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3 years ago

(a) Use the climatograph above to identify the following.

Advanced Placement (AP)

1 answer:

Alex73 [517]3 years ago

3 0

Answer:

(i) Identify the month with the least amount of rainfall in this temperate grassland.

The month with the least amount of rainfall is January, when only around 20 mm of rain fall in the month. In the climatograph, rainfall is represented by the bars, while temperature is represented by the line.

(ii) Identify the maximum mean monthly temperature in this temperate grassland.

The maximum mean monthly temperature occurs at the end of June and the start of July. The mean monthly temperature reaches over 25 degrees centigrades over this time period.

(iii) A climatograph for a tropical grassland or savanna would look different from the climatograph shown for a temperate grassland. Describe one difference in the patterns or trends in the temperature data in a tropical grassland compared to those of a temperate grassland.

The climatograph for the tropical grassland would show similar patterns for rainfall, but higher mean monthly temperatures. Tropical grasslands are generally almost as dry as temperate grasslands, but they are much warmer over the year (lacking a proper winter for example).

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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

This is proved as follows;

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

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

Assume n = 3

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

Multiply both sides by 2

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

$2T_{3} = T_1 + T_2$

Assume n = 4

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

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

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 $T_{4} = T_3$

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

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

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

Replace 5 with n

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

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

$T_{n} = T_3$

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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