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a. Length of the fence around the field = perimeter of quarter circle = 892.7 ft.
b. The area of the outfield is about 39,584 sq. ft..
<h3>What is the Perimeter of a Quarter Circle?</h3>Perimeter of circle = 2πr
Perimeter of a quarter circle = 2r + 1/4(2πr).
a. The length of the fence around the field = perimeter of the quarter circle fence
= 2r + 1/4(2πr).
r = 250 ft
Plug in the value
The length of the fence around the field = 2(250) + 1/4(2 × π × 250)
= 892.7 ft.
b. Size of the outfield = area of the full field (quarter circle) - area of the infield (cicle)
= 1/4(πR²) - πr²
R = radius of the full field = 250 ft
r = radius of the infield = 110/2 = 55 ft
Plug in the values
Size of the outfield = 1/4(π × 250²) - π × 55²
= 49,087 - 9,503
= 39,584 sq. ft.
Learn more about perimeter of quarter circle on:
According to the line of best fit, the value of time when the temperature reach 100°c, the boiling point of water is 5.
<h3>What is linear regression?</h3>Linear regression is a type of regression which is used to model the statement in which the growth or decay initially with constant rate, and then slow down with respect to time.
The table shows the temperature of an amount of water set on a stove to boil, recorded every half minute. a 2-row table with 10 columns.
- Time (minutes X) 0, 0.5,1.0, 1.5, 2.0,2.5,3.0, 3.5, 4, 4.5.
- Temperature (° Celsius Y) 75, 79, 83, 86, 89, 91, 93, 94, 95, 95.5.
The sum of time is 22.5 and the sum of temperature value is 880.5. In this table,
- The mean of time value, 2.25.
- The mean of temperature value 88.05
- Sum of squares 20.625
- Sum of products 93.625
The regression equation for this data can be given as,
Put the value of temperature (y) 100 in this equation.
Hence, according to the line of best fit, the value of time when the temperature reach 100°c, the boiling point of water is 5.
Learn more about the regression here;