Crowd Approximation Makeup work

Multiple Choice

A basketball court is 94 feet by 50 feet.  Eight fans can fit next to each other in a 4 ft. by 4 ft. area.  How many kids would you estimate can fit side by side over the entire floor?

Step 1: Determine a ratio (R) of the crowd?

• R = $$\frac{8}{94\times50}$$  

• R = $$\frac{8}{4\times4}$$   

• R = $$\frac{4\times4}{8}$$   

• R = $$\frac{8}{4}$$   

Multiple Choice

Nine people fit comfortably in a 3 ft. by 3 ft. area.  Use this value to estimate the size of a crowd that is 3 yards deep and 3 miles long.

Step 1: Determine the ratio (R):

• R = $$\frac{9}{3\div3}$$  

• R = $$\frac{9}{3}$$  

• R = $$\frac{9}{3\times3}$$  

• R = $$\frac{3\times3}{9}$$  

Multiple Choice

Nine people fit comfortably in a 3 ft. by 3 ft. area.  Use this value to estimate the size of a crowd that is 3 yards deep and 3 miles long.

Step 2: Determine the area of the crowd?

• 3 ft x 3 5280 ft

• 3(3 ft) x 3 (5280 ft)

• $$\frac{9}{3\times3}$$  (ft²)

• $$3\times3\$$  (ft²)

Fill in the Blank

Nine people fit comfortably in a 3 ft. by 3 ft. area.  Use this value to estimate the size of a crowd that is 3 yards deep and 3 miles long.

Step 3: determine the crowd?

​For 2 sides of the crowd, we will do the same but just multiply 2

Multiple Choice

Six people fit comfortably in a 3 ft. by 3 ft. area.  Use this value to estimate the size of a crowd of 2 sides that is 3 yards deep and 1 mile long.

Step 1: Determine the Ratio (R)

• R = 6/3

• R = 6/9

• R = 3/6

• R = 9/6

Multiple Choice

Six people fit comfortably in a 3 ft. by 3 ft. area.  Use this value to estimate the size of a crowd of 2 sides that is 3 yards deep and 1 mile long.

Step 2: Determine the Area (A)

• A = 3 x 3 (ft²)

• A = $$\frac{6}{3\times3}$$  (ft²)

• A = 3(3ft) x (5280 ft)

• A = 3 x 1 (ft²)

Multiple Choice

Six people fit comfortably in a 3 ft. by 3 ft. area.  Use this value to estimate the size of a crowd of 2 sides that is 3 yards deep and 1 mile long.

Step 3: Determine the 2 sides of crowds:

• 2 sides of crowds = (6/9) x [3(3) x 5280]

• 2 sides of crowds = 2 x (6/9) x [3(3) x 5280]

• 2 sides of crowds = (6/3) x [3(3) x 5280]

• 2 sides of crowds = (6/9) x [3 x 5280]