Answer:
A. 45.28
Step-by-step explanation:
Answer:
Lower limit: 113.28
Upper limit: 126.72
Step-by-step explanation:
Problems of normally distributed samples are 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.
In this problem, we have that:
![\mu = 120, \sigma = 8](https://tex.z-dn.net/?f=%5Cmu%20%3D%20120%2C%20%5Csigma%20%3D%208)
Middle 60%
So it goes from X when Z has a pvalue of 0.5 - 0.6/2 = 0.2 to X when Z has a pvalue of 0.5 + 0.6/2 = 0.8
Lower limit
X when Z has a pvalue of 0.20. So X when ![Z = -0.84](https://tex.z-dn.net/?f=Z%20%3D%20-0.84)
![Z = \frac{X - \mu}{\sigma}](https://tex.z-dn.net/?f=Z%20%3D%20%5Cfrac%7BX%20-%20%5Cmu%7D%7B%5Csigma%7D)
![-0.84 = \frac{X - 120}{8}](https://tex.z-dn.net/?f=-0.84%20%3D%20%5Cfrac%7BX%20-%20120%7D%7B8%7D)
![X - 120 = -0.84*8](https://tex.z-dn.net/?f=X%20-%20120%20%3D%20-0.84%2A8)
![X = 113.28](https://tex.z-dn.net/?f=X%20%3D%20113.28)
Upper limit
X when Z has a pvalue of 0.80. So X when ![Z = 0.84](https://tex.z-dn.net/?f=Z%20%3D%200.84)
![Z = \frac{X - \mu}{\sigma}](https://tex.z-dn.net/?f=Z%20%3D%20%5Cfrac%7BX%20-%20%5Cmu%7D%7B%5Csigma%7D)
![0.84 = \frac{X - 120}{8}](https://tex.z-dn.net/?f=0.84%20%3D%20%5Cfrac%7BX%20-%20120%7D%7B8%7D)
![X - 120 = 0.84*8](https://tex.z-dn.net/?f=X%20-%20120%20%3D%200.84%2A8)
![X = 126.72](https://tex.z-dn.net/?f=X%20%3D%20126.72)
The answer to the question
Let h represent the height of the trapezoid, the perpendicular distance between AB and DC. Then the area of the trapezoid is
Area = (1/2)(AB + DC)·h
We are given a relationship between AB and DC, so we can write
Area = (1/2)(AB + AB/4)·h = (5/8)AB·h
The given dimensions let us determine the area of ∆BCE to be
Area ∆BCE = (1/2)(5 cm)(12 cm) = 30 cm²
The total area of the trapezoid is also the sum of the areas ...
Area = Area ∆BCE + Area ∆ABE + Area ∆DCE
Since AE = 1/3(AD), the perpendicular distance from E to AB will be h/3. The areas of the two smaller triangles can be computed as
Area ∆ABE = (1/2)(AB)·h/3 = (1/6)AB·h
Area ∆DCE = (1/2)(DC)·(2/3)h = (1/2)(AB/4)·(2/3)h = (1/12)AB·h
Putting all of the above into the equation for the total area of the trapezoid, we have
Area = (5/8)AB·h = 30 cm² + (1/6)AB·h + (1/12)AB·h
(5/8 -1/6 -1/12)AB·h = 30 cm²
AB·h = (30 cm²)/(3/8) = 80 cm²
Then the area of the trapezoid is
Area = (5/8)AB·h = (5/8)·80 cm² = 50 cm²
Answer:help me with my questions
Step-by-step explanation: