Answer:
The other side was decreased to approximately .89 times its original size, meaning it was reduced by approximately 11%
Step-by-step explanation:
We can start with the basic equation for the area of a rectangle:
l × w = a
And now express the changes described above as an equation, using "p" as the amount that the width is changed:
(l × 1.1) × (w × p) = a × .98
Now let's rearrange both of those equations to solve for a / l. Starting with the first and easiest:
w = a/l
now the second one:
1.1l × wp = 0.98a
wp = 0.98a / 1.1l
1.1 wp / 0.98 = a/l
Now with both of those equalling a/l, we can equate them:
1.1 wp / 0.98 = w
We can then divide both sides by w, eliminating it
1.1wp / 0.98w = w/w
1.1p / 0.98 = 1
And solve for p
1.1p = 0.98
p = 0.98 / 1.1
p ≈ 0.89
So the width is scaled by approximately 89%
We can double check that too. Let's multiply that by the scaled length and see if we get the two percent decrease:
.89 × 1.1 = 0.979
That should be 0.98, and we're close enough. That difference of 1/1000 is due to rounding the 0.98 / 1.1 to .89. The actual result of that fraction is 0.89090909... if we multiply that by 1.1, we get exactly .98.
Answer:
Should be around 4.5
Step-by-step explanation:
If you round 31.53 to the nearest ten dollars, you get 30, 15 percent of 30 is 4.5
-10x + 1.5 because you get rid of the parenthesis so -2 times 5 is -10 ( -10x ) and then -2 times -0.75 equals positive 1.5 due to two negatives canceling out to positive hope this helped!
![\bf \textit{equation of a circle}\\\\ (x- h)^2+(y- k)^2= r^2 \qquad center~~(\stackrel{}{ h},\stackrel{}{ k})\qquad \qquad radius=\stackrel{}{ r}\\\\ -------------------------------\\\\ (x+1)^2+y^2=36\implies [x-(\stackrel{h}{-1})]^2+[y-\stackrel{k}{0}]^2=\stackrel{r}{6^2}~~~~ \begin{cases} \stackrel{center}{(-1,0)}\\ \stackrel{radius}{6} \end{cases}](https://tex.z-dn.net/?f=%5Cbf%20%5Ctextit%7Bequation%20of%20a%20circle%7D%5C%5C%5C%5C%20%0A%28x-%20h%29%5E2%2B%28y-%20k%29%5E2%3D%20r%5E2%0A%5Cqquad%20%0Acenter~~%28%5Cstackrel%7B%7D%7B%20h%7D%2C%5Cstackrel%7B%7D%7B%20k%7D%29%5Cqquad%20%5Cqquad%20%0Aradius%3D%5Cstackrel%7B%7D%7B%20r%7D%5C%5C%5C%5C%0A-------------------------------%5C%5C%5C%5C%0A%28x%2B1%29%5E2%2By%5E2%3D36%5Cimplies%20%5Bx-%28%5Cstackrel%7Bh%7D%7B-1%7D%29%5D%5E2%2B%5By-%5Cstackrel%7Bk%7D%7B0%7D%5D%5E2%3D%5Cstackrel%7Br%7D%7B6%5E2%7D~~~~%0A%5Cbegin%7Bcases%7D%0A%5Cstackrel%7Bcenter%7D%7B%28-1%2C0%29%7D%5C%5C%0A%5Cstackrel%7Bradius%7D%7B6%7D%0A%5Cend%7Bcases%7D)
so, that's the equation of the circle, and that's its center, any point "ON" the circle, namely on its circumference, will have a distance to the center of 6 units, since that's the radius.
![\bf ~~~~~~~~~~~~\textit{distance between 2 points} \\\\ (\stackrel{x_1}{-1}~,~\stackrel{y_1}{0})\qquad A(\stackrel{x_2}{-1}~,~\stackrel{y_2}{1})\qquad \qquad % distance value d = \sqrt{( x_2- x_1)^2 + ( y_2- y_1)^2} \\\\\\ \stackrel{distance}{d}=\sqrt{[-1-(-1)]^2+(1-0)^2}\implies d=\sqrt{(-1+1)^2+1^2} \\\\\\ d=\sqrt{0+1}\implies d=1](https://tex.z-dn.net/?f=%5Cbf%20~~~~~~~~~~~~%5Ctextit%7Bdistance%20between%202%20points%7D%0A%5C%5C%5C%5C%0A%28%5Cstackrel%7Bx_1%7D%7B-1%7D~%2C~%5Cstackrel%7By_1%7D%7B0%7D%29%5Cqquad%20%0AA%28%5Cstackrel%7Bx_2%7D%7B-1%7D~%2C~%5Cstackrel%7By_2%7D%7B1%7D%29%5Cqquad%20%5Cqquad%20%0A%25%20%20distance%20value%0Ad%20%3D%20%5Csqrt%7B%28%20x_2-%20x_1%29%5E2%20%2B%20%28%20y_2-%20y_1%29%5E2%7D%0A%5C%5C%5C%5C%5C%5C%0A%5Cstackrel%7Bdistance%7D%7Bd%7D%3D%5Csqrt%7B%5B-1-%28-1%29%5D%5E2%2B%281-0%29%5E2%7D%5Cimplies%20d%3D%5Csqrt%7B%28-1%2B1%29%5E2%2B1%5E2%7D%0A%5C%5C%5C%5C%5C%5C%0Ad%3D%5Csqrt%7B0%2B1%7D%5Cimplies%20d%3D1)
well, the distance from the center to A is 1, namely is "inside the circle".
