WEBVTT
1
00:00:01.209 --> 00:00:03.580 A:middle L:90%
so to start solving this. The first thing we
2
00:00:03.580 --> 00:00:08.119 A:middle L:90%
notice is that the maximum temperature occurs in July and
3
00:00:08.119 --> 00:00:13.769 A:middle L:90%
that's maximum temperature is 68.5 and July is the seventh
4
00:00:13.779 --> 00:00:17.629 A:middle L:90%
month and then the minimum temperature occurs in January,
5
00:00:17.640 --> 00:00:20.449 A:middle L:90%
which is the first month, and the minimum temperature
6
00:00:20.449 --> 00:00:23.030 A:middle L:90%
is 18 points years. So this is information that
7
00:00:23.030 --> 00:00:26.109 A:middle L:90%
you can get from the till. Now. Let's
8
00:00:26.109 --> 00:00:31.179 A:middle L:90%
calculate a, which would be cool to maximum minus
9
00:00:31.179 --> 00:00:37.570 A:middle L:90%
two minima divided by two. So we have 68.5
10
00:00:37.579 --> 00:00:42.549 A:middle L:90%
, minus 18.0, divided by two, and that's
11
00:00:42.549 --> 00:00:48.939 A:middle L:90%
equal to 25.25 The crease pattern that next day since
12
00:00:48.939 --> 00:00:51.950 A:middle L:90%
it's a year. So the petered is 12 and
13
00:00:51.960 --> 00:00:55.560 A:middle L:90%
B is equal to two pi divided by the period
14
00:00:55.780 --> 00:00:58.380 A:middle L:90%
, so that would be too high, divided by
15
00:00:58.390 --> 00:01:03.379 A:middle L:90%
12 which is power six. Now the main temperature
16
00:01:07.540 --> 00:01:10.989 A:middle L:90%
, which is equal to D, will be equal
17
00:01:10.989 --> 00:01:15.879 A:middle L:90%
to maximum temperature, plus the minimum temperature divided by
18
00:01:15.890 --> 00:01:23.209 A:middle L:90%
two. So that would be 68.5 plus 18.0 divided
19
00:01:23.219 --> 00:01:29.109 A:middle L:90%
by two, and that is equal to 43.25 So
20
00:01:29.120 --> 00:01:32.500 A:middle L:90%
we've got mean temperature, which would give value for
21
00:01:32.670 --> 00:01:36.549 A:middle L:90%
D now again, looking into the table. What
22
00:01:36.560 --> 00:01:47.599 A:middle L:90%
? The finest at the mean temperature first occurs in
23
00:01:47.609 --> 00:01:55.500 A:middle L:90%
April, which is teas equal to four so we
24
00:01:55.500 --> 00:02:00.010 A:middle L:90%
can rank our equation. Why is equal to a
25
00:02:00.319 --> 00:02:07.219 A:middle L:90%
25.25 sign off pi over six times t by his
26
00:02:07.219 --> 00:02:14.680 A:middle L:90%
four class 43.25? Or why is he put to
27
00:02:15.000 --> 00:02:23.150 A:middle L:90%
25.25 Sign or hi over 60 minus two pi over
28
00:02:23.150 --> 00:02:29.250 A:middle L:90%
three plus 43.25