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irga5000 [103]
3 years ago
15

The U.S. Department of Housing and Urban Development​ (HUD) uses the median to report the average price of a home in the United

States. Why do you think HUD uses the​ median?
Mathematics
1 answer:
MatroZZZ [7]3 years ago
3 0

Answer:

Step-by-step explanation:

given that the U.S. Department of Housing and Urban Development​ (HUD) uses the median to report the average price of a home in the United States.

We know that mean, median and mode are measures of central tendency.

Mean is the average of all the prices while median is the middle entry when arranged in ascending order.

Mean has the disadvantage of showing undue figure if extreme entries are there. i.e. outlier affect mean.

Suppose a price goes extremely high, then mean will fluctuate more than median.

So median using gives a reliable estimate since median gives the middle price and equally spread to other sides.

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24: 1,2,3,4,6,8,24
36: 1,2,3,4,6,9,12,36
27:1,3,9,27
7:1,7
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What is the coefficient of x4y4 in the expansion of (x + y)8?
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70

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If you use Pascal's Triangle, you will need 9 lines, since there are 9 terms in a binomial raised to the 8th power.  Those 9 terms are:

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The first expansion through the one we need are below (I hope you know how to use Pascal's Triangle!):

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Factor 64u–40.<br> Write your answer as a product with a whole number greater than 1.
Vlada [557]

Answer: 8(8u-5)

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8 0
2 years ago
If 3x^2 + y^2 = 7 then evaluate d^2y/dx^2 when x = 1 and y = 2. Round your answer to 2 decimal places. Use the hyphen symbol, -,
S_A_V [24]
Taking y=y(x) and differentiating both sides with respect to x yields

\dfrac{\mathrm d}{\mathrm dx}\bigg[3x^2+y^2\bigg]=\dfrac{\mathrm d}{\mathrm dx}\bigg[7\bigg]\implies 6x+2y\dfrac{\mathrm dy}{\mathrm dx}=0

Solving for the first derivative, we have

\dfrac{\mathrm dy}{\mathrm dx}=-\dfrac{3x}y

Differentiating again gives

\dfrac{\mathrm d}{\mathrm dx}\bigg[6x+2y\dfrac{\mathrm dy}{\mathrm dx}\bigg]=\dfrac{\mathrm d}{\mathrm dx}\bigg[0\bigg]\implies 6+2\left(\dfrac{\mathrm dy}{\mathrm dx}\right)^2+2y\dfrac{\mathrm d^2y}{\mathrm dx^2}=0

Solving for the second derivative, we have

\dfrac{\mathrm d^2y}{\mathrm dx^2}=-\dfrac{3+\left(\frac{\mathrm dy}{\mathrm dx}\right)^2}y=-\dfrac{3+\frac{9x^2}{y^2}}y=-\dfrac{3y^2+9x^2}{y^3}

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\dfrac{\mathrm d^2y}{\mathrm dx^2}\bigg|_{x=1,y=2}=-\dfrac{3\cdot2^2+9\cdot1^2}{2^3}=\dfrac{21}8\approx2.63
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3 years ago
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