In American football, the playing field is 53.33 yards (yd) wide by 120 yards (yd) long. For a special game, the field staff wan

Question

2 months ago

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In American football, the playing field is 53.33 yards (yd) wide by 120 yards (yd) long. For a special game, the field staff wan

t to paint the playing field orange. Of course, they will use biodegradable paint available for purchase in 25-gallon (gal) containers. If the paint is applied in a thickness of 1.2 millimeters (mm) in a uniform layer, how many containers of paint will they need to purchase?

Mathematics

2 answers:

babunello [11.8K] · Answer 1 · 2026-03-21T23:02:39+03:00

Answer:

The required number of containers to purchase is $N_V= 67.85$

Step-by-step explanation:

According to the question,

The width of the field is

$w_f = 53.33 \ yard = 53.33*0.9144 = 48.76m$

The length of the field is

$l_f = 120 \ yards = 120 * 0.9144 = 109.728m$

The volume of a single container is

$V= 25 \ gallon = 25 * 0.00378541 = 0.094625m^3$

The paint thickness is

$t = 1.2 \ mm = 1.2 * 0.001 = 0.0012m$

The area of the field is

$A = 48.76 * 109.728$

$=5350.337m^2$

The amount of paint containers necessary is $N_V$ $= \frac{area \ of \ playing \ field(A) * thickness \ of \ paint \ application(t) }{volume\ single \ container(V)}$

=> $N_V = \frac{5350.337 * 0.0012}{0.094625}$

$N_V= 67.85$

PIT_PIT [12.4K] · Answer 2 · 2026-03-21T23:04:50+03:00

Answer:

214

Step-by-step explanation:

The dimensions of the playing field are 53.33 yards wide and 120 yards long, requiring us to compute the area by multiplying 53.33 by 120 yards. This results in 6399.6. The thickness being applied is 1.2 millimeters. To find how many containers are needed, divide the area, 6399.6 by 1.2 which equals 5333. Finally, dividing that by 25 gallons gives us 213.32, so you need to buy 214 rounded.