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1 year ago

15 points please help me ASAP

Mathematics

1 answer:

gayaneshka [121]1 year ago

6 0

The distance between the means is equal to 4.
The mean absolute deviation (MAD) for player A is equal to 2.
The mean absolute deviation (MAD) for player B is equal to 2.
Distance between the means = two (2) times the MAD.
There is some overlap and the distance between the means is between 1 times the MAD and 5 times the MAD.

<h3>What is a number line?</h3>

A number line can be defined as a type of graph with a graduated straight line which contains both positive and negative numerical values that are placed at equal intervals along its length.

A line plot can be defined as a type of graph that is used to graphically represent a data set above a number line, while using crosses, dots, or any other mathematical symbol.

A mean is also referred to as an average and it can be defined as a ratio of the sum of the total number in a data set (population) to the frequency of the data set.

Mathematically, the distance between the means of both players A and B is given by:

Distance = Mean of player B - Mean of player A

Distance = 10 - 6

Distance = 4.

The mean absolute deviation (MAD) for player A is given by:

MAD A = 1/11[2(3 - 6) + 2(4 - 6) + (5 - 6) + (6 - 6) + (7 - 6) + 2(8 - 6) + 2(9 - 6)]

MAD A = 1/11[2(3) + 2(2) + (1) + (0) + (1) + 2(2) + 2(3)]

MAD A = 1/11[6 + 4 + 1 + 0 + 1 + 4 + 6]

MAD A = 1/11 × [22]

MAD A = 22/11

MAD A = 2.

Also, the mean absolute deviation (MAD) for player B is given by:

MAD B = 1/11[2(7 - 10) + 2(8 - 10) + (9 - 10) + (10 - 10) + (11 - 10) + 2(12 - 10) + 2(13 - 10)]

MAD B = 1/11[2(3) + 2(2) + (1) + (0) + (1) + 2(2) + 2(3)]

MAD B = 1/11[6 + 4 + 1 + 0 + 1 + 4 + 6]

MAD B = 1/11 × [22]

MAD B = 22/11

MAD B = 2.

Therefore, distance between the means = two (2) times the MAD.

Distance = Means × MAD

4 = 2 × 2.

In conclusion, we can infer and logically deduce that there is some overlap and the distance between the means of both players A and B is between one (1) times the MAD and five (5) times the MAD.

Read more on line plot here: brainly.com/question/8989301

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To be clear, the given relation between time and female population is an integral:
 $t = \int { \frac{P+S}{P[(r - 1)P - S]} } \,
dP$

The problem says that r = 1.2 and S = 400, therefore substituting:
 $t = \int { \frac{P+400}{P[(1.2 - 1)P - 400]}
} \, dP$ 

= $
\int { \frac{P+400}{P(0.2P - 400)} } \, dP$

In order to evaluate this integral, we need to write this rational function in a simpler way:
 $\frac{P+400}{P(0.2P - 400)} = \frac{A}{P} +
\frac{B}{(0.2P - 400)}$ 

where we need to evaluate A and B. In order to do so, let's calculate the LCD:
 $\frac{P+400}{P(0.2P - 400)} = \frac{A(0.2P -
400)}{P(0.2P - 400)} + \frac{BP}{P(0.2P - 400)}$ 

the denominators cancel out and we get:
P + 400 = 0.2AP - 400A + BP
 = P(0.2A + B) - 400A

The two sides must be equal to each other, bringing the system:
 $\left \{ {{0.2A + B = 1} \atop {-400A =
400}} \right.$ 

Which can be easily solved:
 $\left \{ {{B=1.2} \atop {A=-1}} \right.$ 

Therefore, our integral can be written as:
 $t = \int { \frac{P+400}{P(0.2P - 400)} } \,
dP = \int {( \frac{-1}{P} + \frac{1.2}{0.2P-400} )} \, dP$ 
= $- \int { \frac{1}{P} \, dP +
1.2\int { \frac{1}{0.2P-400} } \, dP$ 
= $- \int { \frac{1}{P} \, dP +
6\int { \frac{0.2}{0.2P-400} } \, dP$ 
= - ln |P| + 6 ln |0.2P - 400| + C

Now, let’s evaluate C by considering that at t = 0 P = 10000:
0 = - ln |10000| + 6 ln |0.2(10000) - 400| + C
C = ln |10000| - 6 ln |1600|
C = ln (10⁴) - 6 ln (2⁶·5²)
C = 4 ln (10) - 36 ln (2) - 12 ln (5) 
 Therefore, the equation relating female population with time requested is:
t = - ln |P| + 6 ln |0.2P - 400| + 4 ln (10) - 36 ln (2) - 12 ln (5)

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