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3 years ago

How do you determine the range and median of a Box And Whisker Plot??

1 answer:

6 0

The median ( Q2 ) divides the data set into two parts, the upper set and the lower set. The lower quartile ( Q1 ) is the median of the lower half, and the upper quartile ( Q3 ) is the median of the upper half. Example: Find Q1 , Q2 , and Q3 for the following data set, and draw a box-and-whisker plot.

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What is the product of Negative StartFraction 2 over 7 EndFraction and Negative StartFraction 3 over 7 EndFraction?

Andrew [12]

Answer:

your answer is d

Step-by-step explanation:

7 0

3 years ago

Read 2 more answers

Solve the equation. 33 = p – 6.71 A. –39.71 B. –26.29 C. 39.71 D. 26.29

jeka57 [31]

Answer

C. 39.71

Explanation

33 = p - 6.71

The first step is to make the like terms to be on the same side.

Add 6.71 on both sides of the eqution

33 + 6.71 = p - 6.71 + 6.71

39.71 = p

∴ p = 39.71

4 0

3 years ago

Carly said that 0.8÷10=0.04. Is she correct? Explain how you know.

faust18 [17]

Answer:

To divide 0.8 divided by 20 she needs to annex a 0 in the hundredths place. So, the quotient is 0.04, and Carly is correct

3 0

3 years ago

Find an asymptote of this conic section. 9x^2-36x-4y^2+24y-36=0

ki77a [65]

We will begin by grouping the x terms together and the y terms together so we can complete the square and see what we're looking at. $(9x^2-36x)-(4y^2+24y)-36=0$ . Now we need to move that 36 over by adding to isolate the x and y terms. $(9x^2-36x)-(4y^2+24y)=36$ . Now we need to complete the square on the x terms and the y terms. Can't do that, though, til the leading coefficients on the squared terms are 1's. Right now they are 9 and 4. Factor them out: $9(x^2-4x)-4(y^2-6y)=36$ . Now let's complete the square on the x's. Our linear term is 4. Half of 4 is 2, and 2 squared is 4, so add it into the parenthesis. BUT don't forget about the 9 hanging around out front there that refuses to be forgotten. It is a multiplier. So we are really adding in is 9*4 which is 36. Half the linear term on the y's is 3. 3 squared is 9, but again, what we are really adding in is -4*9 which is -36. Putting that altogether looks like this thus far: $9(x^2-4x+4)-4(y^2-6y+9)=36+36-36$ . The right side simplifies of course to just 36. Since we have a minus sign between those x and y terms, this is a hyperbola. The hyperbola has to be set to equal 1. So we divide by 36. At the same time we will form the perfect square binomials we created for this very purpose on the left: $\frac{(x-2)^2}{4}- \frac{(y-3)^2}{9}=1$ . Since the 9 is the bigger of the 2 values there, and it is under the y terms, our hyperbola has a horizontal transverse axis. a^2=4 so a=2; b^2=9 so b=3. Our asymptotes have the formula for the slope of $m=+/- \frac{b}{a}$ which for us is a slope of negative and positive 3/2. Using the slope and the fact that we now know the center of the hyperbola to be (2, 3), we can solve for b and rewrite the equations of the asymptotes. $3= \frac{3}{2}(2)+b$ give us a b of 0 so that equation is y = 3/2x. For the negative slope, we have $3=- \frac{3}{2}(2)+b$ which gives us a b value of 6. That equation then is y = -3/2x + 6. And there you go!

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3 years ago

Based on housing data below which equation can be used to calculate fair housing prices?

VMariaS [17]

Answer:

B. y = 0.097x + 11.142

Step-by-step explanation:

In order to find the best equation to descibe the housing price, we have to calculate the slope of the line first.

The equation of the line can be written as

$y=mx+q$

where m is the slope and q the y-intercept.

In order to have the most accurate estimate for the slope, we use the first point and the last point, that are:

(1900, 196)

and

(2200, 225)

The slope of the line is calculated as:

$m=\frac{y_2-y_1}{x_2-x_1}=\frac{225-196}{2200-1900}=0.097$

So, the slope of the line is 0.097.

Therefore, the correct option is the only one having a slope of 0.097, which is:

B. y = 0.097x + 11.142

5 0

3 years ago