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3 years ago

What is the difference in minutes of median collection times before and after the well was installed?

Mathematics

2 answers:

stiks02 [169]3 years ago

6 0

The difference is the outlier

Svetach [21]3 years ago

6 0

The outlier bro
Hahaha lol

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Factor out the greatest common factor. 10a3b−6a2b2

podryga [215]

Answer:

2a^2b(5a-3b)

If not then the other way is

5a-3b

Step-by-step explanation:

The GCF of 10a^3b and -6a^2b^2 is 2a^2b

So divide by that you will get

2a^2b(5a-3b)

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3 years ago

Solve for m <br>-2(m +5) +14 =-6

Westkost [7]

Answer:

m=5

Step-by-step explanation:

-2m-10+14=-6

-2m+4=-6

-2m=-10

m=5

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2 years ago

Solve each equation & show<br> 3) 36 = 1+ 7a

Sladkaya [172]

Answer:

a=5

The drawing will help

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3 years ago

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$210 at 8% for 7 years

Genrish500 [490]

The answer is $327.60

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3 years ago

The population of a town grows at a rate proportional to the population present at time t. The initial population of 500 increas

Mazyrski [523]

Answer:

The population in 40 years will be 1220.

Step-by-step explanation:

The population of a town grows at a rate proportional to the population present at time t.

This means that:

$P(t) = P(0)e^{rt}$

In which P(t) is the population after t years, P(0) is the initial population and r is the growth rate.

The initial population of 500 increases by 25% in 10 years.

This means that $P(0) = 500, P(10) = 1.25*500 = 625$

We apply this to the equation and find t.

$P(t) = P(0)e^{rt}$

$625 = 500e^{10r}$

$e^{10r} = \frac{625}{500}$

$e^{10r} = 1.25$

Applying ln to both sides

$\ln{e^{10r}} = \ln{1.25}$

$10r = \ln{1.25}$

$r = \frac{\ln{1.25}}{10}$

$r = 0.0223$

$P(t) = 500e^{0.0223t}$

What will be the population in 40 years

This is P(40).

$P(t) = 500e^{0.0223t}$

$P(40) = 500e^{0.0223*40} = 1220$

The population in 40 years will be 1220.

7 0

3 years ago