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3 years ago

12

There are four spheres of earth. These include ______, _, and _____. They are important because when th

ey work together they help keep plants and animals alive on earth.

1 answer:

ololo11 [35]3 years ago

4 0

Answer:

"lithosphere" , "hydrosphere" , "biosphere" , "atmosphere"

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i need help with this asap plz help me i rlly need help plzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzz

Mama L [17]

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3 years ago

Read 2 more answers

You push a 10 n object 10 m how much work was done on the object

dezoksy [38]

Answer: the answer is

Work= Force * Distance

Work= 10N * 10m

Work= 100J

Explanation:

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3 years ago

Read 2 more answers

PythonA group of statisticians at a local college has asked you to create a set of functionsthat compute the median and mode of

skelet666 [1.2K]

Answer:

def median(l):
if(len(l) == 0):
return 0
else:
l.sort()
if(len(l)%2 == 0):
index = int(len(l)/2)
mid = (l[index-1] + l[index]) / 2
else:
mid = l[len(l)//2]
return mid
def mode(l):
if(len(l)==0):
return 0
mode = max(set(l), key=l.count)
return mode
def mean(l):
if(len(l)==0):
return 0
sum = 0
for x in l:
sum += x
mean = sum / len(l)
return mean
lst = [5, 7, 10, 11, 12, 12, 13, 15, 25, 30, 45, 61]
print(mean(lst))
print(median(lst))
print(mode(lst))

Explanation:

Firstly, we create a median function (Line 1). This function will check if the the length of list is zero and also if it is an even number. If the length is zero (empty list), it return zero (Line 2-3). If it is an even number, it will calculate the median by summing up two middle index values and divide them by two (Line 6-8). Or if the length is an odd, it will simply take the middle index value and return it as output (Line 9-10).

In mode function, after checking the length of list, we use the max function to estimate the maximum count of the item in list (Line 17) and use it as mode.

In mean function, after checking the length of list, we create a sum variable and then use a loop to add the item of list to sum (Line 23-25). After the loop, divide sum by the length of list to get the mean (Line 26).

In the main program, we test the three functions using a sample list and we shall get

20.5

12.5

12

3 0

3 years ago

Consider that a system has two entities, Students, Instructors and Course. The Student has the following properties: student nam

tekilochka [14]

Answer:

There's no answer ?

Explanation:

5 0

3 years ago

Consider the following hypothetical scenario for Jordan Lake, NC. In a given year, the average watershed inflow to the lake is 9

dybincka [34]

Answer:

The lake can withdraw a maximum of $1.464\times 10^{10}$ cubic feet per year to provide water supply for the Triangle area.

Explanation:

The maximum amount of water that can be withdrawn from the lake is represented by the following formula:

$V = V_{in}+V_{p}-V_{e}-V_{out}$ (Eq. 1)

Where:

$V$ - Available amount of water for water supply in the Triangle area, measured in cubic feet per year.

$V_{in}$ - Inflow amount of water, measured in cubic feet per year.

$V_{out}$ - Amount of water released for the benefit of fish and downstream water users, measured in cubic feet per year.

$V_{p}$ - Amount of water due to precipitation, measured in cubic feet per year.

$V_{e}$ - Amount of evaporated water, measured in cubic feet per year.

Then, we can expand this expression as follows:

$V = f_{in}\cdot \Delta t+h_{p}\cdot A_{l}-h_{e}\cdot A_{l}-f_{out}\cdot \Delta t$

$V = (f_{in}-f_{out})\cdot \Delta t +(h_{p}-h_{e})\cdot A_{l}$ (Eq. 2)

Where:

$f_{in}$ - Average watershed inflow, measured in cubic feet per second.

$f_{out}$ - Average flow to be released, measured in cubic feet per second.

$\Delta t$ - Yearly time, measured in seconds per year.

$h_{p}$ - Change in lake height due to precipitation, measured in feet per year.

$h_{e}$ - Change in lake height due to evaporation, measured in feet per year.

$A_{l}$ - Surface area of the lake, measured in square feet.

If we know that $f_{in} = 900\,\frac{ft^{3}}{s}$ , $f_{out} = 300\,\frac{ft^{3}}{s}$ , $\Delta t = 31,536,000\,\frac{second}{yr}$ , $h_{p} = 32\,\frac{in}{yr}$ , $h_{e} = 55\,\frac{in}{yr}$ and $A_{l} = 47,000\,acres$ , the available amount of water for supply purposes in the Triangle area is:

$V = \left(900\,\frac{ft^{2}}{s}-300\,\frac{ft^{3}}{s} \right)\cdot \left(31,536,000\,\frac{s}{yr} \right) +\left(32\,\frac{in}{yr}-55\,\frac{in}{yr} \right)\cdot \left(\frac{1}{12}\,\frac{ft}{in}\right)\cdot (47000\,acres)\cdot \left(43560\,\frac{ft^{2}}{acre} \right)$ $V = 1.464\times 10^{10}\,\frac{ft^{3}}{yr}$

The lake can withdraw a maximum of $1.464\times 10^{10}$ cubic feet per year to provide water supply for the Triangle area.

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3 years ago