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Pergunta 1 Relatório
1. The sum of the 2nd and 5th terms of an arithmetic progression (A.P) is 42. If the difference between the 6th and 3rd terms is 12, find:
a. the common difference
b. the first term
c. the 20th term.
a. fifth term, T\(_5\) =? But the nth term of an A.P. T\(_n\) = a (n - 1) d
T\(_2\) = a + d
T\(_3\) = a + 2d
T\(_5\) = a + 4d
T\(_2\) + T\(_5\) = a + d + a + 4d = 2a + 5d = 42 .....(i)
T\(_6\) = a + 5d
T\(_3\) = a + 2d
T\(_6\) - T\(_3\) = (a + 5d) - (a + 2d) = 3d = 12
3d = 12
d = 4
b. put d = 4 into equation (i)
2a + 5(4) = 42
2a = 42 - 20
2a = 22
a = 11 first term,
a = 11
c. T\(_{20}\) = a + 19d
= 11 + 19 x 4
T\(_{20}\) = 87.
Detalhes da Resposta
a. fifth term, T\(_5\) =? But the nth term of an A.P. T\(_n\) = a (n - 1) d
T\(_2\) = a + d
T\(_3\) = a + 2d
T\(_5\) = a + 4d
T\(_2\) + T\(_5\) = a + d + a + 4d = 2a + 5d = 42 .....(i)
T\(_6\) = a + 5d
T\(_3\) = a + 2d
T\(_6\) - T\(_3\) = (a + 5d) - (a + 2d) = 3d = 12
3d = 12
d = 4
b. put d = 4 into equation (i)
2a + 5(4) = 42
2a = 42 - 20
2a = 22
a = 11 first term,
a = 11
c. T\(_{20}\) = a + 19d
= 11 + 19 x 4
T\(_{20}\) = 87.
Pergunta 2 Relatório
2. If 2\(^{2x -2y}\) = 32 and log\(_y\) x = 2, find the values of x and y
Leave your answer in this format "+ x,- y"
2\(^{2x -2y}\) = 32 = 2\(^{2x -2y}\) = 2\(^5\)
2x - 2y = 5 - - - - - - - -- (1)
log\(_y\) x = 2,
x = y\(^2\) - - - - - - - -(2)
Put x = y\(^2\) into equation (i) 2(y\(^2\)) - 3y = 5
2\(y^2\) - 3y - 5 = 0
2y\(^2\) + 2y - 5y - 5 = 0
2y(y + 1) - 5(y + 1) = 0
(2y - 5)(y + 1) = 0
2y - 5 = 0 or y + 1 = 0
2y = 5 or y = -1
y = \(\frac{5}{2}\) or y = -1
y = \(\frac{5}{2}\) or y = -1
When y = -1
x = (-1)\(^2\)
x = 1
When y = \(\frac{5}{2}\)
x = (\(\frac{5}{2}\))\(^2\) = \(\frac{25}{4}\) = 6\(\frac{1}{4}\).
Detalhes da Resposta
2\(^{2x -2y}\) = 32 = 2\(^{2x -2y}\) = 2\(^5\)
2x - 2y = 5 - - - - - - - -- (1)
log\(_y\) x = 2,
x = y\(^2\) - - - - - - - -(2)
Put x = y\(^2\) into equation (i) 2(y\(^2\)) - 3y = 5
2\(y^2\) - 3y - 5 = 0
2y\(^2\) + 2y - 5y - 5 = 0
2y(y + 1) - 5(y + 1) = 0
(2y - 5)(y + 1) = 0
2y - 5 = 0 or y + 1 = 0
2y = 5 or y = -1
y = \(\frac{5}{2}\) or y = -1
y = \(\frac{5}{2}\) or y = -1
When y = -1
x = (-1)\(^2\)
x = 1
When y = \(\frac{5}{2}\)
x = (\(\frac{5}{2}\))\(^2\) = \(\frac{25}{4}\) = 6\(\frac{1}{4}\).
