Answer:
The limit that 97.5% of the data points will be above is $912.
Step-by-step explanation:
Problems of normally distributed samples can be solved using the z-score formula.
In a set with mean
and standard deviation
, the zscore of a measure X is given by:

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.
In this problem, we have that:

Find the limit that 97.5% of the data points will be above.
This is the value of X when Z has a pvalue of 1-0.975 = 0.025. So it is X when Z = -1.96.
So




The limit that 97.5% of the data points will be above is $912.
Answer:
28.56 m²
Step-by-step explanation:
What we need to do for the quarter circle is use the following formula: 1/4(πr²)
(4 * 4) + 1/4(3.14 * 4²)
16 + 1/4(3.14 * 16)
16 + 1/4(50.24)
16 + 12.56
28.56 m²
The best approximation for the area of this figure is 28.56 m².
This is the transformations: See the picture, here we find the coordinate:
A' = (2,-3) and A''=(0.5,-2). There is many ways to get back to A from A'', i have showed one way (the red lines). This is done by going 5 units right and then 6 units up.
see image at https://imgur.com/a/fWQl8
a vertical axis, I assume it means a vertical axis of symmetry, thus it'd be a vertical parabola, like the one in the picture below.
![\bf ~~~~~~\textit{parabola vertex form} \\\\ \begin{array}{llll} y=a(x- h)^2+ k\qquad \qquad \leftarrow vertical\\\\ x=a(y- k)^2+ h \end{array} \qquad\qquad vertex~~(\stackrel{}{ h},\stackrel{}{ k}) \\\\[-0.35em] \rule{34em}{0.25pt}\\\\ \begin{cases} h=0\\ k=0 \end{cases}\implies y=a(x-0)^2+0 \\\\\\ \textit{we also know that } \begin{cases} x=-2\\ y=3 \end{cases}\implies 3=a(-2-0)^2+0\implies 3=4a \\\\\\ \cfrac{3}{4}=a~\hspace{10em}y=\cfrac{3}{4}(x-0)^2+0\implies \boxed{y=\cfrac{3}{4}x^2}](https://tex.z-dn.net/?f=%5Cbf%20~~~~~~%5Ctextit%7Bparabola%20vertex%20form%7D%20%5C%5C%5C%5C%20%5Cbegin%7Barray%7D%7Bllll%7D%20y%3Da%28x-%20h%29%5E2%2B%20k%5Cqquad%20%5Cqquad%20%5Cleftarrow%20vertical%5C%5C%5C%5C%20x%3Da%28y-%20k%29%5E2%2B%20h%20%5Cend%7Barray%7D%20%5Cqquad%5Cqquad%20vertex~~%28%5Cstackrel%7B%7D%7B%20h%7D%2C%5Cstackrel%7B%7D%7B%20k%7D%29%20%5C%5C%5C%5C%5B-0.35em%5D%20%5Crule%7B34em%7D%7B0.25pt%7D%5C%5C%5C%5C%20%5Cbegin%7Bcases%7D%20h%3D0%5C%5C%20k%3D0%20%5Cend%7Bcases%7D%5Cimplies%20y%3Da%28x-0%29%5E2%2B0%20%5C%5C%5C%5C%5C%5C%20%5Ctextit%7Bwe%20also%20know%20that%20%7D%20%5Cbegin%7Bcases%7D%20x%3D-2%5C%5C%20y%3D3%20%5Cend%7Bcases%7D%5Cimplies%203%3Da%28-2-0%29%5E2%2B0%5Cimplies%203%3D4a%20%5C%5C%5C%5C%5C%5C%20%5Ccfrac%7B3%7D%7B4%7D%3Da~%5Chspace%7B10em%7Dy%3D%5Ccfrac%7B3%7D%7B4%7D%28x-0%29%5E2%2B0%5Cimplies%20%5Cboxed%7By%3D%5Ccfrac%7B3%7D%7B4%7Dx%5E2%7D)
1) (2i-4)-(6i+9) = 2i-4-6i-9 = -4i-13
2) (-3+8i)+(3-8i) = -3+8i+3-8i = 0+0i = 0