Find the nearest numbers that are greater and less that your number, in this case, 7 and 8 which square to 49 and 64 and average them by adding them together and dividing by two.
7+8=15
15/2=7.5
7.5 is the nearest estimated square root of 50 that I can get, but the real answer is about 7.071067811865475. (I used a calculator)
<span>For f(x)=4x+1 and g(x)=x^2+5
(f+g)(x) = </span>4x +1 + x^2 + 5
(f+g)(x) = x^2 + 4x + 6
hope that helps
Answer:
A. 110 pounds,
C. 281 pounds
Step-by-step explanation:
Problems of normally distributed samples can be solved using the z-score formula.
In a set with mean
and standard deviation
, the zscore of a measure X is given by:
![Z = \frac{X - \mu}{\sigma}](https://tex.z-dn.net/?f=Z%20%3D%20%5Cfrac%7BX%20-%20%5Cmu%7D%7B%5Csigma%7D)
The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.
A measure is said to be an outlier if it has a pvalue lesser than 0.05 or higher than 0.95.
In this problem, we have that:
![\mu = 173, \sigma = 30](https://tex.z-dn.net/?f=%5Cmu%20%3D%20173%2C%20%5Csigma%20%3D%2030)
A. 110 pounds,
![Z = \frac{X - \mu}{\sigma}](https://tex.z-dn.net/?f=Z%20%3D%20%5Cfrac%7BX%20-%20%5Cmu%7D%7B%5Csigma%7D)
![Z = \frac{110 - 173}{30}](https://tex.z-dn.net/?f=Z%20%3D%20%5Cfrac%7B110%20-%20173%7D%7B30%7D)
![Z = -2.1](https://tex.z-dn.net/?f=Z%20%3D%20-2.1)
has a pvalue of 0.0179. So a weight of 110 pounds is an outlier.
B. 157 pounds,
![Z = \frac{X - \mu}{\sigma}](https://tex.z-dn.net/?f=Z%20%3D%20%5Cfrac%7BX%20-%20%5Cmu%7D%7B%5Csigma%7D)
![Z = \frac{157 - 173}{30}](https://tex.z-dn.net/?f=Z%20%3D%20%5Cfrac%7B157%20-%20173%7D%7B30%7D)
![Z = 0.53](https://tex.z-dn.net/?f=Z%20%3D%200.53)
has a pvalue of 0.702.
So a weight of 157 is not an outlier.
C. 281 pounds
![Z = \frac{X - \mu}{\sigma}](https://tex.z-dn.net/?f=Z%20%3D%20%5Cfrac%7BX%20-%20%5Cmu%7D%7B%5Csigma%7D)
![Z = \frac{281 - 173}{30}](https://tex.z-dn.net/?f=Z%20%3D%20%5Cfrac%7B281%20-%20173%7D%7B30%7D)
![Z = 3.6](https://tex.z-dn.net/?f=Z%20%3D%203.6)
has a pvalue of 0.9988.
So a weight of 281 is an outlier.
Answer:
Terry's yard is greater than Todd's yard by 90 square meters.
Step-by-step explanation:
We are given the following in the question:
Terry's yard dimensions:
Length,l = 18 meters
Width,w = 15 meters
Area of Terry's yard =
![A = l\times w\\A = 18\times 15\\A=270\text{ square meters}](https://tex.z-dn.net/?f=A%20%3D%20l%5Ctimes%20w%5C%5CA%20%3D%2018%5Ctimes%2015%5C%5CA%3D270%5Ctext%7B%20square%20meters%7D)
Todd's yard dimensions:
Length,l = 20 meters
Width,w = 9 meters
Area of Todd's yard =
![A = l\times w\\A = 20\times 9\\A=180\text{ square meters}](https://tex.z-dn.net/?f=A%20%3D%20l%5Ctimes%20w%5C%5CA%20%3D%2020%5Ctimes%209%5C%5CA%3D180%5Ctext%7B%20square%20meters%7D)
Thus, Terry's yard is greater than Todd's yard.
Difference in area of yards
= Terry's yard area - Todd's yard area
![=270-180\\=90\text{ squarde meters}](https://tex.z-dn.net/?f=%3D270-180%5C%5C%3D90%5Ctext%7B%20squarde%20meters%7D)
Thus, Terry's yard is greater than Todd's yard by 90 square meters.