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777dan777 [17]
3 years ago
6

PLZ ANSWER ILL GIVE BRAINILEST2 box plots. The number line goes from 26 to 36. For Pleasantville Middle School, the whiskers ran

ge from 26 to 34, and the box ranges from 29.5 to 33. A line divides the box at 31. For Grandview Middle School, the whiskers range from 29.3 to 35.3, and the box ranges from 30.5 to 33. A line divides the box at 31.8.
PLZ ANSWER ILL GIVE BRAINILEST

Why should you use the mean and mean absolute deviation (MAD) to compare populations with symmetrical distributions?
The mean and MAD will make graphs of the data appear asymmetrical.
The mean and MAD eliminate the influence of outliers.
The mean and MAD can accurately describe the “typical” value in the symmetric data set.
The mean and MAD make the data distribute more unevenly.
Mathematics
2 answers:
Svetach [21]3 years ago
6 0

Answer:

I believe the answer is- The mean and MAD can accurately describe the "typical" value in the symmetric data set.

Step-by-step explanation:

The other answers don't make sense because the mean and MAD are being used for symmetrical distributions and asymmetrical means uneven distributions.

allsm [11]3 years ago
3 0

Answer:

1- mean

2- MAD

Step-by-step explanation:

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Answer:

∴(x+5)(x+3) =x²+8x+15

Step-by-step explanation:

Standard form of a polynomial:

Standard of polynomial means that the terms ordered from biggest exponent of variable to lowest exponent of variable.

Example:

a_nx^n+a_{n-1}x^{n-1}+a_{n-2}x^{n-2}+......

Given,

(x+5)(x+3)

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=x²+8x+15

∴(x+5)(x+3) =x²+8x+15

The degree of the polynomial is 2 and is a one variable polynomial.

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Suppose after 2500 years an initial amount of 1000 grams of a radioactive substance has decayed to 75 grams. What is the half-li
krok68 [10]

Answer:

The correct answer is:

Between 600 and 700 years (B)

Step-by-step explanation:

At a constant decay rate, the half-life of a radioactive substance is the time taken for the substance to decay to half of its original mass. The formula for radioactive exponential decay is given by:

A(t) = A_0 e^{(kt)}\\where:\\A(t) = Amount\ left\ at\ time\ (t) = 75\ grams\\A_0 = initial\ amount = 1000\ grams\\k = decay\ constant\\t = time\ of\ decay = 2500\ years

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\frac{1}{2} A_0 = A_0 e^{(-0.001036t)}\\Dividing\ both\ sides\ by\ A_0\\ \frac{1}{2} = e^{-0.001036t}\\taking\ natural\ logarithm\ of\ both\ sides\\In(0.5) = In (e^{-0.001036t})\\-0.6931 = -0.001036t\\t = \frac{-0.6931}{-0.001036} \\t = 669.02 years\\\therefore t\frac{1}{2}  \approx 669\ years

Therefore the half-life is between 600 and 700 years

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3 years ago
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