The load, p (in pounds), that can be safely supported by a horizontal beam varies jointly as the product of the width, w (in feet), of the beam and the square of the depth, d (in feet), and inversely as its length, l (in feet). if a beam has a width of 2 feet, a depth of 1 foot and a length of 4 feet, the load it can support is 10 pounds. how much could the beam hold if the beam had a length of 6 feet, a depth of 2 feet, and a width of 3 feet?

The formula would be

wd^2c/l = p

(2*1^2c)/4 = 10

(2*1c)/4 = 10

2c/4 = 10

c = 20

3*2^2*20/6 = p

3*4*20/6 = p

240 pounds

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.

step-by-step explanation:

i think something is missing in your question because y-1=(x-12) would not pass through the point (-4,-3). because if you replace x by -4 your are supposed to find y=-3 which is not the case her. if you replace x by -4 you find =-15.

i assume the right equation is y-1=1/4*(x-12) because this one is passing through both your points (if you are interested to know how i found that out i can tell you).

in this equation if i replace x by -4 i find y=-3 and if i replace x by 12, i find y=1 so the line representing this equation is passing through those 2 points.

then the standard form of the equation is finding an expression of y in function of x

y-1=1/4*(x-12)

y=x/4-3+1

y=x/4-2

hope that .

do you confirm there was a mistake in the question?