Mathematics, 17.07.2020 05:01 la200564

A model for the number of lobsters caught per year is based on the assumption that the number of lobsters caught in a year is the average of the number caught in two previous years. (Show your work, circle or put in bold each intermediate and final answer.) a. Find a recurrence relation for Ln -- the number of lobsters caught in year n.
b. Solve this recurrence relation (find Ln), if 100,000 lobsters were caught in year 1 and 300,000 lobsters were caught in year 2.

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Answer from: ybetancourt1

a. Ln = $\frac{1}{2} (L_{n-1} + L_{n-2})$

b. $L_{n} = 233333.4(1)^{n}+266666.7(-\frac{1}{2} )^{n}$

Step-by-step explanation: Recurrence Relation is a polynomial that relates a term in a sequence with its previous terms.

a. The number of lobster is the average of two previous years. Average is the sum of the elements of a set divided by the total number of the set, so for Ln:

Ln = $\frac{L_{n-1}+L_{n-2}}{2}$ = $\frac{1}{2}(L_{n-1}+L_{n-2})$

where $L_{n-1} and L_{n-2}$ are the two previous years.

b) Ln = $\frac{1}{2}(L_{n-1}+L_{n-2})$

$\frac{1}{2}(L_{n-1})+\frac{1}{2} (L_{n-2}) - L_{n} = 0$

To solve this recurrence, find the characteristic polynomial:

$r^{2} - \frac{1}{2}r - \frac{1}{2} = 0$

Solve for r:

The polynomial can be rewritten as:

$(r-1)(r+\frac{1}{2} )=0$

so, r = 1 and r = -1/2

The expression for the recurrence relation will be:

$L_{n} = \alpha_{1} .(1)^{n} + \alpha_{2}.(-\frac{1}{2} )^{n}$

In year 1, there were 100,000 lobsters, so, when n=1:

$\alpha_{1} .(1)^{1} + \alpha_{2}.(-\frac{1}{2} )^{1} = 100,000$

In year 2, when n=2, Ln = 300,000:

$\alpha_{1} .(1)^{2} + \alpha_{2}.(-\frac{1}{2} )^{2} = 300,000$

Solving the system of equations:

$\alpha_{1} .(1)^{1} + \alpha_{2}.(-\frac{1}{2} )^{1} = 100,000$

$\alpha_{1} = 100,000 + \frac{1}{2}.\alpha_{2}$

$\alpha_{1} .(1)^{2} + \alpha_{2}.(-\frac{1}{2} )^{2} = 300,000$

$\alpha_{1} = 300,000 - \frac{1}{4}.\alpha_{2}$

Finding $\alpha _{2}$ :

$100,000+\frac{1}{2}.\alpha_{2} = 300,000 - \frac{1}{4}.\alpha_{2}$

$\frac{3}{4}.\alpha_{2} = 200,000$

$\alpha_{2} =$ 266666.7

With $\alpha _{2}$ , plug in an equation to find $\alpha_{1}$ :

$\alpha_{1} = 100,000 + \frac{1}{2}.\alpha_{2}$

$\alpha_{1} = 100,000 + \frac{266666.7}{2}$

$\alpha_{1} =$ 233333.4

The solved equation for this recurrence relation is:

$L_{n} = 233333.4(1)^{n} + 266666.7(-\frac{1}{2})^{n}$

Answer from: Quest

step-by-step explanation:

hello and welcome!

so, when we see the words "proportional" along with a table, we need to know that we're looking numbers that divide by the same thing.

looking at this table, we simply have to do a trial and error, and divide to see which numbers have the same quotient. if you look at choice a & c, they both have equal quotients for three of the numbers, with one exception, so those can't be the answer. looking at b, there is no similarity, so that can't be it. then, we see d; if you divide all the numbers in the table, you'll see that they all divide by two, which is a proportional relationship, and our answer!

$4/2 = 2\\7/3.5=2\\8/4=2\\10/5=2$

also, always remember that the x will always be the number that goes into y, and somewhere behind this table, is a formula. for this specific table, the formula would be:

$y=\frac{x}{2}$

since our x-value is being divided by two to get the y-value. and if they were to ask for more numbers in the series, this is the formula you would use to find them.

Answer from: Quest

$given : \frac{3}{9} = \frac{x}{5}$

cross multiplying, we get :

⇒ (3) × (5) = (9) × (x)

⇒ 15 = 9x

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