Archive for the ‘General history’ Category

Problem 429


Fiona kept the following records of her phone bills for 12 months:

55,  50,  64,  54,  71,  57,  55,  48,  66,  71,  59,  70


Find the mean and median of Fiona’s monthly phone bills.




First arrange the values in ascending order which is given bellow

48,  50,  54,  55, 55, 57, 59, 64, 66, 70, 71, 71

Mean=sum of terms /number of terms

Sum of terms=48+50+54+55+55+57+59+64+66+70+71+17=720

Number of terms=12

So mean=\frac{720}{12}= 60

When the number of values of even then medain =avarge of = \frac{n}{2} th term  and (\frac{n}{2})+1 th term

Where n=number of terms

So here medain=[\frac{12}{2}   + (\frac{12}{2}  )+1]/2=\frac{(6 th term+7 th term)}{2}

6 th term is  57

7 th term is  59

So median=\frac{(57+59) }{2}   = 58

Maximum–Minimum Problems

Maximum–Minimum Problems


We have seen that for any quadratic function f(x) = ax2 + bx + c, value of f(x) at the vertex is either a maximum or a minimum, meaning that either all outputs are smaller than that value for a maximum or larger than that value for a minimum.

 minimum and maximum

There are many types of applied problems in which we want to find a maximum or minimum value of a quadratic function can be used as a model, we can find such maximums or minimums by finding coordinates of the vertex.


Problem 359

Problem 359

The Rental Depreciation Problem. The owner of a rental house can depreciate its value over a period of  27\tfrac{1}{2}  years, meaning that the  value of the house declines at an even rate over that period of time until the value is  0.     a. By what fraction does the value of the house depreciate the first year?  b. If the house is judged to be worth85,000, what is the value of the first year’s depreciation?



a.  \tfrac{1}{27\tfrac{1}{2}}= \frac{1}{\frac{55}{2}}=\frac{2}{55}

So, the depreciation is \frac{2}{55} each year.


b. Value of the first year’s depreciation = \frac{2}{55}*85000 = $3090.91


Problem 293

Problem 293

Decide whether the following statements are true or false and justify your answer:

a. If n(A) < n(B), then A \subset B.

b. If n(A) \leq n(B) , then A  \subseteq B.

c. If A  \subset B,  then n(A)< n(B).

d. If A \subseteq B, then n(A) \leq n(B).




a. False, elements are not same in both sets.

b. False , not equal or equivalent

c. True , equal or equivalent

d. True , equal or equivalent


Problem 254





The sum of the horizontal and vertical distances is called the taxicab distance in above

graph taxicab distance

= total horizontal distance+ total vertical distance

horizontal distance =1+1+1+1+1+1=6

vertical distance =1+1+1+1+1+1+1=7

so total taxicab distance = 6+7=13


Problem 220

Problem 220

A cylindrical can of cleaner holds 1 L of liquid. What is its height if the diameter of the base is 10 cm?



1L=1000 cm 3  ,r = \frac{10}{2} = 5cm

Volume = πr 2 h

1000 =  π(5) 2 (h)

1000 =  π25(h)

\frac{1000}{25π} =  (h)

h=12.73 cm


Problem 189

Problem 189

Tell if the figure represented by {\farc{9}{3}}  is a star polygon, a polygon, or other, and how many sides it has.



Yes, Star polygon and it has a total of  9 sides.


Problem 166

Problem 166

  If the triangles are congruent, write a congruence statement and give a reason why they are congruent.

triangles are congruen



if three sides of one triangle are congruent to the corresponding three sides of the second triangle, then the two triangles are congruent Triangle ΔHAJ is congruent with triangle ΔOGP because the length of HA is equal to the length of OG, the length of AJ is equal to the length of GP, and the length of JH is the same as the length of PO


Problem 142

Problem 142

A student asks, “What’s wrong with the argument that the probability of rolling a double 6 in two rolls of a die is \frac{1}{3}because \frac{1}{6} + \frac{1}{6} = \frac{1}{3}?”

Write an explanation of your understanding of the student’s misconception.



We can’t add the probabilities, we need to multiply them to get the resultant, if we rolled more than 6 times then we will get probability more than 1 which is not possible.




Find the domain and range of the function that assigns to each positive integer the largest integer not exceeding the square root of the integer.



The domain is Z+ and the range is Z+.