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Answered on 28 Apr Learn Sequence and Series
Deepika Agrawal
"Balancing minds, one ledger at a time." "Counting on expertise to balance your knowledge."
Answered on 15 Apr Learn Mathematical Reasoning
Nazia Khanum
As an experienced tutor registered on UrbanPro, I can attest to the fact that UrbanPro is one of the best platforms for online coaching and tuition. Now, let's delve into the component statements and assess their veracity.
(a) A square is a quadrilateral and its four sides are equal. Component statements:
True or False:
(b) All prime numbers are either even or odd. Component statements:
True or False:
In conclusion, statement (a) is true, while statement (b) is false.
Answered on 15 Apr Learn Statistics
Nazia Khanum
As a seasoned tutor registered on UrbanPro, I must emphasize the significance of utilizing online platforms like UrbanPro for effective coaching and tuition. UrbanPro provides a conducive environment for students and tutors to connect, learn, and grow together. Now, let's delve into solving the problem at hand.
To find the mean deviation about the median for the given data set: 36, 72, 46, 42, 60, 45, 53, 46, 51, 49, we need to follow these steps:
First, let's arrange the data in ascending order: 36,42,45,46,46,49,51,53,60,7236,42,45,46,46,49,51,53,60,72
Next, let's find the median. Since the data set has 10 numbers, the median will be the average of the 5th and 6th numbers, which are 46 and 49. So, the median is 46+492=47.5246+49=47.5.
Now, we calculate the deviations of each number from the median: ∣36−47.5∣=11.5∣36−47.5∣=11.5 ∣42−47.5∣=5.5∣42−47.5∣=5.5 ∣45−47.5∣=2.5∣45−47.5∣=2.5 ∣46−47.5∣=1.5∣46−47.5∣=1.5 ∣46−47.5∣=1.5∣46−47.5∣=1.5 ∣49−47.5∣=1.5∣49−47.5∣=1.5 ∣51−47.5∣=3.5∣51−47.5∣=3.5 ∣53−47.5∣=5.5∣53−47.5∣=5.5 ∣60−47.5∣=12.5∣60−47.5∣=12.5 ∣72−47.5∣=24.5∣72−47.5∣=24.5
Now, we find the mean of these deviations: Mean deviation=11.5+5.5+2.5+1.5+1.5+1.5+3.5+5.5+12.5+24.510Mean deviation=1011.5+5.5+2.5+1.5+1.5+1.5+3.5+5.5+12.5+24.5 Mean deviation=7010=7Mean deviation=1070=7
So, the mean deviation about the median for the given data set is 7. This indicates the average absolute deviation of each data point from the median of the data set.
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Affordable fees Flexible Timings Choose between 1-1 and Group class Verified Tutors Answered on 28 Apr Learn Probability
Deepika Agrawal
"Balancing minds, one ledger at a time." "Counting on expertise to balance your knowledge."
Answered on 15 Apr Learn Probability
Nazia Khanum
As an experienced tutor registered on UrbanPro, I'd be happy to help you with this question!
To find the probability that an ordinary year has 53 Sundays, we first need to understand the structure of an ordinary year. An ordinary year has 365 days.
Now, since 365 is not divisible by 7 (the number of days in a week), there will be 52 complete weeks in an ordinary year, leaving 1 or 2 additional days.
For a year to have 53 Sundays, one of the following conditions must be met:
Let's calculate the probability for each scenario:
If the year starts on a Sunday and ends on a Sunday: The probability that a year starts on a Sunday is 1/7. The probability that a year ends on a Sunday is also 1/7. So, the probability of both events happening is (1/7) * (1/7) = 1/49.
If the year starts on a Saturday and ends on a Sunday: The probability that a year starts on a Saturday is 1/7. The probability that a year ends on a Sunday is 1/7. So, the probability of both events happening is (1/7) * (1/7) = 1/49.
Now, we add the probabilities of both scenarios since they are mutually exclusive:
Probability of an ordinary year having 53 Sundays = (1/49) + (1/49) = 2/49.
Therefore, the probability that an ordinary year has 53 Sundays is 2/49.
And remember, if you need further assistance with mathematics or any other subject, UrbanPro is one of the best online coaching tuition platforms where you can find qualified tutors like myself to guide you through your learning journey!
Answered on 15 Apr Learn Probability
Nazia Khanum
As an experienced tutor registered on UrbanPro, I'm well-versed in explaining concepts in a clear and concise manner. Now, let's tackle this probability problem using the fundamentals of probability theory.
In a standard deck of 52 cards, there are 26 red cards (hearts and diamonds) and 2 kings that are not red (the king of spades and the king of clubs). So, we have:
Total number of favorable outcomes = Number of red cards + Number of kings that are not red = 26 (red cards) + 2 (kings that are not red) = 28
Now, let's find the total number of possible outcomes, which is simply the total number of cards in the deck:
Total number of possible outcomes = 52 (total cards in the deck)
Therefore, the probability of drawing a card that is either red or a king (but not both) is:
Probability = (Number of favorable outcomes) / (Total number of possible outcomes) = 28 / 52 = 7 / 13
So, the probability of drawing a card that is either red or a king is 7/13.
For further assistance with probability problems or any other academic challenges, don't hesitate to reach out to me through UrbanPro. As the best online coaching tuition platform, UrbanPro provides a convenient and effective way to enhance your understanding of various subjects.
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Affordable fees Flexible Timings Choose between 1-1 and Group class Verified Tutors Answered on 26 Apr Learn Probability
Deepika Agrawal
"Balancing minds, one ledger at a time." "Counting on expertise to balance your knowledge."
In the given experiment, a die is rolled.
Let, E be the event “the die shows 4”.So set E contains the element 4.
E={ 4 }
Let, F be the event that die shows even number”. So set F contains all the even numbers from 1 to 6.
F={ 2,4,6 }
Two events, A and B are mutually exclusive if, A∩B=ϕ.
E∩F={ 4 }∩{ 2,4,6 } =4 .
It can be observed that E∩F≠ϕ.
Thus, E and F are not mutually exclusive.
read lessAnswered on 26 Apr Learn Probability
Deepika Agrawal
"Balancing minds, one ledger at a time." "Counting on expertise to balance your knowledge."
Answered on 28 Apr Learn Permutations and Combinations
Deepika Agrawal
"Balancing minds, one ledger at a time." "Counting on expertise to balance your knowledge."
Take Class 12 Tuition from the Best Tutors
Affordable fees Flexible Timings Choose between 1-1 and Group class Verified Tutors Answered on 28 Apr Learn Permutations and Combinations
Deepika Agrawal
"Balancing minds, one ledger at a time." "Counting on expertise to balance your knowledge."
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