Pergunta 3 Relatório
SECTION B
9a. Simplify \(\frac{\sqrt{75} - 3}{\sqrt{3} + 1}\), leaving your answers in the form a + b\(\sqrt{c}\), where a, b, and c are rational numbers.
bi. The points (7,3), (2,8), and (-3,3) lie on a circle. Find the equation
bii. Find the radius of the circle.
a. \(\frac{\sqrt{75}-3}{\sqrt{3}+1}\)
First, simplify the radical:
\(\sqrt{75} = \sqrt{25 \cdot 3} = 5\sqrt{3}\)
So the expression becomes:
\(\frac{5\sqrt{3}-3}{\sqrt{3}+1}\)
Now rationalize the denominator by multiplying top and bottom by \(\sqrt{3}\) - 1):
\(\frac{(5\sqrt{3}-3)(\sqrt{3}-1)}{(\sqrt{3}+1)(\sqrt{3}-1)}\)
Denominator: 3 - 1 = 2
Numerator: 15 - 5\(\sqrt{3} - 3\sqrt{3} + 3 = 18 - 8\sqrt{3}\)
So: \(\frac{18 - 8\sqrt{3}}{2} = 9 - 4\sqrt{3}\).
bi. The general equation of a circle is \(x^2 + y^2 + 2gx + 2fy + c = 0\)
Substitute the three points: Point (7, 3):
\(49 + 9 + 14g + 6f + c = 0 \implies 14g + 6f + c = -58 \quad \text{(1)}\)
Point (2, 8):
\(4 + 64 + 4g + 16f + c = 0 \implies 4g + 16f + c = - 68 \quad \text{(2)}\)
Point (-3, 3):
\(9 + 9 - 6g + 6f + c = 0 \implies -6g + 6f + c = -18 \quad \text{(3)}\)
Subtract (2) from (1):
\(10g - 10f = 10 \implies g - f = 1 \implies g = f + 1 \quad \text{(4)}\)
Subtract (3) from (2):
\(10g + 10f = -50 \implies g + f = -5 \quad \text{(5)}\)
Solve (4) and (5):
\((f + 1) + f = -5 \implies 2f + 1 = -5 \implies 2f = -6 \implies f = -3\)
g = -3 + 1 = -2
Substitute \( g = -2 \), \( f = -3 \) into (3):
\(-6(-2) + 6(-3) + c = -18 \implies 12 - 18 + c = -18 \implies c = -12\)
The equation of the circle is: \(x^2 + y^2 - 4x - 6y - 12 = 0\)
bii. The centre is -g, -f = (2, 3).
The radius \( r \) is given by \(r = \sqrt{g^2 + f^2 - c} = \sqrt{(-2)^2 + (-3)^2 - (-12)} = \sqrt{4 + 9 + 12} = \sqrt{25} = 5\)
Radius of the circle = 5 units.
Detalhes da Resposta
a. \(\frac{\sqrt{75}-3}{\sqrt{3}+1}\)
First, simplify the radical:
\(\sqrt{75} = \sqrt{25 \cdot 3} = 5\sqrt{3}\)
So the expression becomes:
\(\frac{5\sqrt{3}-3}{\sqrt{3}+1}\)
Now rationalize the denominator by multiplying top and bottom by \(\sqrt{3}\) - 1):
\(\frac{(5\sqrt{3}-3)(\sqrt{3}-1)}{(\sqrt{3}+1)(\sqrt{3}-1)}\)
Denominator: 3 - 1 = 2
Numerator: 15 - 5\(\sqrt{3} - 3\sqrt{3} + 3 = 18 - 8\sqrt{3}\)
So: \(\frac{18 - 8\sqrt{3}}{2} = 9 - 4\sqrt{3}\).
bi. The general equation of a circle is \(x^2 + y^2 + 2gx + 2fy + c = 0\)
Substitute the three points: Point (7, 3):
\(49 + 9 + 14g + 6f + c = 0 \implies 14g + 6f + c = -58 \quad \text{(1)}\)
Point (2, 8):
\(4 + 64 + 4g + 16f + c = 0 \implies 4g + 16f + c = - 68 \quad \text{(2)}\)
Point (-3, 3):
\(9 + 9 - 6g + 6f + c = 0 \implies -6g + 6f + c = -18 \quad \text{(3)}\)
Subtract (2) from (1):
\(10g - 10f = 10 \implies g - f = 1 \implies g = f + 1 \quad \text{(4)}\)
Subtract (3) from (2):
\(10g + 10f = -50 \implies g + f = -5 \quad \text{(5)}\)
Solve (4) and (5):
\((f + 1) + f = -5 \implies 2f + 1 = -5 \implies 2f = -6 \implies f = -3\)
g = -3 + 1 = -2
Substitute \( g = -2 \), \( f = -3 \) into (3):
\(-6(-2) + 6(-3) + c = -18 \implies 12 - 18 + c = -18 \implies c = -12\)
The equation of the circle is: \(x^2 + y^2 - 4x - 6y - 12 = 0\)
bii. The centre is -g, -f = (2, 3).
