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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?

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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Zigmanuir [339]

The answer is 17

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7 0

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

11Alexandr11 [23.1K]

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

Each morning, Hungry Harry eats some eggs. On any given morning, the number of eggs he eats is equally likely to be 1, 2, 3, 4,

mamaluj [8]

Answer:

Mean = 35

Variance = 291.7

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

E[X²] = $\frac{1+4+9+16+25+36}{\textup{6}}$

or

E[X²] = $\frac{91}{\textup{6}}$

or

E[X²] = 15.167

Therefore,

variance = E[X²] - ( E[X] )²

or

variance = 15.167 - 3.5²

or

Variance = 2.917

For 10 days = Variance × Days²

= 2.917 × 10²

= 291.7

3 0

3 years ago