Expand the expression?
Perform indicated multiplication:
7a(2a)+7a(-5)-4(2a)-4(-5)
14a^2-35a-8a+20 combine like terms, ie (-35a-8a)=-43a
14a^2-43a+20
First of all, you need to come to an understanding of what you mean by "compare that score to the population." Often, that will mean determining the percentile rank of the score.
To determine the percentile rank of a raw score, you first nomalize it by determining the number of standard deviations it lies from the mean. That is, you subtract the population mean and divide that difference by the population standard deviation. Now, you have what is referred to as a "z-score".
Using a table of standard normal probability functions (or an equivalent calculator or app), you look up the cumulative distribution value corresponding to the z-score you have. This number between 0 and 1 (0% and 100%) will be the percentile rank of the score, the fraction of the population that has raw scores below the raw score you started with.
The slope-intercept form:
m - slope
b - y-intercept
The formula of a slope:
We have the points (-3, 1) and (2, -4). Substitute:
Therefore we have the equation of a line
Put the coordinates of the point (-3, 1) to the equation:
<em>subtract 3 from both sides</em>
<h3>Answer: y = -x - 2</h3>
<span>it depends how the interest is calculated, but there's not much of a difference
assuming its continuously compouned, you use this formula: A(t)=Pe^(rt), where A is the final amount, P is the initial investment, r is the interest, and t is the time in years
you want to find t such that A(t)=18,600 so 18,600=1000e^(.0675t)
you need to use logarithm to figure it out, take the natural log of both sides
the following properties will come into use:
ln(a*b)=ln(a)+ln(b)
ln(a^b)=bln(a)
ln(e)=1
taking the natural log
ln(18,600)=ln(1000e^(.0675t))
ln(18,600)=ln(1000)+ln(e^.0675t)
ln(18600)=ln(1000) + .0675t
now solve for t: t= (ln(18600)-ln(1000))/.0675
t=43.31</span>
0.02% of 40 million pounds is 8000 (pounds)
