S.6 math (thanks)

呢啲加奶

2013-12-28 11:35 pm

Q1. The following are the scores that 15 students got in the first term
Mathmatics examination. (The full mark is 100).

59 59 60 62 63 63 63 65
70 71 72 73 73 75 77

(a) Find the mean and the standard deviation of the scores.
我知道這條是這樣做: mean= 67, standard deviation = 6

(b)(i) Sandy got 73 in the examination. Find the standard score of Sandy.
我知道這條是這樣做: (73-67)/6 =1

(ii) Sandy performs better in the second term examination. If the mean and the
standard deviation of the second term examination are 52 and 11 respectively,
find the range of score that Sandy gets. 我不知道這條如何做，可否列式。
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我不知道這條如何做，可否列式。
Q2. The number of a certain endangered specied of insects decreases at a
constant rate of ｒ% every year. In 2007 and 2012, the numbers of that species of insects were 490 and 360 respectively.

(a) Find, correct to the nearest integer, the value of ｒ.

(b) Will the number of that species of insects be less than 300 in 2017? Explain your answer.

(c) In which year will the number of that species of insects be just less than
200?
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回答 (1)

fooks

2013-12-29 12:40 am

✔ 最佳答案

Q.1
(a)
Mean
= ∑xi/15
= 1005/15
= 67

Standard deviation
= √[∑(xi - 67)²/15]
= √[540/15]
= √36
= 6

(b)(i)
Standard score of Sandy
= (73 - 67)/6
= 6/6
= 1

(b)(ii)
Standard score > 1
(x - 52)/11 > 1
x - 52 > 11
x > 63
Score ofSandy in second term examination > 63

Q.2
(a)
360 = 490 * (1 - r%)^(2012 - 2007)
360 = 490 * (1 - r%)^5
(1 - r%)^4 = 360/490
1 - r% = (36/49)^(1/5)
r% = 1 - (36/49)^(1/5)
r = [1 - (36/49)^(1/5)] * 100
r = 6 (to thenearest integer)

(b)
Number of that species of insects in 2017
= 490 * (1 - 6%)^(2017 - 2007)
= 490 * 0.94^10
= 264 <300

(c)
Let n be the number of years required.

490 * (1 - 6%)^n < 200
0.94^n < 200/490
n log0.94 < log(200/490)
n > log(200/490) / log0.94 ...... (as log0.94 < 1)
n > 14.5

The required year
= 2007 + 15
= 2022

3(a)
Number of cans of food :
in the first year = 30 * 12 = 360
in the second year = (30 + 5) * 12 = 360 + 60
in the third year = (30 + 5*2) * 12 = 360 + 60 * 2
in the nth year = [30 + 5(n - 1)] * 12 = 360 + 60 * (n - 1)

Number of cans of food she buys in n years
= 360 + (360 + 60) + (360 + 60 * 2) + ...... + [360 + 60 * (n - 1)]
= 360 * n + 60 * [1 + 2 + 3 + ...... + (n - 1)]
= 360n + 60 * (n - 1) * [1 + (n - 1)] / 2
= 360n + 30n(n - 1)
= 360n + 30n² - 30n
= 30n² + 330n

參考: fooks

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