ANSWER:
X is -5 and 1.
1) cross multiply
2) subtract five from both sides
3) factor
4) check for extraneous
5) ask me if you are still confused.
She has $5.33 left
$9.00
-$3.67
————
$5.33
15 feet martins family work for with the dodo code for a mask is a horrible thing to ware a little kid in a sweat room or the room with the kids and then they have a white black brown brown green green orange brown orange green green orange orange brown orange orange orange
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.
We know from physics class that the formula for distance of
a linear motion is given as:
d = v t
Where,
d = distance travelled
v = average velocity
t = time it took to reach the destination
Since the distance going to the office and back is just
similar, therefore we can simply equate the two:
v1 t1 = v2 t2
Where 1 signifies going to the office and 2 signifies going
back from the office. Therefore this yields to:
v1 * 5 hours = 65 mph * 4 hours
v1 = 52 mph
<span>Answer: The average
speed going to the office is 52 mph.</span>