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lesya [120]
2 years ago
10

Suppose a population of honey bees in Ephraim, UT has an initial population of 2300 and 12 years later the population reaches 13

000. Use an explicit exponential model to find the rate of growth and common ratio for the honey bee population. Express the rate of growth as a percentage. Round to the nearest tenth. Express the common ratio as a decimal. Round to the nearest thousandth.
Mathematics
1 answer:
Rama09 [41]2 years ago
6 0

Answer:

common ratio: 1.155

rate of growth: 15.5 %

Step-by-step explanation:

The model for exponential growth of population P looks like: P(t)=P_i(1+r)^t

where P(t) is the population at time "t",

P_i is the initial (starting) population

(1+r) is the common ratio,

and r is the rate of growth

Therefore, in our case we can replace specific values in this expression (including population after 12 years, and  initial population), and solve for the unknown common ratio and its related rate of growth:

P(t)=P_i(1+r)^t\\13000=2300*(1+r)^{12}\\\frac{13000}{2300} = (1+r)^12\\\frac{130}{23} = (1+r)^{12}\\1+r=\sqrt[12]{\frac{130}{23} } =1.155273\\

This (1+r) is the common ratio, that we are asked to round to the nearest thousandth, so we use: 1.155

We are also asked to find the rate of increase (r), and to express it in percent form. Therefore we use the last equation shown above to solve for "r" and express tin percent form:

1+r=1.155273\\r=1.155273-1=0.155273

So, this number in percent form (and rounded to the nearest tenth as requested) is: 15.5 %

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A particular sale involves four items randomly selected from a large lot that is known to contain 9% defectives. Let X denote th
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Answer:

The expected repair cost is $3.73.

Step-by-step explanation:

The random variable <em>X</em> is defined as the number of defectives among the 4 items sold.

The probability of a large lot of items containing defectives is, <em>p</em> = 0.09.

An item is defective irrespective of the others.

The random variable <em>X</em> follows a Binomial distribution with parameters <em>n</em> = 9 and <em>p</em> = 0.09.

The repair cost of the item is given by:

C=3X^{2}+X+2

Compute the expected cost of repair as follows:

E(C)=E(3X^{2}+X+2)

        =3E(X^{2})+E(X)+2

Compute the expected value of <em>X</em> as follows:

E(X)=np

         =4\times 0.09\\=0.36

The expected value of <em>X</em> is 0.36.

Compute the variance of <em>X</em> as follows:

V(X)=np(1-p)

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The variance of <em>X</em> is 0.3276.

The variance can also be computed using the formula:

V(X)=E(Y^{2})-(E(Y))^{2}

Then the formula of E(Y^{2}) is:

E(Y^{2})=V(X)+(E(Y))^{2}

Compute the value of E(Y^{2}) as follows:

E(Y^{2})=V(X)+(E(Y))^{2}

          =0.3276+(0.36)^{2}\\=0.4572

The expected repair cost is:

E(C)=3E(X^{2})+E(X)+2

         =(3\times 0.4572)+0.36+2\\=3.7316\\\approx 3.73

Thus, the expected repair cost is $3.73.

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