![\bf ~~~~~~~~~~~~\textit{distance between 2 points} \\\\ (\stackrel{x_1}{-1}~,~\stackrel{y_1}{0})\qquad B(\stackrel{x_2}{-1}~,~\stackrel{y_2}{6})\\\\\\ \stackrel{distance}{d}=\sqrt{[-1-(-1)]^2+(6-0)^2}\implies d=\sqrt{(-1+1)^2+6^2} \\\\\\ d=\sqrt{0+36}\implies d=6](https://tex.z-dn.net/?f=%5Cbf%20~~~~~~~~~~~~%5Ctextit%7Bdistance%20between%202%20points%7D%0A%5C%5C%5C%5C%0A%28%5Cstackrel%7Bx_1%7D%7B-1%7D~%2C~%5Cstackrel%7By_1%7D%7B0%7D%29%5Cqquad%20%0AB%28%5Cstackrel%7Bx_2%7D%7B-1%7D~%2C~%5Cstackrel%7By_2%7D%7B6%7D%29%5C%5C%5C%5C%5C%5C%0A%5Cstackrel%7Bdistance%7D%7Bd%7D%3D%5Csqrt%7B%5B-1-%28-1%29%5D%5E2%2B%286-0%29%5E2%7D%5Cimplies%20d%3D%5Csqrt%7B%28-1%2B1%29%5E2%2B6%5E2%7D%0A%5C%5C%5C%5C%5C%5C%0Ad%3D%5Csqrt%7B0%2B36%7D%5Cimplies%20d%3D6)
notice, the distance to B is exactly 6, and you know what that means.
![\bf ~~~~~~~~~~~~\textit{distance between 2 points} \\\\ (\stackrel{x_1}{-1}~,~\stackrel{y_1}{0})\qquad C(\stackrel{x_2}{4}~,~\stackrel{y_2}{-8}) \\\\\\ \stackrel{distance}{d}=\sqrt{[4-(-1)]^2+[-8-0]^2}\implies d=\sqrt{(4+1)^2+(-8)^2} \\\\\\ d=\sqrt{25+64}\implies d=\sqrt{89}\implies d\approx 9.43398](https://tex.z-dn.net/?f=%5Cbf%20~~~~~~~~~~~~%5Ctextit%7Bdistance%20between%202%20points%7D%0A%5C%5C%5C%5C%0A%28%5Cstackrel%7Bx_1%7D%7B-1%7D~%2C~%5Cstackrel%7By_1%7D%7B0%7D%29%5Cqquad%20%0AC%28%5Cstackrel%7Bx_2%7D%7B4%7D~%2C~%5Cstackrel%7By_2%7D%7B-8%7D%29%0A%5C%5C%5C%5C%5C%5C%0A%5Cstackrel%7Bdistance%7D%7Bd%7D%3D%5Csqrt%7B%5B4-%28-1%29%5D%5E2%2B%5B-8-0%5D%5E2%7D%5Cimplies%20d%3D%5Csqrt%7B%284%2B1%29%5E2%2B%28-8%29%5E2%7D%0A%5C%5C%5C%5C%5C%5C%0Ad%3D%5Csqrt%7B25%2B64%7D%5Cimplies%20d%3D%5Csqrt%7B89%7D%5Cimplies%20d%5Capprox%209.43398)
notice, C is farther than the radius 6, meaning is outside the circle, hiking about on the plane.
Adding (or subtracting) a constant to every data value adds (or subtracts) the same constant to measures of position such as center,percentiles, max or min.
Its shape and spread such as range, IQR, standard deviation remain unchanged.
When we multiply (or divide) all the data values by any constant, all measures of position (such as the mean, median, and percentiles) and measures of spread (such as the range, the IQR, and the standard deviation) are multiplied (or divided) by that same constant.
Part A:
The lowest score is a measure of location, so both addition and multiplying the lowest score of test B by 40 and adding 50 to the result will affect the lowest score of test A.
Thus, the lowest score of test A is given by 40(21) + 50 = 890
Therefore, the lowest score of test A is 890.
Part B:
The mean score is a measure of location, so both
addition and multiplying the mean score of test B by 40 and adding 50
to the result will affect the lowest score of test A.
Thus, the mean score of test A is given by 40(29) + 50 = 1,210
Therefore, the mean score of test A is 890.
Part C:
The standard deviation is a measure of spread, so multiplying the standard deviation of test B by 40 will affect the standard deviation but adding 50
to the result will not affect the standard deviation of test A.
Thus, the standard deviation of test A is given by 40(2) = 80
Therefore, the standard deviation of test A is 80.
Part D
The Q3 score is a measure of location, so both
addition and multiplying the Q3 score of test B by 40 and adding 50
to the result will affect the Q3 score of test A.
Thus, the Q3 score of test A is given by 40(28) + 50 = 1,170
Therefore, the Q3 score of test a is 1,170.
Part E:
The median score is a measure of location, so both
addition and multiplying the median score of test B by 40 and adding 50
to the result will affect the median score of test A.
Thus, the median score of test A is given by 40(26) + 50 = 1,090
Therefore, the median score of test A is 1,090.
Part F:
The IQR is a measure of spread, so multiplying the IQR of test B by 40 will affect the IQR but adding 50
to the result will not affect the IQR of test A.
Thus, the IQR of test A is given by 40(6) = 240
Therefore, the IQR of test A is 240.
