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Likurg_2 [28]
3 years ago
7

The histogram below displays the distribution of 50 ages at dHere are the number of hours that 9 students spend on the computer

on a typical day:_______.
3 3 5 7 7 9 9 9 12
What is the mode number of hours spent on the computer?
Mathematics
1 answer:
andreyandreev [35.5K]3 years ago
6 0

Answer:

9

Step-by-step explanation:

There are 9 students who spend the following number of hours in a typical day on the computer.

3,3,5,7,7,9,9,9,12

The mode is the number that is of the highest frequency in any given set of values.

The mode number of hours is the highest number of hours spent on the computer.

The number 9 occured the most times. Therefore the mode hour is 9

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What number is greater 0.111 or 1.001. also 1.11 or 1.01
Paul [167]
1.001 is greater than 0.111 because the 1 is automatically greater than zero.  1.11 is greater than 1.01 because there is a place in the tenths place in 1.11 but none for 1.01.
4 0
3 years ago
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A large tank is filled to capacity with 700 gallons of pure water. Brine containing 3 pounds of salt per gallon is pumped into t
inn [45]

Answer:

The number A(t) of pounds of salt in the tank at time 't' is;

{A(t)}= -2,100 \times e^{\dfrac{-t}{100} } + 2,100

Step-by-step explanation:

In the question, we have;

The volume of pure water initially in the tank = 700 gal

The concentration of brine pumped into the tank = 3 pounds per gallon

The rate at which the brine is pumped into the tank,  = 7 gal/min

The rate at which the well mixed solution is pumped out = The same 7 gal/min

The number of pounds of salt in the tank at time 't' is found as follows;

The rate of change in A(t) with time = The rate of salt input - The rate of salt output

The \ rate \  of  \ change \  in  \ A(t)  \ with \  time = \dfrac{dA}{dt}

The rate of salt input = 7 gal/min × 3 lbs/gal = 21 lbs/min

The rate of salt output = (A(t)/700) lb/gal × 7 gal/min = (A(t)/100) lb/min

Therefore, we have;

\dfrac{dA}{dt} = 21 - \dfrac{A(t)}{100}

Therefore;

\dfrac{dA}{dt} + \dfrac{A(t)}{100}= 21

The integrating factor is e^{\int\limits {\frac{1}{100} } \, dx } = e^{\frac{x}{100} }

e^{\dfrac{t}{100} } \times \dfrac{dA}{dt} + e^{\dfrac{t}{100} } \times\dfrac{A(t)}{100}= e^{\dfrac{t}{100} } \times21

\dfrac{d}{dt} \left[ e^{\dfrac{t}{100} } \times{A(t)}{} \right]= e^{\dfrac{t}{100} } \times21

Using an online tool, we get;

{A(t)}= c_1 \times e^{\dfrac{-t}{100} } + 2,100

At time t = 0, A(t) = 0

We get;

0= c_1 \times e^{\dfrac{-0}{100} } + 2,100

c₁ = -2,100

Therefore, the number A(t) of pounds of salt in the tank at time 't' is given as follows;

{A(t)}= -2,100 \times e^{\dfrac{-t}{100} } + 2,100

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Step-by-step explanation:

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