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schepotkina [342]
3 years ago
6

A used boat dealership buys a boat for $2990 and sells it for $4200. what is percent increase?

Mathematics
1 answer:
maxonik [38]3 years ago
4 0
It is a 40.5% increase, how I got my answer.

It is really simple, all I done was added $2,990 and on a caculater started at 100% increase would be 5,980, the reason I started out with 100% is because it is much easyer to judge how much I need to go down or up in percentage, so I knew that I needed to find $4,200, so I moved it down to 45% which equaled $4,335.5 so I brought it down to 41% which equals $4,215.9, so I brought it down to 40% which equals $4,186, then I brought it up to 40.5% which equals $4,200.95, but the .95 does not matter.

I hope this helps!
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3 years ago
A colony of bacteria is growing at a rate of 0.2 times its mass. Here time is measured in hours and mass in grams. The mass of t
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Answer:

  • <u>Question 1:</u>      dm/dt=0.2m<u />

<u />

  • <u>Question 2:</u>     m=Ae^{(0.2t)}<u />

<u />

  • <u>Question 3:</u>      m=10e^{(0.2t)}<u />

<u />

  • <u>Question 4:</u>      m=10g<u />

Explanation:

<u>Question 1: Write down the differential equation the mass of the bacteria, m, satisfies: m′= .2m</u>

<u></u>

a) By definition:  m'=dm/dt

b)  Given:  rate=0.2m

c) By substitution:  dm/dt=0.2m

<u>Question 2: Find the general solution of this equation. Use A as a constant of integration.</u>

a) <u>Separate variables</u>

     dm/m=0.2dt

b)<u> Integrate</u>

           \int dm/m=\int 0.2dt

            ln(m)=0.2t+C

c) <u>Antilogarithm</u>

       m=e^{0.2t+C}

       m=e^{0.2t}\cdot e^C

         e^C=A\\\\m=Ae^{(0.2t)}

<u>Question 3. Which particular solution matches the additional information?</u>

<u></u>

Use the measured rate of 4 grams per hour after 3 hours

            t=3hours,dm/dt=4g/h

First, find the mass at t = 3 hours

            dm/dt=0.2m\\\\4=0.2m\\\\m=4/0.2\\\\m=20g

Now substitute in the general solution of the differential equation, to find A:

          m=Ae^{(0.2t)}\\\\20=Ae^{(0.2\times 3)}\\\\A=20/e^{(0.6)}\\\\A=10.976

Round A to 1 significant figure:

  • A = 10.

<u>Particular solution:</u>

           

             m=10e^{(0.2t)}

<u>Question 4. What was the mass of the bacteria at time =0?</u>

Substitute t = 0 in the equation of the particular solution:

         m=10e^{0}\\\\m=10g

3 0
3 years ago
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mamaluj [8]

Answer:

Mean = 35

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Step-by-step explanation:

Data provided in the question:

X : 1, 2, 3, 4, 5, 6

All the data are independent

Thus,

The mean for this case will be given as:

Mean, E[X] = \frac{\textup{Sum of all the observations}}{\textup{Total number of observations}}

or

 E[X] = \frac{\textup{1+2+3+4+5+6}}{\textup{6}}

or

E[X] = 3.5

For 10 days, Mean = 3.5 × 10 = 35

And,

variance = E[X²] - ( E[X] )²

Now, for this case of independent value,

E[X²] = \frac{1^2+2^2+3^2+4^2+5^2+6^2}{\textup{6}}

or

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Therefore,

variance = E[X²] - ( E[X] )²

or

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or

Variance = 2.917

For 10 days = Variance × Days²

= 2.917 × 10²

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3 0
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