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Free_Kalibri [48]
3 years ago
8

Can anyone solve this?

Mathematics
1 answer:
matrenka [14]3 years ago
7 0
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.
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(b)

\dfrac{dP(t)}{P(t)} =k dt\\$Take the integral of both sides\\\int \dfrac{dP(t)}{P(t)} =\int k dt\\\ln P(t)=kt+C, $C a constant of integration\\Take the exponential of both sides\\e^{\ln P(t)}=e^{kt+C}\\P(t)=e^{kt}\cdot e^C  $, (Since e^C$ is a constant, we then have:)\\P(t)=Ce^{kt}

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Answer: Choice A
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------------------------------------------------------------------

Explanation:

The domain is the set of x values, or the set of inputs, of a function or relation. All you have to do is list off the values in the "input" oval. Those values from top to bottom are: -3, 2, -5, 1, 6. Sort them out from smallest to largest and you get -5, -3, 1, 2, 6 which points us to choice A.

Another route you could go is eliminating choices B, C and D because -9 is not in the "input" oval. 
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