There can be only one diagram of a net associated with each three-dimensional figure. Group of answer choices True False

1a) The trend line is linear, so we just need to find the slope and y-intercept to find an equation for it. Our y-intercept is (0,90), or 90, and our slope is -10/3. Our equation is now y=-10/3x+90.

1b) To find when x=21, we plug 21 into our equation where the x is. Now we do the math.

y=-10/3(21)+90 (plug in)

y=-70+90 (simplify by multiplying -10/3 by 21)

y=20 (simplify by adding -70 to 90)

Therefore, we can predict that when x is 21, y is 20.

1c) To find when y=150, we plug 150 into our equation where the y is. Now we do some more math.

150=-10/3x+90 (plug in)

60=-10/3x (subtract 90 from both sides

-18=x (divide both sides by -10/3)

Therefore, we can predict that when y is 150, x is -18.

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2a) The mean absolute deviation (or MAD for short) of a data set is calculated as such:

Step 1) Find the mean (average) by finding the sum of the data values, then dividing the sum of the data values by the number of data values. In this case, we have the numbers 65, 58, 64, 61, and 67, which add up to 315. The data set has 5 numbers, so we divide our sum of 315 by 5 to get 63. Therefore, our mean is 63.

Step 2) Find the absolute value of the distance between each data value and the mean. In this case, we find out how far away each data value is from 63, our mean. To do this, we subtract 63 from each number.

65-63=2

58-63=-5

64-63=1

61-63=-2

67-63=4

Some of these values are negative, but we're using absolute value so they all become positive. We now have a new set of values: 2, 5, 1, 2, and 4.

Step 3) Finally, we calculate the mean of our new set of values. In this case, we will add up 2, 5, 1, 2, and 4 to get 14 and divide by 5 to get our MAD of 2.8. Therefore, the MAD (and the answer to problem 2a) is 2.8.

2b) Now we just find out what the MAD means in this context. The MAD always is a measure of variance in a data set. In this context, it's describing how much the heights (in inches) of five people on a basketball team vary from the average height.

Hope this helps!

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2 years ago