CPhill

We can represent this with a sine function [ cosine is also a possibility ].... we have the form

y = Asin (Bx + C) + D

The midline is at  [ 1.8 - 1.4] / 2  = 0.4/2 =   0.2 ft =  D

The amplitude is 1.6  =  A

The period is 14 days

B =   2pi/period  = 2pi/14  = pi/ 7

C  is the phase shift......this is the most difficult

We need to solve this  for C

Since we have a max at day 4, 

pi (4)/7 +   C  =  pi/2

C = pi/2 - pi (4)/7    =    [7pi - 8pi  ] / 14  = -pi/14

So.....the function will be

y = 1.6sin ( pi * x / 7   - pi / 14) + 0.2

Where y is the tide height  on the " x th " day of the month

Here is a graph : https://www.desmos.com/calculator/pcvovyj5gs

Angle EAB  = Angle CFD

Angle ABE  = Angle FCD

So, by AA congruency, Triangle CFD  s similar to Triangle BAE

So

BA / EB = CF / DC      which means that

BA * DC   =  EB * CF    but EB = CF

So

BA * DC =  EB^2

28 * 58  = EB^2   =  area of square BCFE  = 1624 units^2

We have this relationship

Angle DAB  = Angle CBD

Angle ADB = Angle CDB

So...by AA congruency......Triangle ADB  is similar to Triangle BDC

And AD  = AC - DC  = 13 - 2   = 11

So

AD / DB = BD / DC

So

AD * DC  =  BD^2

11 * 2  =  BD^2

BD^2 = 22

BD =  sqrt (22)

Very nice, Omi  !!!

a)  is correct   !!!

b) Probability  that student  has cycling as a hobby  [ not, holiday...LOL!!!]

There  must be 10 students that like cycling since this sector angle is twice that of those that favor walking

So...the probability  is     10 / 60   =    1 / 6  

51 / 9    =    45/ 9  +  6/9   =  5  + 6/9   =  5 + 2/3

Sorry...just a typo....supposed to be 2.5.....thanks for catching that, EP

First one.....add the data values

Each sector of the circle will occupy this many degrees  =   

Single Data Value /  Sum of Data Values  x 360

So...for example....the Magpie would occupy  15 /72 x 360  =  5 / 24 x 360  = 75°

It should look like this  when you finish:

OK, ant  !!!!

THX, ant  !!!!