Problem Solving

Can anyone explain to me how to do the following questions?

1.The function f is defined for all positive integers n by the following rule:f(n) is the number of positive integers each of which is less than n and has no positive factor in common with n other than 1. if p is any prime number then f(p) =

a)p-1
b)p-2
c)p+1/2
d)p-1/2
e)2


2. A thin piece of wire 40 meters long is cut into 2 pieces. One piece is used to form a circle with radius r and the other is used to form a square, no wire is left over. Which of the following represents the total area in square meters of the circular and the square regions in terms of r

a)IIr sq
b)IIr sq +10
c)II r sq + 1/4 II sq r sq
d)IIr sq + (40-2IIr) sq
e)IIs sq +(10- 1/2IIr) sq


3. A vending machine is designed to dispense a ounces of coffee into a cup. after a test that recorded the number of ounce of coffee in each of 1000 cups dispensed by the vending machine the 12 listed amounts below, in ounces were selected from the data. If the 1000 recorded amounts have a mean of 8.1 ounces and standard deviation of 0.3 ounce, how many of the 12 listed amounts are within 1.5 standard deviation of the mean?
7.51 8.22 7.86 8.36
8.09 7.83 8.30 8.01
7.73 8.25 7.96 8.53

a) four
b) six
c)nine
d)ten
e)eleven

janvierek wrote:Can anyone explain to me how to do the following questions?

1.The function f is defined for all positive integers n by the following rule:f(n) is the number of positive integers each of which is less than n and has no positive factor in common with n other than 1. if p is any prime number then f(p) =

a)p-1
b)p-2
c)p+1/2
d)p-1/2
e)2

Take n = 5. The positive integers that are less than 5 and that have
no positive factor in common with 5 are 1, 2, 3 and 4.

For a prime number p, we know there are 1 ... p-1 are the numbers
that don't have a factor in common with p. So, f(p) = p-1.

janvierek wrote:
2. A thin piece of wire 40 meters long is cut into 2 pieces. One piece is used to form a circle with radius r and the other is used to form a square, no wire is left over. Which of the following represents the total area in square meters of the circular and the square regions in terms of r

a)IIr sq
b)IIr sq +10
c)II r sq + 1/4 II sq r sq
d)IIr sq + (40-2IIr) sq
e)IIs sq +(10- 1/2IIr) sq

Circle with radius r will have a perimeter of 2*pi*r. The rest i.e. 40-2*pi*r
will be the perimeter of the square.

Side of the square = (40-2*pi*r) / 4 = 10-pi*r/2

So, total area pi*r^2 + (10-pi*r/2)^2

janvierek wrote:
3. A vending machine is designed to dispense a ounces of coffee into a cup. after a test that recorded the number of ounce of coffee in each of 1000 cups dispensed by the vending machine the 12 listed amounts below, in ounces were selected from the data. If the 1000 recorded amounts have a mean of 8.1 ounces and standard deviation of 0.3 ounce, how many of the 12 listed amounts are within 1.5 standard deviation of the mean?
7.51 8.22 7.86 8.36
8.09 7.83 8.30 8.01
7.73 8.25 7.96 8.53

a) four
b) six
c)nine
d)ten
e)eleven

1.5 standard deviations = 1.5 * .3 = .45

1.5 sd from the mean (+ve) = 8.1 + 0.45 = 8.55
1.5 sd from the mean (-ve) = 8.1 - 0.45 = 7.65

Find the values which are in the range 7.65 - 8.55 from the list
and you have the answer...

Thank for your help. However i still don't get how you got p-1. i do understand the first part (Take n = 5. The positive integers that are less than 5 and that have no positive factor in common with 5 are 1, 2, 3 and 4)
could you please explain to me how you got p-1?
thanks

1.The function f is defined for all positive integers n by the following rule:f(n) is the number of positive integers each of which is less than n and has no positive factor in common with n other than 1. if p is any prime number then f(p) =

a)p-1
b)p-2
c)p+1/2
d)p-1/2
e)2

Anonymous wrote:Jayhawk.... why did you multiply .3 * 1.5 for question #2

Std dev measure how far data is from mean. Let the mean = m and std dev = s. A value that is one std dev away from the mean has to lie between (m-s) and (m+s). Similary, a value that is 2 std dev away will lie between (m-2s) and (m+2s).