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slavikrds [6]
3 years ago
5

Both Country X and Country Y can produce cell phones and video games. The opportunity costs per video game for X and Y are 2 and

5 cell phones, respectively. Which country has comparative advantage to make video games?
a. Country X, and Country Y has comparative advantage to make cell phones
b. Country Y, and Country X has comparative advantage to make cell phones
c. Both Country X and Country Y have comparative advantage to make cell phones and video games.
d. Country X. There is not enough information to find out which country has comparative advantage to make cell phones ​
Advanced Placement (AP)
1 answer:
LiRa [457]3 years ago
4 0

Answer:

D

Explanation:

We need to know about materials cost and how things are produced how much they pay their workers things like that to understand the comparative advantages

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Answer:

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Explain in detail, how you solved the following problem: The first two terms of a sequence are 10 and 20. If each term after the
KIM [24]

Answer:

T_{2020} = 15

Explanation:

Given

T_1 = 10

T_2 = 20

Each term after the second term is the average of all of the preceding terms

Required:

Explain how to solve the 2020th term

Solve the 2020th term

Solving the 2020th term of a sequence using conventional method may be a little bit difficult but in questions like this, it's not.

The very first thing to do is to solve for the third term;

The value of the third term is the value of every other term after the second term of the sequence; So, what I'll do is that I'll assign the value of the third term to the 2020th term

<em>This is proved as follows;</em>

From the question, we have that "..... each term after the second term is the average of all of the preceding terms", in other words the MEAN

T_{n} = \frac{\sum T{k}}{n-1} ; where: k = 1 .... n -1

<em>Assume n = 3</em>

T_{3} = \frac{T_1 + T_2}{2}

<em>Multiply both sides by 2</em>

2 * T_{3} = \frac{T_1 + T_2}{2} * 2

2T_{3} = T_1 + T_2

<em>Assume n = 4</em>

T_{4} = \frac{T_1 + T_2 + T_3}{3}

T_{4} = \frac{(T_1 + T_2) + T_3}{3}

Substitute 2T_{3} = T_1 + T_2

T_{4} = \frac{2T_3 + T_3}{3}

T_{4} = \frac{3T_3}{3}

T_{4} = T_3

Assume n = 5

T_{5} = \frac{T_1 + T_2 + T_3 +T_4}{4}

T_{5} = \frac{(T_1 + T_2) + T_3 +(T_4)}{4}

Substitute 2T_{3} = T_1 + T_2 and T_{4} = T_3

T_{5} = \frac{2T_3 + T_3 +T_3}{4}

T_{5} = \frac{4T_3}{4}

T_{5} = \frac{(5-1)T_3}{5-1}

<em>Replace 5 with n</em>

T_{n} = \frac{(n-1)T_3}{n-1}

<em>(n-1) will definitely cancel out (n-1); So, we're left with</em>

T_{n} = T_3

Hence,

T_{2020} = T_3

Calculating T_3

T_{3} = \frac{10 + 20}{2}

T_{3} = \frac{30}{2}

T_{3} = 15

Recall that T_{2020} = T_3

T_{2020} = 15

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Answer:

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