Answer:
1999
Step-by-step explanation:
y = 0.2313(y - 1992)^2 + 2.600(y - 1992) + 35.17
verify with known data point
y = 0.2313(2004 - 1992)^2 + 2.600(2004 - 1992) + 35.17
y = 0.2313(12)^2 + 2.600(12) + 35.17
y = 27.9873 + 28.6 + 35.17
y = 99.6772 which verifies our equation
65 = 0.2313x² + 2.6x + 35.17
0 = 0.2313x² + 2.6x - 29.83
quadratic formula
x = (-2.6 ±√(2.6² - 4(0.2313)(-29.83))) / (2(0.2313))
x = (-2.6 + 5.86) / 0.4626 = 7.05 years
7.05 = y - 1992
y = 1999.05
Since beta is in the first quadrant, the final answer will be positive.
To find cos(beta) so we can use the half angle identity, we can substitute into the Pythagorean identity. Doing so gives us that

So, this means that

Answer:
x = {1}{4}
Step-by-step explanation:
Area of the rectangular mat = 12 x 4 = 48 in²
Area of the 3 circles = 3( πr²) = 3(3.14 x 2²) = 37.68 in²
P(landing in a circle) = 37.68/48 = 157/200 = 0.79 (nearest hundredth)
Answer: The probability of landing in one of the circles is 0.79.