A
900 a piece
negative 12.50 because it's a loss of money
Answer:
No, the Roger’s claim is not correct.
Step-by-step explanation:
We are given that Roger claims that the two statistics most likely to change greatly when an outlier is added to a small data set are the mean and the median.
This statement by Roger is incorrect because the median is unaffected by the outlier value and only the mean value gets affected by the outlier value.
As the median represents the middlemost value of our dataset, so any value which is an outlier will be either at the start or at the end will not the median value. So, the median will not likely change when an outlier is added to a small data set.
Now, the mean is the average of all the data set values, that is the sum of all the observations divided by the number of observations. The mean will get affected by the outlier value because it take into account each and every value of the data set.
Hence, the mean will likely to change greatly when an outlier is added to a small data set.
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Explanation:</h2><h2>
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Hello! Remember you have to write clear questions in order to get good and exact answers. Here I'll assume the data given in a comment above. So:
We know that the area (A) of any rectangle is given by:
Total depth of the bottom of the plate is 4 + 1 = 5
Force = limit(5,1) 62.5 *7* x * dx
= 437.5. Lim(5,1) x*dx
= 437.5(x^2/2)^5 , 1
= 437.5 x (5^2/2 - 1/2)
= 437.5 x 12
= 5,250 pounds
a)
(59, 76) and (65, 100)
y - y1 = m(x - x1)
y - 100 = (100-76)/(65 - 59)(x - 65)
y - 100 = (24/6)(x - 65)
y - 100 = 4(x - 65)
y - 100 = 4x - 260
y = 4x - 160
b)
x = 90
y = 4x - 160
y = 4(90) - 160
y = 360 - 160
y = 200
200 chirps per minute
c)
y = 120
y = 4x - 160
120 = 4x - 160
280 = 4x
x = 70
70 degrees