The radius \( r \) is given by \(r = \sqrt{g^2 + f^2 - c} = \sqrt{(-2)^2 + (-3)^2 - (-12)} = \sqrt{4 + 9 + 12} = \sqrt{25} = 5\)
Radius of the circle = 5 units.
Pergunta 4 Relatório
5a. There are 6 points in a plane. How many triangles can be formed with the points?
b. A family of 6 is to be seated in a row. In how many ways can this be done if the father and mother are not to sit together?
Leave your answer in whole numbers " abc."
a. Assuming no three points are collinear (as is standard unless specified otherwise), the number of triangles is the number of ways to choose 3 points out of 6, which form a triangle.
I.e \(^6C_3\) = \(\frac{6!}{(6-3)!3!}\) = \(\frac{6!}{3!3!}\)
= \(\frac{6 \times 5 \times 4 \times 3!}{3! \times 3 \times 2 \times 1}\) = \(\frac{6 \times 5 \times 4}{3 \times 2 \times 1}\) = 5 x 4 = 20 triangles.
b. Total number of ways to seat 6 people in a row (no restrictions): 6! =720
Number of ways where father and mother sit together: Treat father and mother as a single unit (they can switch places within the unit: father-mother or mother-father). This gives 5 units to arrange: 5! x 2 = 120 x 2 = 240ways.
Number of ways where they are not together: Total - Together = 720 − 240 = 480 ways
Detalhes da Resposta
a. Assuming no three points are collinear (as is standard unless specified otherwise), the number of triangles is the number of ways to choose 3 points out of 6, which form a triangle.
I.e \(^6C_3\) = \(\frac{6!}{(6-3)!3!}\) = \(\frac{6!}{3!3!}\)
= \(\frac{6 \times 5 \times 4 \times 3!}{3! \times 3 \times 2 \times 1}\) = \(\frac{6 \times 5 \times 4}{3 \times 2 \times 1}\) = 5 x 4 = 20 triangles.
b. Total number of ways to seat 6 people in a row (no restrictions): 6! =720
Number of ways where father and mother sit together: Treat father and mother as a single unit (they can switch places within the unit: father-mother or mother-father). This gives 5 units to arrange: 5! x 2 = 120 x 2 = 240ways.
Number of ways where they are not together: Total - Together = 720 − 240 = 480 ways
Pergunta 5 Relatório
17a. A body of mass 5 kg is placed on a smooth plane inclined at an angle of 30º to the horizontal. Find: the magnitude of the force acting parallel to the plane.
bi. A uniform plank PQ of length 10m and mass m kg rests on two support A and B. Where \PA\ = \BQ\ = 1m. A load of mass 8kg is placed on the plank at point C such that \AC\ = 3.5m, if the reaction at B is 100N. Calculate the value of m
bii. the reaction at A [ take g = 10m/s\(^2\)].
17a. Since the plane is smooth (frictionless), the only force acting parallel to the inclined plane is the component of the weight of the body down the plane.
F = mg sin 30\(^\circ\)
Given m = 5 kg and taking g = 10 m/s\(^2\).
F = 5 \(\times 10 \times \sin 30^\circ\) = 5 \(\times 10 \times\) 0.5 = 25 N.
The magnitude of the force acting parallel to the plane is 25 N.
bi. The uniform plank PQ is 10 m long with supports at A and B such that PA = 1 m and BQ = 1 m. Thus, AB = 8 m. The plank’s centre of mass G is at its midpoint (5 m from P), so AG = 4 m. The 8 kg load is placed at C, where AC = 3.5 m.
