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evablogger [386]
3 years ago
9

What did India's approach to limiting population growth focus on in the 1970s?

Advanced Placement (AP)
2 answers:
Arte-miy333 [17]3 years ago
6 0

Answer:

Permanent birth control through sterilization

Explanation:

This answer makes the most sense.

I also took the quiz and that was the right answer.

Hope this Helps ;P

finlep [7]3 years ago
3 0
India had introduced modern contraceptive during this time. The government wanted to reduce the rapidly increasing fertility rate amongst married women in India. Though the population continues to increase they were able to reduce abortion and unwanted pregnancies.
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The next letter would be S.
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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
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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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Explanation:

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