Answer:
D. 5x^2 + 2y^2 + 12x + 5y
Step-by-step explanation:
6x + 2y + y^2 + 6x + 2y + y^2 + 4x^2 + x^2 + y
6x + 6x = 12x
12x + 2y + y^2 + 2y + y^2 + 4x^2 + x^2 + y
2y + 2y + y = 5y
12x + 5y + y^2 + y^2 + 4x^2 + x^2
y^2 + y^2 = 2y^2
2y^2 + 4x^2 + x^2 + 12x + 5y
4x^2 + x^2 = 5x^2
2y^2 + 5x^2 + 12x + 5y
5x^2 + 2y^2 + 12x + 5y
Answer:
x = 5
Step-by-step explanation:
First you distribute
9x - 21 = 24
Add 21 to both sides
9x = 45
Divide both sides by 9
x = 5
Ava's sister paid a 4.5% annual interest rate
Her sister paid $54 more than she borrowed, divide that by 3 (years till she paid)
54/3 = 18
That's $18 for every year she'd had the $400 loan out, divide it by 4 to get the percentage
18/4 = 4.5/100 = 4.5%
Answer:
how is your mom doing tell her last night was crazy :D
Step-by-step explanation:
The first part where he mentions "The distribution is skewed right" is correct. The tail on the right hand side is pulled longer than it should be, or longer compared to the left tail. "Right skew" or "skewed right" is the same as "positively skewed". They all mean the same thing.
The portion where Maurice says "This implies that the mean is less than the median" is incorrect. This is where his error is located. The reason for the error is because it should be "The mean is greater than the median" when we have right-skewed data like this.
Here's one example to see why this is the case. Consider the distribution {1,2,3,4,5}. We can see that the median is 3 since it's directly in the center. The mean is also 3 since we add up all the values and divide by 5. Symmetric data has mean = median. Now, we'll replace the '5' with something much larger to get a large outlier. Let's say we have the set {1,2,3,4,55}. The median is still 3. The outlier doesn't pull on the median. However, the mean has changed and it is (1+2+3+4+55)/5 = 65/5 = 13. The mean went from 3 to 13. Therefore, mean > median here. The mean is affected by outliers. Think of it like the outlier is pulling on the mean in a sort of magnetic way. This outlier makes the mean larger than it should be. If the outlier is extreme enough, then it sometimes makes sense to consider the trimmed mean as opposed to the regular arithmetic mean. The trimmed mean is the mean of the main cluster of values ignoring the outlier(s).
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To recap:
If we have right skewed data, then mean > median
As you can probably guess, if we had left-skewed data, then mean < median. In symmetric data, mean = median.
