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Lesson Posted on 21 May Learn Mathematics

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

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

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

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

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 19 Apr

To find the radius of the circle formed by bending a wire into an arc subtending an angle of 60∘, we can use the formula for the length of an arc of a circle: read more

To find the radius of the circle formed by bending a wire into an arc subtending an angle of 60, we can use the formula for the length of an arc of a circle:
$$&space;\text{Length&space;of&space;arc}&space;=&space;\frac{\theta}{360^\circ}&space;\times&space;2\pi&space;r&space;$&space;Given:&space;Length&space;of&space;wire&space;($$&space;l&space;$$)&space;=&space;22&space;cm&space;Angle&space;subtended&space;($$&space;\theta&space;$$)&space;=&space;$$&space;60^\circ&space;$$&space;We&space;can&space;substitute&space;these&space;values&space;into&space;the&space;formula:&space;$&space;22&space;=&space;\frac{60}{360}&space;\times&space;2\left(\frac{22}{7}\right)&space;r&space;$$

$Let's&space;solve&space;for&space;$$&space;r&space;$$:&space;$&space;22&space;=&space;\frac{1}{6}&space;\times&space;\frac{44}{7}&space;r&space;$&space;$&space;22&space;=&space;\frac{22}{7}&space;r&space;$&space;Now,&space;let's&space;solve&space;for&space;$$&space;r&space;$$:&space;$&space;r&space;=&space;\frac{22&space;\times&space;7}{22}&space;$&space;$&space;r&space;=&space;7&space;\,&space;\text{cm}&space;$&space;So,&space;the&space;radius&space;of&space;the&space;circle&space;is&space;$$&space;7&space;\,&space;\text{cm}&space;$$.$

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Answered on 19 Apr

To solve this problem, we first find the area the calf can graze with a rope of length 6 m and then find the area it can graze with a rope of length 11.5 m. The difference between these areas will give us the increase in the area of the grassy lawn the calf can graze. read more

To solve this problem, we first find the area the calf can graze with a rope of length 6 m and then find the area it can graze with a rope of length 11.5 m. The difference between these areas will give us the increase in the area of the grassy lawn the calf can graze.
$1.&space;**Area&space;the&space;calf&space;can&space;graze&space;with&space;a&space;rope&space;of&space;length&space;6&space;m:**&space;When&space;the&space;calf&space;is&space;tied&space;with&space;a&space;rope&space;of&space;length&space;6&space;m,&space;it&space;can&space;graze&space;within&space;a&space;circle&space;whose&space;radius&space;is&space;6&space;m.&space;This&space;forms&space;a&space;circular&space;grazing&space;area.&space;The&space;area&space;of&space;the&space;circular&space;grazing&space;area&space;is&space;given&space;by&space;$$&space;\pi&space;r^2&space;$$,&space;where&space;$$&space;r&space;$$&space;is&space;the&space;radius&space;of&space;the&space;circle&space;(6&space;m).&space;Therefore,&space;the&space;area&space;the&space;calf&space;can&space;graze&space;with&space;a&space;rope&space;of&space;length&space;6&space;m&space;is&space;$$&space;\pi&space;\times&space;6^2&space;=&space;36\pi&space;$$&space;square&space;meters.&space;2.&space;**Area&space;the&space;calf&space;can&space;graze&space;with&space;a&space;rope&space;of&space;length&space;11.5&space;m:**&space;When&space;the&space;length&space;of&space;the&space;rope&space;is&space;increased&space;to&space;11.5&space;m,&space;the&space;radius&space;of&space;the&space;circular&space;grazing&space;area&space;also&space;increases&space;to&space;11.5&space;m.&space;The&space;area&space;of&space;the&space;circular&space;grazing&space;area&space;with&space;a&space;radius&space;of&space;11.5&space;m&space;is&space;$$&space;\pi&space;\times&space;11.5^2&space;$$&space;square&space;meters.&space;Therefore,&space;the&space;area&space;the&space;calf&space;can&space;graze&space;with&space;a&space;rope&space;of&space;length&space;11.5&space;m&space;is&space;$$&space;\pi&space;\times&space;11.5^2&space;$$&space;square&space;meters.$

