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jek_recluse [69]

4 years ago

8

This picture shows the United States Arctic National wildlife refuge, a large portion of Alaska that is protected from human act

ivities that would affect living things in an area as well as non living resources that support living things. Even though oil and other resources in the area could be collected in the area and sold by humans to increase American wealth l, construction is prohibited in the area. The most complex level of organizational hierarchy in the Arctic National Wildlife Refuge that is preserved is...
A) the population of brown bears in the area
B) every species of fish in the area
C) every ecosystem in the area
D) the biosphere that includes Alaskan life
A. The population

2 answers:

Cloud [144]4 years ago

9 0

Answer:

C!

Explanation:

Edge 2020

took the test :)

finlep [7]4 years ago

4 0

Answer:C) every ecosystem in the area

Explanation:

i took the test

trust me

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A bacteria culture starts with 460 bacteria and grows at a rate proportional to its size. After 2 hours there will be 920 bacter

ad-work [718]

Answer: Check explanation.

Explanation:

Important things to note from the question: the total number the bacteria stated with is =460. After time(t), which is two hours passed, the numbers of bacteria increased to= 920.

Now, let us delve into the solution.

The number of bacteria at any time,t= N(t). And N(t) = e^Xt + V. Where V = ln 460 and X can be calculated below;

total number of bacteria remaining after the two hours, 920 = e^2X. e^460.

So, we make Exponential 2X the subject of the formula;

Therefore, e^2X = 920/460.

====> e^2X = 2.

The exponential goes into the right side, turns into 'ln'.

====> 2X = ln 2.

Therefore, X= 1/2 ×ln 2.

OR

X= 0.5 × 0.69315 = 0.35.

Therefore, Expressing the population after t hours as a function of t;

N(t) = 460 × (e^1/2 ln 2)t.

OR

N(t)= 460× e^0.35 t.

CONCLUSION: With this we can slot in any value of time,t to determine the value of the number of bacteria at any time,t(N)

7 0

3 years ago

In the circulatory system nutrients such as glucose and amino acids are transported by

katen-ka-za [31]

In the circulatory system, nutrients such as amino acids and glucose are transported by the PLASMA.
The plasma is the liquid portion of the blood and one of its most important function is to transport materials such as glucose, amino acids, hormones, enzymes, water, etc.

3 0

3 years ago

3. Were there any stages of the cell cycle that you did not observe? How can you explain this using evidence from the cell cycle

mario62 [17]

Answer:

The surface area of the cell increases to volume ratio

Explanation:

As a cell grows, the surface area of the cell increases to volume ratio. When cell division occurs, the surface area is distributed and each daughter cell have less surface area but with the passage of time, the surface area increases. The main function of mitosis is to increase the population of the cell, the cell must be divided into two daughter cells in the process of mitosis. Tumors occurs when the process of mitosis is uncontrollable. One observation that is found interesting while looking at the onion root tip cells is the multiplication of the cells by the process of mitosis.

5 0

3 years ago

A fossil is discovered that has only 12.5% of the carbon-14 that it would have had originally (when the animal was alive). If th

mariarad [96]

Answer:

The fossil is 17,100 years old.

Explanation:

The decay equation:

$\frac{dN}{dt}\propto -N$

$\Rightarrow \frac{dN}{dt}= -\lambda N$

$\Rightarrow \frac{dN}{N}= -\lambda dt$

Integrating both sides

$\Rightarrow \int\frac{dN}{N}= \int-\lambda dt$

$\Rightarrow ln |N|=-\lambda t+c$

When t=0, N= $N_0$ = initial amount

$ln |N_0|=-\lambda .0+c$

$\Rightarrow ln |N_0|=c$

$\therefore ln |N|=-\lambda t+ln|N_0|$

$\Rightarrow ln |N|-ln|N_0|=-\lambda t$

$\Rightarrow ln |\frac {N}{N_0}|=-\lambda t$

$\Rightarrow \frac {N}{N_0}=e^{-\lambda t}$

$\Rightarrow N=N_0e^{-\lambda t}$

The decay equation is

$N=N_0e^{-\lambda t}$

Given that,

The half life of carbon - 14 is 5700 years.

For half life, $N=\frac{1}{2} N_0$

To find the value of $\lambda$ , we need to put the value of N and t in the decay equation.

$\frac12N_0=N_0e^{-\lambda \times 5700}$

$\Rightarrow \frac12=e^{-\lambda \times 5700}$ [ Divided $N_0$ both sides]

Taking ln both sides

$\Rightarrow ln| \frac12|=ln|e^{-\lambda \times 5700}|$

$\Rightarrow ln| \frac12|={-\lambda \times 5700}$

$\Rightarrow \lambda= \frac{ln| \frac12|}{-5700}$

$\Rightarrow \lambda= \frac{ln|1|-ln|2|}{-5700}$ [ $ln|\frac mn|= ln |m|-ln |n|$ ]

$\Rightarrow \lambda= \frac{ln|2|}{5700}$ [ln 1= 0]

The fossil has only 12.5% of the carbon carbon-14 that it would have had originally.

So, $N=\frac{12.5}{100} N_0$

Then,

$\frac{12.5}{100} N_0=N_0e^{-\frac{ln|2|}{5700}t$

$\Rightarrow \frac{12.5}{100} =e^{-\frac{ln|2|}{5700}t$

Taking ln both sides

$\Rightarrow ln|\frac{12.5}{100} |=ln|e^{-\frac{ln|2|}{5700}t}|$

$\Rightarrow ln|\frac{12.5}{100} |={-\frac{ln|2|}{5700}t}$

$\Rightarrow t=\frac{ ln|\frac{12.5}{100}|} {-\frac{ln|2|}{5700}}$

$\Rightarrow t=\frac{ ln|\frac{12.5}{100}|\times 5700} {-{ln|2|}}$

$\Rightarrow t=17,100$

The fossil is 17,100 years old.

7 0

4 years ago

What are the inputs and outputs of a cell system?

Kamila [148]

Answer:

The inputs cellular respiration are glucose and oxygen. The outputs, or products, of cellular respiration are Carbon Dioxide, water, energy

Explanation:

8 0

3 years ago

Read 2 more answers