The image is not attached with, but by reading the question it is obvious that the blue region lies inside the larger square and outside the smaller square. That is the region between the two squares is the blue region.
We know the dimensions of both squares, using which we can find the area of both squares. Subtracting the area of smaller square from larger one, we can find the area of blue square and further we can find the said probability.
Area of larger square = 8 x 8 = 64 in²
Area of smaller square = 2 x 2 = 4 in²
Area of blue region = 64 - 4 = 60 in²
The probability that a randomly chosen point lies within the blue region = Area of blue region/Total area available
Therefore, the probability that a point chosen at random is in the blue region = 60/64 = 0.9375
Answer:
Q1
cos 59° = x/16
x = 16 cos 59°
x = 8.24
Q2
BC is given 23 mi
Maybe AB is needed
AB = √34² + 23² = 41 (rounded)
Q3
BC² = AB² - AC²
BC = √(37² - 12²) = 35
Q4
Let the angle is x
cos x = 19/20
x = arccos (19/20)
x = 18.2° (rounded)
Q5
See attached
Added point D and segments AD and DC to help with calculation
BC² = BD² + DC² = (AB + AD)² + DC²
Find the length of added red segments
AD = AC cos 65° = 14 cos 65° = 5.9
DC = AC sin 65° = 14 sin 65° = 12.7
Now we can find the value of BC
BC² = (19 + 5.9)² + 12.7²
BC = √781.3
BC = 28.0 yd
All calculations are rounded
Our two binomials are (X-5)(X-7)
If we FOIL those two terms [x*x = x2; x*-7 = -7x; x*-5 = -5x; -5*-7 = 35] and combine the like terms, we're left with one of many quadratic function
Answer
x2-12x+35
