Supposing, for the sake of illustration, that the mean is 31.2 and the std. dev. is 1.9.
This probability can be calculated by finding z-scores and their corresponding areas under the std. normal curve.
34 in - 31.2 in
The area under this curve to the left of z = -------------------- = 1.47 (for 34 in)
1.9
32 in - 31.2 in
and that to the left of 32 in is z = ---------------------- = 0.421
1.9
Know how to use a table of z-scores to find these two areas? If not, let me know and I'll go over that with you.
My TI-83 calculator provided the following result:
normalcdf(32, 34, 31.2, 1.9) = 0.267 (answer to this sample problem)
You will need this formula:
Years = ln (Total / Principal) / rate
(where "ln" means natural logarithm)
and we'll use $100 and $200 for beginning and ending amount
Years = ln (200 / 100) / rate
Years = 0.69314718056 / .052
Years =
<span>
<span>
<span>
13.3297534723
</span>
</span>
</span>
Rounding to the nearest tenth of a year:
Years =
<span>
<span>
<span>
13.3
Source:
http://www.1728.org/rate2.htm
</span></span></span>
Answer:
A. 2x + 9 + 11 over x - 2
Step-by-step explanation:
2x^2+5x-7 over x-2
divide - 2x^2 + 5x - 7 over x - 2 = 2x + 9x - 7 over x - 2
= 2x + 9x - 7 over x - 2
divide - 9x - 7 over x - 2 = 9 + 11 over x - 2
= 2x + 9 + 11 over x - 2
Answer: it obvious that you are not a good non algebraic
Step-by-step explanation:
Lollllll
A prime number can <span>be divided evenly only by 1, or itself. </span>
<span>And it must be a whole number greater than 1.
Therefore, the prime numbers between 0, 1, 2, 3 are
2 and 3
and the only even one among those two is two, so your answer would be One. </span>
