Int1 4.1.1 line of best fit (classwork)

in particular, 4-3 onwards i am struggling with. i included the picture of the the first page as background information.

due tomorrow, any and all is appreciated, you!

a. 2401.06

b. 37.54%

c. 56.3%

Step-by-step explanation:

hopefully this is right

*note: I think you forgot to convert 4-2 back into yards.

a. Robbie's field of view to the North end includes parts not on the football field. so to find the area of the football field he can see, we need to find:

total area of what Robbie sees - area of non football field Robbie sees

what you shaded represents the total of what Robbie sees. it's a triangle. area of a triangle is 1/2(b)(h) where h is distance away and b is width of view.

total area = 1/2(170)(30.92) = 2628.56 yd

area of non football field (pipe to South end)

= 1/2(50)(9.1) = 227.5

so 2628.56 - 227.5 = 2401.06

b. area found in part a / total area of football field

2401.06 / (120*53.3) = .3754

.3754 * 100 = 37.54%

c. the chance of Robbie seeing the touchdown depends on how much of the (North) endzone he can see.

area of north endzone is

10 * 53.3 = 533.

area Robbie sees in endzone is

2628.56 - 2328.48 = 300.08

(found by total area Robbie sees - area of non endzone Robbie sees)

300.08 / 533 = 0.563

= 56.3%

option (b) is correct

step-by-step explanation:

to find the solution of this system, we need to convert it into reduced row-echelon form. we have,

after applying row operations r2 -> r2/2 and r3 -> r3/3, we get,