Weight of plank = \( 10m \) N
Weight of load = \( 8 \times 10 = 80 \) N
Reaction at B, \( R_B = 100 \) N (given).
For rotational equilibrium, take moments about A (clockwise = anticlockwise):
\((10m) \times 4 + 80 \times 3.5 = 100 \times 8\)
40m + 280 = 800
40m = 520 \(\implies m = \frac{520}{40}\) = 13
Thus, m = 13 kg.
bii. For vertical equilibrium: \(R_A + R_B = \text{total weight}\)
\(R_A + 100 = (10 \times\) 13) + 80
R\(_A\) + 100 = 130 + 80 = 210
R\(_A\) = 210 - 100 = 110 N.
The reaction at A is 110 N.
Detalhes da Resposta
17a. Since the plane is smooth (frictionless), the only force acting parallel to the inclined plane is the component of the weight of the body down the plane.
F = mg sin 30\(^\circ\)
Given m = 5 kg and taking g = 10 m/s\(^2\).
F = 5 \(\times 10 \times \sin 30^\circ\) = 5 \(\times 10 \times\) 0.5 = 25 N.
The magnitude of the force acting parallel to the plane is 25 N.
bi. The uniform plank PQ is 10 m long with supports at A and B such that PA = 1 m and BQ = 1 m. Thus, AB = 8 m. The plank’s centre of mass G is at its midpoint (5 m from P), so AG = 4 m. The 8 kg load is placed at C, where AC = 3.5 m.
Weight of plank = \( 10m \) N
Weight of load = \( 8 \times 10 = 80 \) N
Reaction at B, \( R_B = 100 \) N (given).
For rotational equilibrium, take moments about A (clockwise = anticlockwise):
\((10m) \times 4 + 80 \times 3.5 = 100 \times 8\)
40m + 280 = 800
40m = 520 \(\implies m = \frac{520}{40}\) = 13
Thus, m = 13 kg.
bii. For vertical equilibrium: \(R_A + R_B = \text{total weight}\)
\(R_A + 100 = (10 \times\) 13) + 80
R\(_A\) + 100 = 130 + 80 = 210
R\(_A\) = 210 - 100 = 110 N.
The reaction at A is 110 N.
Pergunta 6 Relatório
13a. The table below shows the distribution of hours spent at work by the employees of a factory in a week
| Time(hours) | 20 - 29 | 30 - 39 | 40 - 49 | 50 - 59 | 60 - 69 | 70 - 79 |
| No. of persons | 8 | 11 | 23 | 25 | 8 | 5 |
Draw an Ogive for the distribution
b. Using your graph, estimate
i. the median.
ii. estimate the lower quartile
iii. 40th percentile
iv. number of employees that spent at least 50 hours 30 mins.
| Time(hours) | Frequency | Cumulative frequency | Upper boundary |
| 20-29 | 8 | 8 | 29.5 |
| 30-39 | 11 | 19 | 39.5 |
| 40-49 | 23 | 42 | 49.5 |
| 50-59 | 25 | 67 | 59.5 |
| 60-69 | 8 | 75 | 69.5 |
| 70-79 | 5 | 80 | 79.5 |
b. Median = (\(\frac{ n + 1}{2}\))\(^{th}\) data
Median = \(\frac{80 + 1}{2}\) = \(\frac{81}{2}\) = 40.5th
So, from the graph, the median is 48.8 hours.
ii. Lower quartile, Q\(_1\) = (\(\frac{n + 1}{4}\))\(^{th}\) data
Q\(_1\) = \(\frac{80 + 1}{4}\) = \(\frac{81}{4}\) = 20.25th data
Q\(_1\) = 40 hours from the graph
iii. 40th percentile
= \(\frac{40}{100}\) x 80 = 32\(^{nd}\) data.
From the graph, 40th percentile = 45.5 hours.
iv. The number of employees who spent at least 50 hours 30 mins = 32.
