Answer:
a) 0.4
b) 0.133
c)
Step-by-step explanation:
We are given the following information in the question:
The load is said to be uniformly distributed over that part of the beam between 90 and 105 pounds per linear foot.
a = 90 and b = 105
Thus, the probability distribution function is given by

a) P( beam load exceeds 99 pounds per linear foot)
P( x > 99)
![=\displaystyle\int_{99}^{105} f(x) dx\\\\=\displaystyle\int_{99}^{105} \frac{1}{15} dx\\\\=\frac{1}{15}[x]_{99}^{105} = \frac{1}{15}(105-99) = 0.4](https://tex.z-dn.net/?f=%3D%5Cdisplaystyle%5Cint_%7B99%7D%5E%7B105%7D%20f%28x%29%20dx%5C%5C%5C%5C%3D%5Cdisplaystyle%5Cint_%7B99%7D%5E%7B105%7D%20%5Cfrac%7B1%7D%7B15%7D%20dx%5C%5C%5C%5C%3D%5Cfrac%7B1%7D%7B15%7D%5Bx%5D_%7B99%7D%5E%7B105%7D%20%3D%20%5Cfrac%7B1%7D%7B15%7D%28105-99%29%20%3D%200.4)
b) P( beam load less than 92 pounds per linear foot)
P( x < 92)
c) We have to find L such that
![\displaystyle\int_{L}^{105} f(x) dx\\\\=\displaystyle\int_{L}^{105} \frac{1}{15} dx\\\\=\frac{1}{15}[x]_{L}^{105} = \frac{1}{15}(105-L) = 0.4\\\\\Rightarrow L = 99](https://tex.z-dn.net/?f=%5Cdisplaystyle%5Cint_%7BL%7D%5E%7B105%7D%20f%28x%29%20dx%5C%5C%5C%5C%3D%5Cdisplaystyle%5Cint_%7BL%7D%5E%7B105%7D%20%5Cfrac%7B1%7D%7B15%7D%20dx%5C%5C%5C%5C%3D%5Cfrac%7B1%7D%7B15%7D%5Bx%5D_%7BL%7D%5E%7B105%7D%20%3D%20%5Cfrac%7B1%7D%7B15%7D%28105-L%29%20%3D%200.4%5C%5C%5C%5C%5CRightarrow%20L%20%3D%2099)
The beam load should be greater than or equal to 99 such that the probability that the beam load exceeds L is 0.4.
Answer:
Answer
Step-by-step explanation:
y=-4x+22
x=(11/2,0)
y=(0,22)
y=3x+10
x=(-10/3,0)
y=(0,10)
Answer:
3/4
Step-by-step explanation:
Answer:
x=10t+25
Step-by-step explanation:
The delivery fee is 25 dollars and each shirt is 10 dollars