$Now,&space;let's&space;find&space;the&space;difference&space;between&space;the&space;areas&space;the&space;calf&space;can&space;graze&space;with&space;these&space;two&space;rope&space;lengths:&space;$&space;\text{Increase&space;in&space;area}&space;=&space;(\pi&space;\times&space;11.5^2)&space;-&space;(\pi&space;\times&space;6^2)&space;$&space;$&space;\text{Increase&space;in&space;area}&space;=&space;\pi&space;\times&space;(11.5^2&space;-&space;6^2)&space;$&space;$&space;\text{Increase&space;in&space;area}&space;=&space;\pi&space;\times&space;(132.25&space;-&space;36)&space;$&space;$&space;\text{Increase&space;in&space;area}&space;=&space;\pi&space;\times&space;96.25&space;$&space;Using&space;$$&space;\pi&space;\approx&space;\frac{22}{7}&space;$$:&space;$&space;\text{Increase&space;in&space;area}&space;\approx&space;\frac{22}{7}&space;\times&space;96.25&space;$&space;$&space;\text{Increase&space;in&space;area}&space;\approx&space;303.75&space;\,&space;\text{square&space;meters}&space;$&space;So,&space;the&space;increase&space;in&space;the&space;area&space;of&space;the&space;grassy&space;lawn&space;in&space;which&space;the&space;calf&space;can&space;graze&space;is&space;approximately&space;$$&space;303.75&space;\,&space;\text{square&space;meters}&space;$$.$

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Answered on 29 Apr

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

To find the probability of not drawing a white marble, we need to consider the total number of marbles in the box and the number of marbles that are not white. Given: 3 blue marbles 2 white marbles 4 red marbles read more

To find the probability of not drawing a white marble, we need to consider the total number of marbles in the box and the number of marbles that are not white.

Given:

• 3 blue marbles
• 2 white marbles
• 4 red marbles
$1.&space;**Total&space;number&space;of&space;marbles:**&space;There&space;are&space;$$3&space;+&space;2&space;+&space;4&space;=&space;9$$&space;marbles&space;in&space;total.&space;2.&space;**Number&space;of&space;marbles&space;that&space;are&space;not&space;white:**&space;There&space;are&space;$$3&space;+&space;4&space;=&space;7$$&space;marbles&space;that&space;are&space;not&space;white&space;(3&space;blue&space;marbles&space;+&space;4&space;red&space;marbles).&space;Now,&space;let's&space;calculate&space;the&space;probability&space;of&space;not&space;drawing&space;a&space;white&space;marble:&space;$&space;\text{Probability}&space;=&space;\frac{\text{Number&space;of&space;marbles&space;that&space;are&space;not&space;white}}{\text{Total&space;number&space;of&space;marbles}}&space;$&space;$&space;\text{Probability}&space;=&space;\frac{7}{9}&space;$&space;So,&space;the&space;probability&space;of&space;not&space;drawing&space;a&space;white&space;marble&space;from&space;the&space;box&space;is&space;$$&space;\frac{7}{9}&space;$$.$
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Answered on 19 Apr

In a standard deck of 52 playing cards, there are 2 black queens: the queen of spades and the queen 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 queens): There are 2 black queens in the deck. Now,... read more

In a standard deck of 52 playing cards, there are 2 black queens: the queen of spades and the queen 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 queens): There are 2 black queens in the deck.

Now, let's calculate the probability of drawing a black queen:

Probability=Number of favorable outcomesTotal number of possible outcomes

$$&space;\text{Probability}&space;=&space;\frac{\text{Number&space;of&space;favorable&space;outcomes}}{\text{Total&space;number&space;of&space;possible&space;outcomes}}&space;$&space;$&space;\text{Probability}&space;=&space;\frac{2}{52}&space;$&space;$&space;\text{Probability}&space;=&space;\frac{1}{26}&space;$&space;So,&space;the&space;probability&space;of&space;drawing&space;a&space;black&space;queen&space;from&space;a&space;well-shuffled&space;deck&space;of&space;52&space;cards&space;is&space;$$&space;\frac{1}{26}&space;$$.$

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