Two control towers are located at points Q(500, 0) and R(500, 0), on a straight shore where the x-axis runs through (all distances are in meters). At the same moment, both towers sent a radio signal to a ship out at sea, each traveling at 300 m/µs. The ship received the signal from Q 3 µs (microseconds)before the message from R. Find the equation of the curve containing the possible location of the ship.

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  x^2/81 -y^2/19 = 2500

Step-by-step explanation:

The time difference between signals is 3 µs, so the distance difference is ...

  (300 m/µs)(3 µs) = 900 m

If we assume the coordinates of Q are (-500, 0), then the distances from point X to the control towers are ...

  XQ = √((x +500)^2 +y^2)

  XR = √((x -500)^2 +y^2)

We want the difference in distances to be 900, so we have the equation ...

  |XQ -XR| = |√((x +500)^2 +y^2) -√((x -500)^2 +y^2)| = 900

Squaring both sides gives ...

  (x +500)^2 +2(x +500)y +y^2 -2√(((x +500)^2 +y^2)((x -500)^2 +y^2)) +((x -500)^2 +y^2) = 810000

Separating the root from the rest of the equation, squaring again, and simplifying the rather messy expression, we can arrive at the equation ...

  x^2/81 -y^2/19 = 2500 . . . . . a hyperbola opening horizontally

solution:

as, mean is defined as sum of all the observations divided by total number of observations.

and , median is mid value of the data whether arranged in ascending order or descending order.

consider, roadsters:

2,3,4,5,6,7,8,9

mean =

also, median= , because number of observations is even, so median is mean of two mid values.

now, coming to bandits

2,3,4,5,6

mean =  

median = 4, as number of observations is odd.

so, roadsters has greater median as well mean than bandits.

→option a is true about the repairs performed on 2 types of cars:

bandits have lower median and mean than roadsters.