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andreyandreev [35.5K]
2 years ago
7

Question 5

Mathematics
1 answer:
Levart [38]2 years ago
4 0

Answer:

7.8

Step-by-step explanation:

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Identify the variable and the constant in the expression below. n + 3 variable: ___________________________ constant: __________
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N is the variable and 3 is the constant.
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Which algebraic expressions are polynomials? Check all that apply. PLEASE HELP
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Selections 2, 3, 5, 6 are polynomials.

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3 0
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A population has the following characteristics. (a) A total of 75% of the population survives the first year. Of that 75%, 25% s
Artemon [7]

Answer:

= \left[\begin{array}{ccc}1344\\84\\28\end{array}\right]  \left \begin{array}{ccc}{0 \  \leq  age   \leq  1 }\\{ 1 \  \leq  age   \leq  2 }\\{2 \  \leq  age  \leq 3}\end{array}\right

i.e after the first year ;

there 1344 members in the first age class

84 members for the second age class; and

28 members for the third age class

Step-by-step explanation:

We can deduce that the age distribution vector x represents the number of population members for each age class; Given that in each class of age there are 112 members present.

The current age distribution vector is as follows:

x = \left[\begin{array}{ccc}1&1&2\\1&1&2\\1&1&2\end{array}\right] \left[\begin{array}{ccc}{0 \  \leq  age   \leq  1 }\\{ 0 \  \leq  age   \leq  2 }\\{0 \  \leq  age   \leq 3}\end{array}\right]

Also , the age transition matrix is as follows:

L = \left[\begin{array}{ccc}3&6&3\\0.75&0&0 \\0&0.25&0\end{array}\right]

After 1 year ; the age distribution vector will be :

x_2 =Lx_1 = \left[\begin{array}{ccc}3&6&3\\0.75&0&0 \\0&0.25&0\end{array}\right]  \left[\begin{array}{ccc}1&1&2\\1&1&2\\1&1&2\end{array}\right]

= \left[\begin{array}{ccc}1344\\84\\28\end{array}\right]  \left \begin{array}{ccc}{0 \  \leq  age   \leq 1 }\\{ 1 \  \leq  age   \leq  2 }\\{2 \  \leq  age   \leq  3}\end{array}\right

6 0
3 years ago
The standard deviation of a sample taken from population A is 17.6 for a sample of 25.
lawyer [7]

Answer:

The standard deviation of the sample mean differences is _5.23_

Step-by-step explanation:

We have a sample of a population A and a sample of a population B.

For the sample of population A, the standard deviation \sigma_A is

\sigma_A = 17.6

The sample size n_A is:

n_A = 25.

For the sample of population B, the standard deviation \sigma_B is

\sigma_B = 21.2

The sample size n_B is:

n_B = 30.

Then the standard deviation for the difference of means has the following form:

\sigma=\sqrt{\frac{\sigma_A^2}{n_A}+\frac{\sigma_B^2}{n_B}}

Finally

\sigma=\sqrt{\frac{17.6^2}{25}+\frac{21.2^2}{30}}\\\\\sigma= 5.23

7 0
3 years ago
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