What is the complete question?
For the answer to the question above asking <span>what is the probability that a randomly selected student will be a boy that brings his lunch to school? I'll provide a solution for the following problem below.
</span>P(boy |lunch) = P(boy and lunch)/P(lunch) = 0.40
P(boy and lunch) = P(lunch)*0.4 = 0.3*0.4 = 0.12
Answer:
1. -4/7
2. 4/3
3. x^2 + 16x + 63
4. x^2 + 19x + 90
Step-by-step explanation:
1. (1, 8) and (8, 4)
slope = m = (8 - 4)/(1 - 8) = -4/7
2. (2, 4) and (5, 8)
slope = m = (8 - 4)/(5 - 2) = 4/3
3. (x + 9)(x + 7) =
= x^2 + 7x + 9x + 63
= x^2 + 16x + 63
4. (x + 10)(x + 9) =
= x^2 + 9x + 10x + 90
= x^2 + 19x + 90
After plotting the quadrilateral in a Cartesian plane, you can see that it is not a particular quadrilateral. Hence, you need to divide it into two triangles. Let's take ABC and ADC.
The area of a triangle with vertices known is given by the matrix
M =
![\left[\begin{array}{ccc} x_{1}&y_{1}&1\\x_{2}&y_{2}&1\\x_{3}&y_{3}&1\end{array}\right]](https://tex.z-dn.net/?f=%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D%20x_%7B1%7D%26y_%7B1%7D%261%5C%5Cx_%7B2%7D%26y_%7B2%7D%261%5C%5Cx_%7B3%7D%26y_%7B3%7D%261%5Cend%7Barray%7D%5Cright%5D%20)
Area = 1/2· | det(M) |
= 1/2· | x₁·y₂ - x₂·y₁ + x₂·y₃ - x₃·y₂ + x₃·y₁ - x₁·y₃ |
= 1/2· | x₁·(y₂ - y₃) + x₂·(y₃ - y₁) + x₃·(y₁ - y₂) |
Therefore, the area of ABC will be:
A(ABC) = 1/2· | (-5)·(-5 - (-6)) + (-4)·(-6 - 7) + (-1)·(7 - (-5)) |
= 1/2· | -5·(1) - 4·(-13) - 1·(12) |
= 1/2 | 35 |
= 35/2
Similarly, the area of ADC will be:
A(ABC) = 1/2· | (-5)·(5 - (-6)) + (4)·(-6 - 7) + (-1)·(7 - 5) |
= 1/2· | -5·(11) + 4·(-13) - 1·(2) |
= 1/2 | -109 |
<span> = 109/2</span>
The total area of the quadrilateral will be the sum of the areas of the two triangles:
A(ABCD) = A(ABC) + A(ADC)
= 35/2 + 109/2
= 72
The perimeter would be 3/8+3/8+4/5+4/5= 2.35 or 2 7/20 square foot.
I hope this helps.
YOU'RE WELCOME :D