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Aleks [24]
3 years ago
8

Jeff jones is paid $60 per day at his job at J. C. Penny. Jeff became ill on Monday and had to leave after 2/5 of a day. What di

d he earn on Monday?
Mathematics
2 answers:
sukhopar [10]3 years ago
7 0

To solve this, we simply multiply his usual pay by the amount of "day" he actually worked.

60 * (1 - 2/5) = 60 * (3/5)

180/5 = 36

Jeff jones is paid $36 on Monday

mel-nik [20]3 years ago
6 0

5 \sqrt{60}
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Mike ran 1 mile in 4 minutes. How far did he run in 10seconds?
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Answer:

0.015 miles

Step-by-step explanation:

1mile = 4min \\ 1mile = 240sec \\ 0.609km = 240sec \\  \frac{0.609}{24}  =  \frac{240}{24}  \\ 10sec = 0.025km \\ 10sec = 0.015miles

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Mrs Hom bought 44 pizzas which consist of regular size and large size pizzas to be distributed equally to 96 workers. Each regul
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Answer:

I came up with 412cm^2

Step-by-step explanation:

10*12+(12*2)2+(10*2)2+6*8+(8*3)2+(6*3)2+(12*10-6*8)

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If a tennis ball has a diameter of 14, What is the volume of the tennis ball in terms of pie?
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2 years ago
Read 2 more answers
A tank initially contains 60 gallons of brine, with 30 pounds of salt in solution. Pure water runs into the tank at 3 gallons pe
adoni [48]

Answer:

the amount of time until 23 pounds of salt remain in the tank is 0.088 minutes.

Step-by-step explanation:

The variation of the concentration of salt can be expressed as:

\frac{dC}{dt}=Ci*Qi-Co*Qo

being

C1: the concentration of salt in the inflow

Qi: the flow entering the tank

C2: the concentration leaving the tank (the same concentration that is in every part of the tank at that moment)

Qo: the flow going out of the tank.

With no salt in the inflow (C1=0), the equation can be reduced to

\frac{dC}{dt}=-Co*Qo

Rearranging the equation, it becomes

\frac{dC}{C}=-Qo*dt

Integrating both sides

\int\frac{dC}{C}=\int-Qo*dt\\ln(\abs{C})+x1=-Qo*t+x2\\ln(\abs{C})=-Qo*t+x\\C=exp^{-Qo*t+x}

It is known that the concentration at t=0 is 30 pounds in 60 gallons, so C(0) is 0.5 pounds/gallon.

C(0)=exp^{-Qo*0+x}=0.5\\exp^{x} =0.5\\x=ln(0.5)=-0.693\\

The final equation for the concentration of salt at any given time is

C=exp^{-3*t-0.693}

To answer how long it will be until there are 23 pounds of salt in the tank, we can use the last equation:

C=exp^{-3*t-0.693}\\(23/60)=exp^{-3*t-0.693}\\ln(23/60)=-3*t-0.693\\t=-\frac{ln(23/60)+0.693}{3}=-\frac{-0.959+0.693}{3}=  -\frac{-0.266}{3}=0.088

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3 years ago
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