Answer:
with what you didnt put a question anywhere
Answer:
3.2 ft
Step-by-step explanation:
The initial distance from the base of the ladder to the wall is:
![20^2=16^2+d^2\\d=12\ ft](https://tex.z-dn.net/?f=20%5E2%3D16%5E2%2Bd%5E2%5C%5Cd%3D12%5C%20ft)
After moving the ladder, the distance from the base to the wall is 12 - 2x while the height of the ladder is 16 + x. The distance x is given by:
![20^2 = (12-2x)^2+(16+x)^2\\400 = 144-48x+4x^2+256+32x+x^2\\5x^2-16x=0\\5x=16\\x=3.2\ ft](https://tex.z-dn.net/?f=20%5E2%20%3D%20%2812-2x%29%5E2%2B%2816%2Bx%29%5E2%5C%5C400%20%3D%20144-48x%2B4x%5E2%2B256%2B32x%2Bx%5E2%5C%5C5x%5E2-16x%3D0%5C%5C5x%3D16%5C%5Cx%3D3.2%5C%20ft)
Therefore, the top of the ladder is 3.2 ft higher now.
Formula for the area of a circle<span>2
A= pi r</span>
$5.36 per hamburger. Remember that in this scenario, they added the whole chicken sandwich info even though it does not apply to the situation.
Answer:
(a) The standard deviation of your waiting time is 4.33 minutes.
(b) The probability that you will have to wait more than 2 standard deviations is 0.4227.
Step-by-step explanation:
Let <em>X</em> = the waiting time for the bus at the parking lot.
The random variable <em>X</em> is uniformly distributed with parameters <em>a</em> = 0 to <em>b</em> = 15.
The probability density function of <em>X</em> is given as follows:
![f_{X}(x)=\frac{1}{b-a};\ a](https://tex.z-dn.net/?f=f_%7BX%7D%28x%29%3D%5Cfrac%7B1%7D%7Bb-a%7D%3B%5C%20a%3CX%3Cb%2C%5C%20a%3Cb)
(a)
The standard deviation of a Uniformly distributed random variable is given by:
![SD=\sqrt{\frac{(b-a)^{2}}{12}}](https://tex.z-dn.net/?f=SD%3D%5Csqrt%7B%5Cfrac%7B%28b-a%29%5E%7B2%7D%7D%7B12%7D%7D)
Compute the standard deviation of the random variable <em>X</em> as follows:
![SD=\sqrt{\frac{(b-a)^{2}}{12}}](https://tex.z-dn.net/?f=SD%3D%5Csqrt%7B%5Cfrac%7B%28b-a%29%5E%7B2%7D%7D%7B12%7D%7D)
![=\sqrt{\frac{(15-0)^{2}}{12}}](https://tex.z-dn.net/?f=%3D%5Csqrt%7B%5Cfrac%7B%2815-0%29%5E%7B2%7D%7D%7B12%7D%7D)
![=\sqrt{\frac{225}{12}}](https://tex.z-dn.net/?f=%3D%5Csqrt%7B%5Cfrac%7B225%7D%7B12%7D%7D)
![=\sqrt{18.75}\\=4.33](https://tex.z-dn.net/?f=%3D%5Csqrt%7B18.75%7D%5C%5C%3D4.33)
Thus, the standard deviation of your waiting time is 4.33 minutes.
(b)
The value representing 2 standard deviations is:
![X=2\times SD=2\times4.33=8.66](https://tex.z-dn.net/?f=X%3D2%5Ctimes%20SD%3D2%5Ctimes4.33%3D8.66)
Compute the value of P (X > 8.66) as follows:
![P(X>8.66)=\int\limits^{10}_{8.66} {\frac{1}{15-0}}\, dx\\](https://tex.z-dn.net/?f=P%28X%3E8.66%29%3D%5Cint%5Climits%5E%7B10%7D_%7B8.66%7D%20%7B%5Cfrac%7B1%7D%7B15-0%7D%7D%5C%2C%20dx%5C%5C)
![=\frac{1}{15}\times \int\limits^{10}_{8.66} {1}\, dx\\](https://tex.z-dn.net/?f=%3D%5Cfrac%7B1%7D%7B15%7D%5Ctimes%20%5Cint%5Climits%5E%7B10%7D_%7B8.66%7D%20%7B1%7D%5C%2C%20dx%5C%5C)
![=\frac{1}{15}\times |x|^{15}_{8.66}\\](https://tex.z-dn.net/?f=%3D%5Cfrac%7B1%7D%7B15%7D%5Ctimes%20%7Cx%7C%5E%7B15%7D_%7B8.66%7D%5C%5C)
![=\frac{15-8.66}{15}\\](https://tex.z-dn.net/?f=%3D%5Cfrac%7B15-8.66%7D%7B15%7D%5C%5C)
![=0.4227](https://tex.z-dn.net/?f=%3D0.4227)
Thus, the probability that you will have to wait more than 2 standard deviations is 0.4227.