Detalhes da Resposta
| Time(hours) | Frequency | Cumulative frequency | Upper boundary |
| 20-29 | 8 | 8 | 29.5 |
| 30-39 | 11 | 19 | 39.5 |
| 40-49 | 23 | 42 | 49.5 |
| 50-59 | 25 | 67 | 59.5 |
| 60-69 | 8 | 75 | 69.5 |
| 70-79 | 5 | 80 | 79.5 |
b. Median = (\(\frac{ n + 1}{2}\))\(^{th}\) data
Median = \(\frac{80 + 1}{2}\) = \(\frac{81}{2}\) = 40.5th
So, from the graph, the median is 48.8 hours.
ii. Lower quartile, Q\(_1\) = (\(\frac{n + 1}{4}\))\(^{th}\) data
Q\(_1\) = \(\frac{80 + 1}{4}\) = \(\frac{81}{4}\) = 20.25th data
Q\(_1\) = 40 hours from the graph
iii. 40th percentile
= \(\frac{40}{100}\) x 80 = 32\(^{nd}\) data.
From the graph, 40th percentile = 45.5 hours.
iv. The number of employees who spent at least 50 hours 30 mins = 32.
Pergunta 7 Relatório
6. The table shows the distribution of the ages of a group of people in a village.
| Ages(in years) | 15-18 | 19-22 | 23-26 | 27-30 | 31-34 | 35-38 |
| Frequency | 40 | 33 | 25 | 10 | 8 | 4 |
Using an assumed mean of 24.5. Calculate the mean distribution.
Assumed mean, Am = 24.5
| Ages | f | x | d = x - Am | fd |
| 15 - 18 | 40 | 16.5 | -8.0 | -320 |
| 19 - 22 | 33 | 20.5 | -4.0 | -132 |
| 23 - 26 | 25 | 24.5 | 0.0 | 0 |
| 27 - 30 | 10 | 28.5 | 4.0 | 40 |
| 31 - 34 | 8 | 32.5 | 8.0 | 64 |
| 35 - 38 | 4 | 36.5 | 12.0 | 48 |
| \(\sum\)f = 120 | \(\sum\)fd = - 300 |
Mean (\(\overline{x}\)) = Am + \(\frac{\sum fd}{\sum f}\) = 24.5 + \(\frac{(-300)}{120}\)
Mean (\(\overline{x}\)) = 24.5 - 2.5 = 22.0
Note:(x = sum of upper and lower limit divided by 2)
Detalhes da Resposta
Assumed mean, Am = 24.5
| Ages | f | x | d = x - Am | fd |
| 15 - 18 | 40 | 16.5 | -8.0 | -320 |
| 19 - 22 | 33 | 20.5 | -4.0 | -132 |
| 23 - 26 | 25 | 24.5 | 0.0 | 0 |
| 27 - 30 | 10 | 28.5 | 4.0 | 40 |
| 31 - 34 | 8 | 32.5 | 8.0 | 64 |
| 35 - 38 | 4 | 36.5 | 12.0 | 48 |
| \(\sum\)f = 120 | \(\sum\)fd = - 300 |
Mean (\(\overline{x}\)) = Am + \(\frac{\sum fd}{\sum f}\) = 24.5 + \(\frac{(-300)}{120}\)
Mean (\(\overline{x}\)) = 24.5 - 2.5 = 22.0
Note:(x = sum of upper and lower limit divided by 2)
Pergunta 8 Relatório
3. If (x + 2) and (x -1) are factors of f(x) = 6x\(^4\) + mx\(^3\) - 13x\(^2\) + n x + 14, find the:
a. value of m and n. Leave your answer in this format 'm,n.'
b. remainder when f(x) is divided by (x + 1)
a. (x + 2) and (x -1) are factors of f(x) = 6x\(^4\) + mx\(^3\) - 13x\(^2\) + n x + 14
Then, f(-2) and f(1) are = 0
f(-2) = 6(-2)\(^4\) + m(-2)\(^3\) - 13(-2)\(^2\) + n (-2) + 14 = 0
= 6 x 16 - 8m - 13 x 2 - 2n + 14 = 0 = 96 - 8m - 52 - 2n + 14 = 0
= 8m + 2n = 58: divide through by 2
= 4m + n = 29 - - -- - - - - - -(i)
f(1) = 6x\(^4\) + mx\(^3\) - 13x\(^2\) + n x + 14 = 0
f(1) = 6(1)\(^4\) + m(1)\(^3\) - 13(1)\(^2\) + n (1) + 14 = 0
= 6 + m - 13 + n + 14 = 0
= m + n = -7 - - - - - - - - -(ii)
Solving eqn i and ii simultaneously
from eqn i - - - - - n = 58 - 4m
put n = 58 - 4m into eqn ii
m + n = -7 = m + 29 - 4m = -7
- 3m = -36
m = \(\frac{36}{3}\) = 12
put m = 12 into eqn ii
m + n = -7 = 12 + n = -7
n = -7 - 12 = -19.
