Whats the problem? just substitute. For example: 3x-5 if x=4 so what you do is this; 3(4)-5=7
We are given
![tan(A)=\frac{7}{24}](https://tex.z-dn.net/?f=%20tan%28A%29%3D%5Cfrac%7B7%7D%7B24%7D%20%20)
we know that
![tan(A)=\frac{opposite}{adjacent}](https://tex.z-dn.net/?f=%20tan%28A%29%3D%5Cfrac%7Bopposite%7D%7Badjacent%7D%20%20)
so, we get
opposite =7
adjacent=24
now, we can find hypotenuse
![hypotenuse=\sqrt{7^2+24^2}](https://tex.z-dn.net/?f=%20hypotenuse%3D%5Csqrt%7B7%5E2%2B24%5E2%7D%20%20)
![hypotenuse=25](https://tex.z-dn.net/?f=%20hypotenuse%3D25%20)
now, we can draw triangle and then switch vertices accordingly
we can find cos(B) using second triangle
![cos(B)=\frac{adjacent}{hypotenuse}](https://tex.z-dn.net/?f=%20cos%28B%29%3D%5Cfrac%7Badjacent%7D%7Bhypotenuse%7D%20%20)
In second triangle:
adjacent=7
hypotenuse =25
so, we get
................Answer
Width: x
Length: 3x
Perimeter=x+x+3x+3x=8x
8x=400
x=50
Dimension is 150×50 feet (length × width)
You would have to make them all one form that is the same, for example if you wanted to find out if 80%, 0.75, 1/2 are the same you would have to change them into the same form. If you wanted to make them all decimals it would be 0.08, 0.75, 0.5, if you wanted to make them all percents it would be 80%, 75%, 50%, you wanted them all of be a fraction it would be 4/5, 3/4, and 1/2. I hope explained this well and this helped :)
-Madi
Answer:
a) ![\bar X = 369.62](https://tex.z-dn.net/?f=%5Cbar%20X%20%3D%20369.62)
b) ![Median=175](https://tex.z-dn.net/?f=Median%3D175%20)
c) ![Mode =450](https://tex.z-dn.net/?f=%20Mode%20%3D450%20)
With a frequency of 4
d) ![MidR= \frac{Max +Min}{2}= \frac{49+3000}{2}= 1524.5](https://tex.z-dn.net/?f=%20MidR%3D%20%5Cfrac%7BMax%20%2BMin%7D%7B2%7D%3D%20%5Cfrac%7B49%2B3000%7D%7B2%7D%3D%201524.5)
<u>e)</u>![s = 621.76](https://tex.z-dn.net/?f=%20s%20%3D%20621.76)
And we can find the limits without any outliers using two deviations from the mean and we got:
![\bar X+2\sigma = 369.62 +2*621.76 = 1361](https://tex.z-dn.net/?f=%20%5Cbar%20X%2B2%5Csigma%20%3D%20369.62%20%2B2%2A621.76%20%3D%201361)
And for this case we have two values above the upper limit so then we can conclude that 1500 and 3000 are potential outliers for this case
Step-by-step explanation:
We have the following data set given:
49 70 70 70 75 75 85 95 100 125 150 150 175 184 225 225 275 350 400 450 450 450 450 1500 3000
Part a
The mean can be calculated with this formula:
![\bar X = \frac{\sum_{i=1}^n X_i}{n}](https://tex.z-dn.net/?f=%20%5Cbar%20X%20%3D%20%5Cfrac%7B%5Csum_%7Bi%3D1%7D%5En%20X_i%7D%7Bn%7D)
Replacing we got:
![\bar X = 369.62](https://tex.z-dn.net/?f=%5Cbar%20X%20%3D%20369.62)
Part b
Since the sample size is n =25 we can calculate the median from the dataset ordered on increasing way. And for this case the median would be the value in the 13th position and we got:
![Median=175](https://tex.z-dn.net/?f=Median%3D175%20)
Part c
The mode is the most repeated value in the sample and for this case is:
![Mode =450](https://tex.z-dn.net/?f=%20Mode%20%3D450%20)
With a frequency of 4
Part d
The midrange for this case is defined as:
![MidR= \frac{Max +Min}{2}= \frac{49+3000}{2}= 1524.5](https://tex.z-dn.net/?f=%20MidR%3D%20%5Cfrac%7BMax%20%2BMin%7D%7B2%7D%3D%20%5Cfrac%7B49%2B3000%7D%7B2%7D%3D%201524.5)
Part e
For this case we can calculate the deviation given by:
![s =\sqrt{\frac{\sum_{i=1}^n (X_i -\bar X)^2}{n-1}}](https://tex.z-dn.net/?f=%20s%20%3D%5Csqrt%7B%5Cfrac%7B%5Csum_%7Bi%3D1%7D%5En%20%28X_i%20-%5Cbar%20X%29%5E2%7D%7Bn-1%7D%7D)
And replacing we got:
![s = 621.76](https://tex.z-dn.net/?f=%20s%20%3D%20621.76)
And we can find the limits without any outliers using two deviations from the mean and we got:
![\bar X+2\sigma = 369.62 +2*621.76 = 1361](https://tex.z-dn.net/?f=%20%5Cbar%20X%2B2%5Csigma%20%3D%20369.62%20%2B2%2A621.76%20%3D%201361)
And for this case we have two values above the upper limit so then we can conclude that 1500 and 3000 are potential outliers for this case