7. scatterplots easy help asap

Mathematics

2 answers:

Lynna [10]3 years ago

8 0

Answer:

300

Step-by-step explanation: 180 in mins is three hours and the numbers going vertical are to be expressed in hundreds and 3 and 180 meet at a point

Minchanka [31]3 years ago

3 0

Answer:

The answer is 180

Step-by-step explanation:

looking at the line plot, if you go up to the 3 and see where the line intersects with it, you see that it's 180.

You might be interested in

If 3500 miles are traveled at a rate of 175 miles per hour, what is the time in hours? (hint: Distance = Rate * Time)

laiz [17]

3500 miles divided by 175 Miles Per Hour. 3500/175=20 Hours.

4 0

3 years ago

Factor by finding the GCF:<br> 8x-10

lys-0071 [83]

Answer:

The GCF is 2

Step-by-step explanation:

Hope this helps!!

3 0

3 years ago

Read 2 more answers

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

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