b. 6x\(^4\) + mx\(^3\) - 13x\(^2\) + n x + 14 = 0 becomes 6x\(^4\) + 12x\(^3\) - 13x\(^2\) - 19x + 14 = 0
f(x) = 6x\(^4\) + 12x\(^3\) - 13x\(^2\) - 19x + 14 = 0
remainder when f(x) is divided by (x + 1)
Let (x + 1) = 0 then, x = -1
f(-1) = 6(-1)\(^4\) + 12(-1)\(^3\) - 13(-1)\(^2\) - 19(-1) + 14
= 6 - 12 - 13 + 19 + 14 = 14
Therefore, the remainder = 14.
Detalhes da Resposta
a. (x + 2) and (x -1) are factors of f(x) = 6x\(^4\) + mx\(^3\) - 13x\(^2\) + n x + 14
Then, f(-2) and f(1) are = 0
f(-2) = 6(-2)\(^4\) + m(-2)\(^3\) - 13(-2)\(^2\) + n (-2) + 14 = 0
= 6 x 16 - 8m - 13 x 2 - 2n + 14 = 0 = 96 - 8m - 52 - 2n + 14 = 0
= 8m + 2n = 58: divide through by 2
= 4m + n = 29 - - -- - - - - - -(i)
f(1) = 6x\(^4\) + mx\(^3\) - 13x\(^2\) + n x + 14 = 0
f(1) = 6(1)\(^4\) + m(1)\(^3\) - 13(1)\(^2\) + n (1) + 14 = 0
= 6 + m - 13 + n + 14 = 0
= m + n = -7 - - - - - - - - -(ii)
Solving eqn i and ii simultaneously
from eqn i - - - - - n = 58 - 4m
put n = 58 - 4m into eqn ii
m + n = -7 = m + 29 - 4m = -7
- 3m = -36
m = \(\frac{36}{3}\) = 12
put m = 12 into eqn ii
m + n = -7 = 12 + n = -7
n = -7 - 12 = -19.
b. 6x\(^4\) + mx\(^3\) - 13x\(^2\) + n x + 14 = 0 becomes 6x\(^4\) + 12x\(^3\) - 13x\(^2\) - 19x + 14 = 0
f(x) = 6x\(^4\) + 12x\(^3\) - 13x\(^2\) - 19x + 14 = 0
remainder when f(x) is divided by (x + 1)
Let (x + 1) = 0 then, x = -1
f(-1) = 6(-1)\(^4\) + 12(-1)\(^3\) - 13(-1)\(^2\) - 19(-1) + 14
= 6 - 12 - 13 + 19 + 14 = 14
Therefore, the remainder = 14.
Pergunta 9 Relatório
14a. Two pupils are chosen at random from a group of 4 boys and 5 girls. Find the probability that the two pupils chosen would be boys. Leave your answer in fraction 'a/b.'
b. Twenty percent of the total production of transistors produced by a machine are below standard. If a random sample of six transistors produced by the machine is taken, what is the probability of getting
i. Exactly 2
ii. Exactly 1. Leave your answer in six decimal places " a.bcdefg."
iii. At least 2. Leave your answer in five decimal places " a. bcdeg."
iv. At most 2 standard transistors
14a. Total number of pupils = 4 boys + 5 girls = 9.
Number of ways to choose 2 boys = \(\binom{4}{2} = 6\).
Total number of ways to choose any 2 pupils = \(\binom{9}{2} = 36\).
Probability that both chosen are boys = \(\dfrac{6}{36} = \dfrac{1}{6}\).
14b. This is a binomial distribution with n = 6 trials and p = 0.80 (probability a transistor is standard), q = 0.2 (Prb. of a transistor below standard)
Let Y = number of standard transistors.
