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Lesson Posted on 21 May Learn Mathematics

Area of shaded region

Sonal Gupta

Hello Learners! I am Sonal. I am MBA | B.ED having 15yrs of exp. as Math Educator. I am in ed-tech Industry...

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Lesson Posted on 29 Apr Learn Mathematics

Algebra Basics - The Product of algebraic expressions

Deeksha

I have approximately 3 years of experience of teaching Mathematics across different boards including...

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Lesson Posted on 21 Apr Learn Mathematics

Probability Brief Lesson plan

Abhishek Sharma

I am a civil engineer graduate currently teaching Mathematics in Cambridge board in Ahmedabad. My key...

Classes explore different scenarios using manipulatives to learn about the difference between independent and dependent probability. Learners experiment with colored chips to model the two types of probabilities. To test their understanding, individuals sort independent and dependent probabilities into... read more

Classes explore different scenarios using manipulatives to learn about the difference between independent and dependent probability. Learners experiment with colored chips to model the two types of probabilities. To test their understanding, individuals sort independent and dependent probabilities into like groups.

 

Real Life Context The following examples could be used to explore real life contexts.

Looking at statistics from the Census, questions like:

• How long will I live?

• Will I get married?

• How many children will I have? These questions can be answered with some degree of certainty based on population statistics. Life assurance companies work out how much to charge for their premiums based on tables of life expectancy. Why are some premiums cheaper than others?

 

The following examples could be used to explore misconceptions:

• What is the most difficult number to get when throwing a fair die?

• Random events should have outcomes which appear random; for example, in the lotto theory tells us that any of the six numbers is equally likely to turn up, yet more people choose randomly spaced numbers than numbers which form a pattern like 1,2,3,4,5,6 etc.

• The likelihood of 2 consecutive numbers appearing in any Lotto draw (which is > 50%) could easily be investigated by reference to a number of recent draws.

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Answered on 16 May Learn Constructions

Pranjal B.

To construct a triangle Δ���ΔABC with ��=6AB=6 cm, ∠�=30∘∠A=30∘, and ∠�=60∘∠B=60∘, follow these steps: Draw a line segment ��AB of length 6 cm. At point �A, construct an angle of 30∘30∘. At point �B, construct an angle of 60∘60∘. Label the intersection of the... read more

To construct a triangle Δ���ΔABC with ��=6AB=6 cm, ∠�=30∘∠A=30∘, and ∠�=60∘∠B=60∘, follow these steps:

  1. Draw a line segment ��AB of length 6 cm.
  2. At point �A, construct an angle of 30∘30∘.
  3. At point �B, construct an angle of 60∘60∘.
  4. Label the intersection of the two constructed angles as �C.
  5. Connect points �A and �C, as well as points �B and �C.

To construct a similar triangle Δ��′�′ΔAB′C′ with base ��′=8AB′=8 cm, follow these steps:

  1. Extend ��AB to the right to create ��′AB′ of length 8 cm.
  2. At point �A, construct an angle of 30∘30∘.
  3. At point �′B′, construct an angle of 60∘60∘.
  4. Label the intersection of the two constructed angles as �′C′.
  5. Connect points �A and �′C′, as well as points �′B′ and �′C′.

Both triangles will be similar because they have corresponding angles of the same measure.

 
 
 
 
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Answered on 29 Apr Learn Areas related to circles

Aniruddha Singh

PCM sream in 11 , Preparing for JEE

154/3 cm2 EXPLANATION:- Minute hand complete full circle in 1 hour. Angle swept by minute in 1 hour (60 min)=360o Angle swept by minute hand in 1 min=36060=6o Angle swept by minute hand in 5 min=6×5=30o θ=30o,r=14 cm Area swept by minute hand = Area of sector =θ360×πr2 ... read more

154/3 cm2

EXPLANATION:-

Minute hand complete full circle in 1 hour.

 
Angle swept by minute in 1 hour (60 min)=360o
 
Angle swept by minute hand in 1 min=36060=6o
 
Angle swept by minute hand in 5 min=6×5=30o
 
θ=30o,r=14 cm
 
Area swept by minute hand = Area of sector
 
=θ360×πr2
 
=30360×227×(14)2
 
=112×227×14×14
 
Area swept by minute hand =1543 cm2
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Answered on 29 Apr Learn Areas related to circles

Aniruddha Singh

PCM sream in 11 , Preparing for JEE

Let the radius of the circle is =r cm Circumference of the circle =22 cm ⇒2πr=22 2×227×r=22 r=72 cm Area of the quadrant =πr24 =227×14×72×72 =778 cm2 read more
Let the radius of the circle is =r cm
 
Circumference of the circle =22 cm
 
2πr=22
 
2×227×r=22
 
r=72 cm
 
Area of the quadrant =πr24
 
=227×14×72×72
 
=778 cm2
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Answered on 19 Apr Learn Probability

Sadika

In a standard deck of 52 playing cards, there are 2 black kings: the king of spades and the king of clubs. Total number of possible outcomes (total number of cards): There are 52 cards in a standard deck. Total number of favorable outcomes (black kings): There are 2 black kings in the deck. read more

In a standard deck of 52 playing cards, there are 2 black kings: the king of spades and the king of clubs.

  1. Total number of possible outcomes (total number of cards): There are 52 cards in a standard deck.

  2. Total number of favorable outcomes (black kings): There are 2 black kings in the deck.

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Answered on 19 Apr Learn Probability

Sadika

To find the probability that two different friends have different birthdays, we can consider the total number of possible combinations of birthdays and the number of combinations where the birthdays are different. In a non-leap year, there are 365 days. Therefore, the probability that two friends have... read more

To find the probability that two different friends have different birthdays, we can consider the total number of possible combinations of birthdays and the number of combinations where the birthdays are different.

In a non-leap year, there are 365 days. Therefore, the probability that two friends have different birthdays is the complement of the probability that they have the same birthday.

  1. Total number of possible combinations of birthdays: Each friend can have their birthday on any of the 365 days of the year. So, the total number of possible combinations of birthdays for two friends is 365×365365×365.

  2. Number of combinations where the birthdays are different: If one friend has their birthday on a specific day, the other friend can have their birthday on any of the other 364 days of the year. So, the number of combinations where the birthdays are different is 365×364365×364.

Now, let's calculate the probability that two different friends have different birthdays:

Probability=Number of combinations where the birthdays are differentTotal number of possible combinations of birthdays

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Answered on 19 Apr Learn Statistics

Sadika

To find the correct mean, we first need to adjust the sum of the observations by adding the difference between the correct value of 125 and the wrongly noted value of 25. Then we can divide this adjusted sum by the total number of observations. read more

To find the correct mean, we first need to adjust the sum of the observations by adding the difference between the correct value of 125 and the wrongly noted value of 25. Then we can divide this adjusted sum by the total number of observations.

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Answered on 19 Apr Learn Statistics

Sadika

To find the mean of the 32 numbers, we can use the concept of weighted averages. Given: The mean of 10 of the numbers is 15. The mean of 20 of the numbers is 11. The last two numbers are 10. Let's denote: read more

To find the mean of the 32 numbers, we can use the concept of weighted averages.

Given:

  1. The mean of 10 of the numbers is 15.
  2. The mean of 20 of the numbers is 11.
  3. The last two numbers are 10.

Let's denote:

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