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What would you like to do differently?
Next time, I would like to design more innovative toys, incorporating different types of simple machines for better functionality. I want to use eco-friendly materials to promote sustainability. Improving accuracy in measurements and assembling parts precisely will be a focus. Additionally, I wouldRead more
Next time, I would like to design more innovative toys, incorporating different types of simple machines for better functionality. I want to use eco-friendly materials to promote sustainability. Improving accuracy in measurements and assembling parts precisely will be a focus. Additionally, I would like to refine my bicycle repair skills, particularly in fixing brakes and gears. Exploring advanced simple machine applications in real-world scenarios would also help me deepen my understanding of their mechanics.
See lessWhat were the challenges you faced?
I faced challenges in gathering the right materials and ensuring that the toys worked as expected. Understanding how different simple machines function required trial and error. Sometimes, parts didn’t fit well, or the models didn’t work correctly. Bicycle repairs were tricky, especially fixing theRead more
I faced challenges in gathering the right materials and ensuring that the toys worked as expected. Understanding how different simple machines function required trial and error. Sometimes, parts didn’t fit well, or the models didn’t work correctly. Bicycle repairs were tricky, especially fixing the chain and aligning the wheels. However, with practice, teamwork, and guidance from my teacher, I overcame these challenges. The experience taught me problem-solving skills, patience, and the importance of careful planning.
See lessFive fair coins are tossed simultaneously. The probability of the events that at least one head comes up is
When five fair coins are tossed simultaneously the probability of getting at least one head is found using the complement rule 1. Total possible outcomes when tossing 5 coins = 2⁵ = 32 2. Only unfavorable outcome is getting all tails which happens in one way: TTTTT 3. Probability of getting all tailRead more
When five fair coins are tossed simultaneously the probability of getting at least one head is found using the complement rule
1. Total possible outcomes when tossing 5 coins = 2⁵ = 32
2. Only unfavorable outcome is getting all tails which happens in one way: TTTTT
3. Probability of getting all tails = 1/32
4. Probability of getting at least one head = 1 – P(all tails) = 1 – 1/32 = 31/32
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See lesshttps://www.tiwariacademy.com/ncert-solutions/class-12/maths/#chapter-13
If for any two events A and B, P(A) = 4/5 and P(A ∩ B) = 7/10, then P(B/A) is equal to
Given: P(A) = 4/5 P(A ∩ B) = 7/10 Using the conditional probability formula: P(B|A) = P(A ∩ B) / P(A) Substituting the values: P(B|A) = (7/10) ÷ (4/5) = (7/10) × (5/4) = 35/40 = 7/8 So the correct answer is 7/8 Click here for more: https://www.tiwariacademy.com/ncert-solutions/class-12/maths/#chapteRead more
Given:
P(A) = 4/5
P(A ∩ B) = 7/10
Using the conditional probability formula:
P(B|A) = P(A ∩ B) / P(A)
Substituting the values:
P(B|A) = (7/10) ÷ (4/5)
= (7/10) × (5/4) = 35/40 = 7/8
So the correct answer is 7/8
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See lesshttps://www.tiwariacademy.com/ncert-solutions/class-12/maths/#chapter-13
Three persons A, B and C, fire at a target in turn, starting with A. Their probability of hitting the target are 0.4, 0.3 and 0.2 respectively. The probability of two hits is
Given: P(A hits) = 0.4 P(B hits) = 0.3 P(C hits) = 0.2 To find the probability of exactly two hits, we consider the cases where exactly two of A, B, and C hit the target: 1. A and B hit, C misses: P(A ∩ B ∩ C') = (0.4) × (0.3) × (1 - 0.2) = 0.4 × 0.3 × 0.8 = 0.096 2. A and C hit, B misses: P(A ∩ B'Read more
Given:
P(A hits) = 0.4
P(B hits) = 0.3
P(C hits) = 0.2
To find the probability of exactly two hits, we consider the cases where exactly two of A, B, and C hit the target:
1. A and B hit, C misses:
P(A ∩ B ∩ C’) = (0.4) × (0.3) × (1 – 0.2)
= 0.4 × 0.3 × 0.8 = 0.096
2. A and C hit, B misses:
P(A ∩ B’ ∩ C) = (0.4) × (1 – 0.3) × (0.2)
= 0.4 × 0.7 × 0.2 = 0.056
3. B and C hit, A misses:
P(A’ ∩ B ∩ C) = (1 – 0.4) × (0.3) × (0.2)
= 0.6 × 0.3 × 0.2 = 0.036
Total probability of exactly two hits:
P = 0.096 + 0.056 + 0.036 = 0.188
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See lesshttps://www.tiwariacademy.com/ncert-solutions/class-12/maths/#chapter-13