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djyliett [7]
3 years ago
15

Residence time is the amount of time in which a substance remains in a reservoir, such as the atmosphere. For example, the resid

ence time of carbon tetrafluoride (CF4) in our atmosphere is more than 50,000 years. In contrast, carbon dioxide (CO2) has a residence time that is much shorter. Which statement best explains how the residence time of a substance affects the atmosphere?
A. Substances that have a longer residence time are more beneficial to the atmosphere.

B. Substances that have a shorter residence time make up a smaller portion of the atmosphere.

C. Substances that have a longer residence time will cycle faster through the atmosphere.

D. Substances that have a shorter residence time do not stay in the atmosphere as long.
Advanced Placement (AP)
2 answers:
Alja [10]3 years ago
5 0

Given the description of the meaning of "residence time," you must conclude ...

... D. Substances that have a shorter residence time do not stay in the atmosphere as long.

xeze [42]3 years ago
5 0

Answer:

The correct answer is D.  Substances that have a shorter residence time do not stay in the atmosphere as long.

I just took the test on plato and it was right.

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Identify and explain 1 reason why the institution of slavery was politically important to the South during the Antebellum period
Ivenika [448]

Might not be a great answer but:

Due to the 3/5 clause, each slave was counted as 3/5 of a person represented in congress. The south's political power relied largely on this fact.

5 0
2 years ago
Which of the following statements is most correct?
r-ruslan [8.4K]
The correct answer for the question that is being presented above is this one: "C) Depression is related to neurotransmitter deregulation, genetics, and cognitive processes." Among the four statements that is presented, the correct statement is that <span>Depression is related to neurotransmitter deregulation, genetics, and cognitive processes.</span>
6 0
3 years ago
The graph of x^2=-2+y+5cosy is shown for y=11
erik [133]

(a)

We have been given an equation x^{2}=-2+y+5cosy. Upon taking derivative of this equation with respect to x on both the sides, we get:

2x=0+\frac{dy}{dx}+5(-siny)\frac{dy}{dx}\\2x=(1-5siny)\frac{dy}{dx}\Rightarrow \frac{dy}{dx}=\frac{2x}{1-5siny}

(b)

In order to write the equation of tangent line, we need to find the slope of the line. We know that point (3,11) lies on the graph. Therefore, we write the slope of the tangent as:

\text{Slope}=\frac{dy}{dx}=\frac{2(3)}{1-5sin(11)}=1

Therefore, equation of tangent line is:

y-y_{1}=m(x-x_{1})\Rightarrow y-11=1(x-3)\\y-11=x-3\\y=x+8

(c)

We know that slope of tangent is given as \frac{dy}{dx}=\frac{2x}{1-5siny}. The tangent will be vertical when denominator of slope is zero, that is:

1-5siny=0\Rightarrow siny=\frac{1}{5}\Rightarrow y=arcsin(\frac{1}{5})\\y=1.732,1.525, 3.063 \text{ etc.}

5 0
2 years ago
Match the values associated with this data set to their correct descriptions. {6, 47, 49, 15, 43, 41, 7, 36}quartile38.5median11
cluponka [151]
Answers:
1) The first quartile (Q₁) = 11 ;  2) The median = 38.5 ; 
3) The third quartile (Q₃) = 45 ;
4) The difference of the largest value and the median = 10.5 .
_______
Explanation: 

Given this data set with 8 (eight) values:  →  {6, 47, 49, 15, 43, 41, 7, 36};
→Rewrite the values in increasing order; to help us find the median, first quartile (Q,) and third quartile (Q₃) : → {6, 7, 15, 36, 41, 43, 47, 49}.
→We want to find; or at least match; the following 4 (four) values [associated with the above data set] — 38.5, 11, 10, 45 ;

1) The first quartile (Q₁);  2) The median;  3) The third quartile (Q₃); & 
4) The difference of the largest value and the median.

