The ages of students in a small primary school were recorded in the table below.
| Age | 5-6 | 7-8 | 9-10 |
| Frequency | 29 | 40 | 38 |
Estimate the median.
7.725
6.225
7.5
6.5
Correct answer is A
\(\frac {N + 1}{2} = \frac {107 + 1}{2}\) th = 54th value i.e the median age is in the interval "7 - 8" (29 + 25 = 54)
Median = \(1_m + (\frac {^{\sum f} /_2 - f_b}{f_m})\)c
\(l_m\)=lower class boundary of the median age = 6.5
\(\sum f\) = 107
\(f_b\) = sum of all frequencies before the median age = 29
\(f_m\)=frequency of the median age = 40
c = class width = 8.5 - 6.5 = 2
Median = 6.5 + \((\frac {^{107}/_2 - 29}{40})\) x 2
= 6.5 + \((\frac {53.5 - 29}{40})\) x 2
= 6.5 + \((\frac {24.5}{40})\) x 2
= 6.5 + 1.225
\(\therefore\) median age = 7.725
\(\frac {1}{3}\)
\(\frac {2}{9}\)
\(\frac {2}{3}\)
\(\frac {8}{33}\)
Correct answer is D
Let the number of white balls be n
The number of red balls = 8
Now, the probability of drawing a white ball = \(\frac {n}{n + 8}\)
The probability of drawing a red ball = \(\frac {8}{n + 8}\)
Since Pr(White ball) = \(\frac{1}{2}\) x Pr(Red ball)
\(\therefore \frac {n}{n + 8} = \frac {1}{2}\times \frac {8}{n + 8}\)
= \(\frac {n}{n + 8} = \frac {4}{n + 8}\)
\(\therefore n = 4\)
Number of white balls = 4
The possible number of outcomes = n + 8 = 4 + 8 = 12
Pr(Red ball and White ball) = Pr(Red ball) x Pr(White ball)
Pr(Red ball) = \(\frac {8}{12}\)
Pr(Red ball) = \(\frac {4}{11}\)
Pr(Red ball and white ball) = \(\frac {8}{12} \times \frac {4}{11}\)
\(\therefore\)Pr(Red ball and white ball) = \(\frac {8}{33}\)
If A = { 1, 2, 3, 4, 5, 6}, B = { 2, 4, 6, 8 }. Find (A – B) ⋃ (B – A).
{1, 3, 5, 8}
{8}
{1, 2, 3, 4, 5, 6, 8}
{1, 3, 5}
Correct answer is A
A = { 1, 2, 3, 4, 5, 6}
B = { 2, 4, 6, 8 }
A - B = { 1, 3, 5 }
B - A = { 8 }
∴ (A - B) ⋃ (B - A) = {1, 3, 5, 8}
\((2x - y)(2x + y)(4x^2 + y^2)\)
\((2x + y)(2x + y)(4x^2 + y^2)\)
\((2x - y)(2x - y)(4x^2 + y^2)\)
\((2x - y)(2x + y)(4x^2 - y^2)\)
Correct answer is A
16\(x^4 - y^4\)
= 2\(^4x^4 - y^4\)
= \((2x)^4 - y^4\)
= \(((2x)^2)^2 - (y^2)^2\)
Using a\(^2 - b^2\) = (a - b)(a + b) identity
= ((2x)\(^2 - y^2)((2x)^2 + y^2)\)
Using the identity one more time
= \((2x - y)(2x + y)((2x)^2 + y^2)\)
∴ \((2x - y)(2x + y)(4x^2 + y^2)\)
750
850
250
150
Correct answer is C
Let F be the set of people who can speak French and E be the set of people who can speak English. Then,
n(F) = 400
n(E) = 350
n(F ∪ E) = 500
We have to find n(F ∩ E).
Now, n(F ∪ E) = n(F) + n(E) – n(F ∩ E)
⇒ 500 = 400 + 350 – n(F ∩ E)
⇒ n(F ∩ E) = 750 – 500 = 250.
∴ 250 people can speak both languages.