You start from the right and work your way to the left.
3/3= 1
1/43= .023255814
.023255814/8= .0029069767
The awnser is 0.0029069767
Answer:
The slope for number 4 is 5/-4 and for number 6 is 4/1
Step-by-step explanation:
5/-4: slope is rise over run so you have to go up (positive) 5 and you have to go to the left (negative) so its 5/-4
4/1: again it's rise over run so you have to go up (positive) 4 and right one (positive)
Answer:
The average speed for the journey is 6.4 mph.
Step-by-step explanation:
Given:
If you walk for 45 minutes at a rate of and then run for 30 minutes at a rate of , then how many miles have you gone at the end of one hour and 15 minutes Find the average speed for the journey
Step 1 of 1
We can compute the walking distance and running distances separately, then add them.
You walk for 45 minutes, which is hours, at a rate of , so the walking distance is miles. You run for 30 minutes, which is , so the running distance is miles.
So the total distance you cover is 3+5=8 miles.
Now we can also compute the average speed for the journey.
You covered 8 miles in hours, so your average speed was
So find the common denominator of all:30
12/30, 9/30 and 8/30 then add them 29/30
These are tough. You really have to pay attention to what the table is telling you.
This table is apparently the result of a survey of 80 students.
The survey asked each student two questions:
-- Do you have a cellphone ?
-- Do you have an .mp3 player ?
The table shows all the results of the survey.
(If you knew any of this, you should have told us.)
The table says that ...
-- 37 of the students surveyed have a cell phone and an .mp3 player too.
-- 21 of the students surveyed have a cell phone but no.mp3 player.
-- 13 of the students surveyed have no cell phone but do have an .mp3 player.
-- 9 of the students surveyed have no cell phone and no .mp3 player either.
Think about the boxes in the table for a second.
-- Either you have a cellphone or you don't.
-- Either you have an .mp3 player or you don't.
-- So the same student is never counted in more than 1 box.
-- So the sum of the 4 boxes is the total number of students surveyed.
That's 80.
OK ?
Now, let's check out the choices and see if any are true:
A). "All students that own cellphones also own .mp3 players."
The upper right box in the table says that 21 of the students surveyed
have a cellphone but no .mp3 player.
This choice is false.
B). "Less than half of the students own a cellphone."
The top row of the table ... the 37 and the 21 ... are
the students who own cellphones.
They add up to 58 ... more than half of the 80 who were surveyed.
This choice is false.
C). "More than half of the students own an .mp3 player."
The left column of the table ... the 37 and the 13 ... are the
students that own an .mp3 player.
They add up to (37 + 13) = 50 ... more than half of the 80.
This choice is true.
D). "Less than 10% of the students do not own either one."
The bottom-right box in the table ... the 9 ... are the students
surveyed who don't own either one.
9 is a little bit more than 11% of the students surveyed.
So this choice doesn't work either.
Choice-C is the only one supported by the numbers in the table.
I told you these are tough.
At least they are for me.
What makes them especially tough may be the fact that they require thinking.