P(Y = k) = C(6, k) × (0.8)\(^k\) × (0.2)\(^{(6−k)}\)
i: k = 2 → 15 × 0.64 × 0.0016 = 0.015360
ii: k = 1 → 6 × 0.8 × 0.00032 = 0.001536
iii: P(Y ≥ 2) = 1 − P(Y = 0) − P(Y = 1) = 1 − 0.000064 − 0.001536 = 0.99840
iv: P(Y ≤ 2) = P(Y = 0) + P(Y = 1) + P(Y = 2) = 0.000064 + 0.001536 + 0.015360 = 0.016960
Detalhes da Resposta
14a. Total number of pupils = 4 boys + 5 girls = 9.
Number of ways to choose 2 boys = \(\binom{4}{2} = 6\).
Total number of ways to choose any 2 pupils = \(\binom{9}{2} = 36\).
Probability that both chosen are boys = \(\dfrac{6}{36} = \dfrac{1}{6}\).
14b. This is a binomial distribution with n = 6 trials and p = 0.80 (probability a transistor is standard), q = 0.2 (Prb. of a transistor below standard)
Let Y = number of standard transistors.
P(Y = k) = C(6, k) × (0.8)\(^k\) × (0.2)\(^{(6−k)}\)
i: k = 2 → 15 × 0.64 × 0.0016 = 0.015360
ii: k = 1 → 6 × 0.8 × 0.00032 = 0.001536
iii: P(Y ≥ 2) = 1 − P(Y = 0) − P(Y = 1) = 1 − 0.000064 − 0.001536 = 0.99840
iv: P(Y ≤ 2) = P(Y = 0) + P(Y = 1) + P(Y = 2) = 0.000064 + 0.001536 + 0.015360 = 0.016960
Pergunta 10 Relatório
15a. A body of mass 15kg is suspended at a point P by two light inextensible strings XP\(^→\) and YP\(^→\). The strings are inclined at 60º and 40º, respectively, to the downward vertical. Find, correct to two decimal places, the tension in the strings (take g = 10m/s\(^2\))
b. The height h metres, of a ball thrown into the air is 2 + 20t + kt\(^2\), after t seconds. If its takes 2 seconds for the ball to reach its height point, Find:
i. the value of k
ii. its highest point from the point of throw.
\(\frac{150}{sin 100} = \frac{T_1}{sin 140}\)
T\(_1\) = \(\frac{150 sin 140}{sin 100}\) = 97.91N
In same manner:
\(\frac{150}{sin 100} = \frac{T_2}{sin 120}\)
T\(_2\) = \(\frac{150 sin 120}{sin 100}\)
T\(_2\) = 131.91N.
i. At the highest point, v = 0 when t = 2 s:
\(20 + 2k \cdot 2 = 0 \)
20 + 4k = 0
k = - 5
ii. The highest point "from the point of throw" is the upward displacement above the release point t = 0, where the variable terms begin.
At t = 2:
2 + 20 x 2 + k x (2)\(^2\) = 2 + 40 + (-5) x 4 = 42 - 20 = 22 m.
Detalhes da Resposta
\(\frac{150}{sin 100} = \frac{T_1}{sin 140}\)
T\(_1\) = \(\frac{150 sin 140}{sin 100}\) = 97.91N
In same manner:
\(\frac{150}{sin 100} = \frac{T_2}{sin 120}\)
T\(_2\) = \(\frac{150 sin 120}{sin 100}\)
T\(_2\) = 131.91N.
i. At the highest point, v = 0 when t = 2 s:
\(20 + 2k \cdot 2 = 0 \)
20 + 4k = 0
k = - 5
ii. The highest point "from the point of throw" is the upward displacement above the release point t = 0, where the variable terms begin.
At t = 2:
2 + 20 x 2 + k x (2)\(^2\) = 2 + 40 + (-5) x 4 = 42 - 20 = 22 m.
