T be the plane-2x-2y+z =-13. Find the shortest distance d from the point Po=(-5,-5,-3) to T, and the point Q in T that is cl
GaryK [48]

Answer:

d=10u

Q(5/3,5/3,-19/3)

Step-by-step explanation:

The shortest distance between the plane and Po is also the distance between Po and Q. To find that distance and the point Q you need the perpendicular line x to the plane that intersects Po, this line will have the direction of the normal of the plane n=(-2,-2,1), then r will have the next parametric equations:

x=-5-2\lambda\\y=-5-2\lambda\\z=-3+\lambda

To find Q, the intersection between r and the plane T, substitute the parametric equations of r in T

-2x-2y+z =-13\\-2(-5-2\lambda)-2(-5-2\lambda)+(-3+\lambda) =-13\\10+4\lambda+10+4\lambda-3+\lambda=-13\\9\lambda+17=-13\\9\lambda=-13-17\\\lambda=-30/9=-10/3

Substitute the value of \lambda in the parametric equations:

x=-5-2(-10/3)=-5+20/3=5/3\\y=-5-2(-10/3)=5/3\\z=-3+(-10/3)=-19/3\\

Those values are the coordinates of Q

Q(5/3,5/3,-19/3)

The distance from Po to the plane

d=\left| {\to} \atop {PoQ}} \right|=\sqrt{(\frac{5}{3}-(-5))^2+(\frac{5}{3}-(-5))^2+(\frac{-19}{3}-(-3))^2} \\d=\sqrt{(\frac{5}{3}+5))^2+(\frac{5}{3}+5)^2+(\frac{-19}{3}+3)^2} \\d=\sqrt{(\frac{20}{3})^2+(\frac{20}{3})^2+(\frac{-10}{3})^2}\\d=\sqrt{\frac{400}{9}+\frac{400}{9}+\frac{100}{9}}\\d=\sqrt{\frac{900}{9}}=\sqrt{100}\\d=10u

7 0
3 years ago
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