Note: Let us start by finding the "median". This will help us find the correct values for the descriptions in "Numbers 2 & 4" above.
The "median" would be the middle number within a data set, when the values are placed in smallest to largest (or, largest to smallest).  However, our data set contains an EVEN number [specifically, "8" (eight)] values. In these cases , we take the 2 (two) numbers closest to the middle, and find the "mean" of those 2 (two) numbers; and that value obtained is the median.  So, in our case, the 2 (two) numbers closest to the middle are:
"36 & 41".  To get the "mean" of these 2 (two) numbers, we add them together to get the sum; and then, we divide that value by "2" (the number of values we are adding):
→  36 + 41 = 77;  → 77/2 = 38.5 ; → which is the median for our data set; and is a listed value.
→Now, examine Description "(#4): The difference of the largest value and the median"—(SEE ABOVE) ;
→ We can calculate this value.  We examine the values within our data set to find the largest value, "49".  Our calculated "median" for our dataset, "38.5".  So, to find the difference, we subtract: 49 − 38.5 = 10.5 ; which is a given value".
→Now, we have 2 (two) remaining values, "11" & "45"; with only 2 (two) remaining "descriptions" to match;
 →So basically we know that "11" would have to be the "first quartile (Q₁)";  & that "45" would have to be the "third quartile (Q₃)".
→Nonetheless, let us do the calculations anyway.
→Let us start with the "first quartile";  The "first quartile", also denoted as Q₁, is the median of the LOWER half of the data set (not including the median value)—which means that about 25% of the numbers in the data set lie below Q₁; & that about 75% lie above Q₁.). 
→Given our data set:   {6, 7, 15, 36, 41, 43, 47, 49};
We have a total of 8 (eight) values; an even number of values. 
The values in the LOWEST range would be:  6, 7, 15, 36.
The values in the highest range would be:  41, 43, 47, 49.
Our calculated median is: 38.5 .  →To find Q₁, we find the median of the numbers in the lower range. Since the last number of the first 4 (four) numbers in the lower range is "36"; and since "36" is LESS THAN the [calculated] median of the data set, "38.5" ; we shall include "36" as one of the numbers in the "lower range" when finding the "median" to calculate Q₁
→ So given the lower range of numbers in our data set:  6, 7, 15, 36 ;
We don't have a given "median", since we have an EVEN NUMBER of values.  In this case, we calculate the MEDIAN of these 4 (four) values, by finding the "mean" of the 2 (two) numbers closest to the middle, which are "7 & 15".  To find the mean of "7 & 15" ; we add them together to get a sum; 
then we divide that sum by "2" (i.e. the number of values added up);
   → 7 + 15 = 22 ;  → 22 ÷ 2 = 11 ;  ↔ Q₁ = 11.
Now, let us calculate the third quartile; also known as "Q₃".
    Q₃ is  the median of the last half of the higher values in the set, not including the median itself.  As explained above, we have a calculated median for our data set, of 38.5; since our data set contains an EVEN number of values.  We now take the median of our higher set of values (which is Q₃). Since our higher set of values are an even number of values; we calculate the median of these 4 (four) values by taking the mean of the 2 (two) numbers closest to the center of the these 4 (four) values.  This value is Q₃.  →Given our higher set of values:  41, 43, 47, 49 ;  → We calculate the "median" of these 4 (four) numbers; by taking the mean of the 2 (two) numbers in the middle; "43 & 47".
 → Method 1): List the integers from "43 to 47" ;  → 43, 44, 45, 46, 47;
→ Since this is an ODD number of integers in sequential order;
→ "45" is not only the "median"; but also the "mean" of (43 & 47); 
thus, 45 = Q₃; 
→ Method 2):  Our higher set of values:  41, 43, 47, 49 ;
→ We calculate the "median" of these 4 (four) numbers; by taking the
"mean" of the 2 (two) numbers in the middle; "43 & 47";  We don't have a given "median", since we have an EVEN NUMBER of values.  In this case, we calculate the MEDIAN of these 4 (four) values, by finding the mean of the 2 (two) numbers closest to the middle, which are "43 & 47."  To find the mean of "43 & 47"; we add them together to get a sum; then we divide that sum by "2" (i.e. the number of values added);
→ 43 + 47 = 90 ;  → 90 ÷ 2 = 45 ;  → 45 = Q₃ .
4 0
3 years ago
Read 2 more answers
Just writing down your grades after each test does not count as monitoring progress.
andreyandreev [35.5K]

Answer:

False.

Explanation:

Writing down your grades after each test gives you an accurate and recent set of data to consistently look at and monitor. Whether you progress or degrees, you are still monitoring the data.

4 0
3 years ago
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