Pergunta 11 Relatório
7a. A body of mass 5 kg resting on a smooth horizontal plane is acted upon by forces 6i + 2j, 5i + 4j, and 4i − j. Calculate: the velocity of the body
b. the magnitude of its velocity, after 4 seconds
a. Mass , m = 5 kg
Resultant force, f = (6i+2j) + (5i+4j) + (4i−j) = (6+5+4)i + (2+4−1)j = F = 15i + 5j
From Newton second law of motion F = \(\frac{m \Delta V}{\text{t}}\)
F = \(\frac{m(v - u)}{\text{t}}\)
where u=0 ms\(^{−1}\)
F = \(\frac{\text{mv}}{\text{t}}\)
v = \(\frac{\text{Ft}}{\text{m}}\)
v = \(\frac{t(15i + 5j)}{5}\)
v = t(3i + j)m/s
b. Velocity, V after 4 s
v = t(3i + j) = (4)(3i + j) = 12i + 4jm/s.
Detalhes da Resposta
a. Mass , m = 5 kg
Resultant force, f = (6i+2j) + (5i+4j) + (4i−j) = (6+5+4)i + (2+4−1)j = F = 15i + 5j
From Newton second law of motion F = \(\frac{m \Delta V}{\text{t}}\)
F = \(\frac{m(v - u)}{\text{t}}\)
where u=0 ms\(^{−1}\)
F = \(\frac{\text{mv}}{\text{t}}\)
v = \(\frac{\text{Ft}}{\text{m}}\)
v = \(\frac{t(15i + 5j)}{5}\)
v = t(3i + j)m/s
b. Velocity, V after 4 s
v = t(3i + j) = (4)(3i + j) = 12i + 4jm/s.
Pergunta 12 Relatório
11a. Using the substitution U = 5 - x\(^2\)
evaluate \(\int _1^2 \frac{\text{x}}{\sqrt{5 - x^2}}\) dx
b. If y = px\(^2\) + qx, \(\frac{\text{dy}}{\text{dx}}\) = 7 and \(\frac{d^2y}{dx^2}\) = 6. Find the values of p and q.
11a. Using the substitution \( u = 5 - x^2 \),
\( du = -2x \, dx \) so \( x \, dx = -\frac{1}{2} du \).
Limits: when \( x = 1 \), \( u = 4 \); when \( x = 2 \), \( u = 1 \).
\(\int_1^2 \frac{x}{\sqrt{5 - x^2}} \, dx = \int_4^1 \frac{1}{\sqrt{u}} \left( -\frac{1}{2} \right) du = \frac{1}{2} \int_1^4 u^{-1/2} \, du\)
\(= \frac{1}{2} \left[ 2u^{1/2} \right]_1^4 = \left[ \sqrt{u} \right]_1^4 = \sqrt{4} - \sqrt{1}\) = 2 - 1 = 1.
11b. Given \( y = p x^2 + q x \),
\( \frac{dy}{dx} = 2px + q \).
But it is given that \( \frac{dy}{dx} = 6x + 7 \).
Equate coefficients of like terms: 2p = 6 ⇒ p = 3, q = 7.
Check: \( \frac{d^2 y}{dx^2} = 2p = 2 \times 3 = 6 \), which matches the given condition.
Thus, p = 3, and q = 7.
Detalhes da Resposta
11a. Using the substitution \( u = 5 - x^2 \),
\( du = -2x \, dx \) so \( x \, dx = -\frac{1}{2} du \).
Limits: when \( x = 1 \), \( u = 4 \); when \( x = 2 \), \( u = 1 \).
\(\int_1^2 \frac{x}{\sqrt{5 - x^2}} \, dx = \int_4^1 \frac{1}{\sqrt{u}} \left( -\frac{1}{2} \right) du = \frac{1}{2} \int_1^4 u^{-1/2} \, du\)
\(= \frac{1}{2} \left[ 2u^{1/2} \right]_1^4 = \left[ \sqrt{u} \right]_1^4 = \sqrt{4} - \sqrt{1}\) = 2 - 1 = 1.
11b. Given \( y = p x^2 + q x \),
\( \frac{dy}{dx} = 2px + q \).
But it is given that \( \frac{dy}{dx} = 6x + 7 \).
Equate coefficients of like terms: 2p = 6 ⇒ p = 3, q = 7.
Check: \( \frac{d^2 y}{dx^2} = 2p = 2 \times 3 = 6 \), which matches the given condition.
Thus, p = 3, and q = 7.
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