Don’t the permeter of the figure (HELPP PLEASEEE)
1 answer:
You might be interested in
Answer:
see the attachment
Step-by-step explanation:
We assume that the question is interested in the probability that a randomly chosen class is a Friday class with a lab experiment (2/15). That is somewhat different from the probability that a lab experiment is conducted on a Friday (2/3).
Based on our assumption, we want to create a simulation that includes a 1/5 chance of the day being a Friday, along with a 2/3 chance that the class has a lab experiment on whatever day it is.
That simulation can consist of choosing 1 of 5 differently-colored marbles, and rolling a 6-sided die with 2/3 of the numbers being designated as representing a lab-experiment day. (The marble must be replaced and the marbles stirred for the next trial.) For our purpose, we can designate the yellow marble as "Friday", and numbers greater than 2 as "lab-experiment".
The simulation of 70 different choices of a random class is shown in the attachment.
_____
<em>Comment on the question</em>
IMO, the use of <em>70 trials</em> is coincidentally the same number as the first <em>70 days</em> of school. The calendar is deterministic, so there will be exactly 14 Fridays in that period. If, in 70 draws, you get 16 yellow marbles, you cannot say, "the probability of a Friday is 16/70." You need to be very careful to properly state the question you're trying to answer.
one way would be to factor
I can't factor it so we will have to use Descartes' Rule of Signs which is helpful for finding how many real roots you have
it goes like this:
for a polynomial with real coefients, consider .
after arranging the terms in decending order in terms of degree, count how many times the signs of the coeffients change direction and minus 2 from that number until you get to 1 or 0. that will be the number of even roots the function can have
We have (-, -, +, +). the signs changed 1 times, so it has 1 real positive root
to get the negative roots, we evaluate f(-x) and see how many times the root changes
signs are (-, -, -, +). there was 1 change in sign
so the function has 1 real negative root
a total of 2 real roots
a function of degree can have at most,
roots
our function is degree 6 so it has 6 roots
if 2 are real, then the others must be complex
6-2=4 so there are 4 complex roots
you can also show that there are only 2 real roots by using a graphing utility to see that there are only 2 real roots