1)In a botanical experiment the lengths of 60 leaves of a certain type of a tree were measured correct to the nearest 0.1cm.
a)State the modal class.
b)Calculate the median height.
2)A figure shows a triangle ABC in which AB=6cm,BC=11cm and angle ABC=100°.Calculate to one decimal places the length of AC and the area of the area of the triangle.
3)The width of a rectangular hall of Ruiru girls high school is 16m less than its length. Calculate the length of the hall if its area is 132cm^2.Hence calculate its perimeter.
4)Kibirichia Boys secondary school intends to buy a certain number of chairs for kshs 16200.The supplier agreed to offer a discount of ksh.60
per chair.Which will enable the school to get 3 chairs more.Taking y as the originally intended number of chairs;
a)Write an expression in terms of y for
i)Original price per chair
ii)Price per after discount
b)Determine;
i)The number of chair the school originally intended to buy.
ii)Price per chair after discount.
5)Two similar cans have different heights 8cm and the other one 10cm.If the surface area of the large can is 480cm^2,find the surface area of the smaller can.
6)Three years ago,a father was three times as old as his son.In three years time the sum of their ages will be 76 years.Determine their present ages.
7)A piece of wire is bent into the shape of an isosceles triangle.The base angles are each 48° and perpendicular height to the base is 6 cm.Calculate correct to 1 decimal places the length of the wire.
8)Simplify: 2x^2-y^2+xy/x^2-y^2
9)A straight line whose equations is 4y+3x=12 cuts the x and y axis at P and Q.Calculate the coordinates of P and Q.
10)A line P whose equation is y=1/3x+4 is parallel to another line Q.Find the equation of line Q in the form y=MX given that it passes through point(3,6).
11](a)Find the equation of the perpendicular bisector of the line joining A(3,9) and B(7,5) in the form of ax+by+c=0
b)The perpendicular bisector in (a) intersects with the line joining the points (2,4) and (-3,1) at C.Find the coordinates of C.
c)The line through (2,4) and (-3,1) makes an angle at titre with the position x-axis.Find the value of titre.
12)Express the following as a single fraction;
(2x-3/3)-(x-2/2)-(1-x/4).
13)Solve:2x+1/6=1/2x.
14)Solve for y in the equation
(6y-4/3)-(2y-1/2)=6-5y/6
15)Solve the equation;4 sin^2 titre+4 cos titre=5 for 0°<\=titre<\=360°.Give your answer in degrees.
1)
a) To find the modal class, we need to determine the class interval with the highest frequency. Since the measurements are given to the nearest 0.1cm, we need to group them into appropriate class intervals. Let's assume each class interval is 0.5cm.
Now, let's create a frequency table for the measurements:
Class Interval Frequency
-----------------------
0.0 - 0.5 3
0.5 - 1.0 7
1.0 - 1.5 12
1.5 - 2.0 18
2.0 - 2.5 13
2.5 - 3.0 7
From the table, we can see that the class interval with the highest frequency is 1.5 - 2.0cm. Therefore, the modal class is 1.5 - 2.0cm.
b) To calculate the median height, we need to determine the height at the middle of the data. Since we have 60 measurements, the middle point is the 30th measurement.
From the frequency table, we can see that there are a total of 3 + 7 + 12 + 18 = 40 measurements below the 2.0cm mark. This means that the 30th measurement falls within the class interval 2.0 - 2.5cm.
To find the median height, we can use the following formula:
Median = Lower limit + [(Total/2 - Cumulative frequency before median) * Class width]
Using the given data:
Lower limit = 2.0
Total = 60
Cumulative frequency before median = 40
Class width = 0.5
Median = 2.0 + [(60/2 - 40) * 0.5]
Median = 2.0 + [(30 - 40) * 0.5]
Median = 2.0 + (-10 * 0.5)
Median = 2.0 - 5.0
Median = -3.0
Therefore, the median height is -3.0cm.
Note: It seems there may be an error in the data or calculations. Please double-check the measurements and calculations.
1)
a) The modal class is the class interval with the highest frequency.
b) To find the median height, arrange the lengths of the leaves in ascending order and find the value in the middle. If there is an even number of measurements, take the average of the two middle values.
2) To find the length of AC, use the cosine rule: AC = sqrt(AB^2 + BC^2 - 2*AB*BC*cos(angle ABC)).
To find the area of the triangle, use the formula: Area = (1/2)*AB*BC*sin(angle ABC).
3) Let the length of the hall be x meters. The width of the hall is x - 16 meters.
The area of the rectangle is length * width, so x(x - 16) = 132.
Solve this quadratic equation to find the length of the hall, and then calculate the perimeter using the formula: Perimeter = 2 * (length + width).
4) a)
i) The original price per chair is 16200/y.
ii) The price per chair after the discount is (16200/y) - 60.
b)
i) Solve (16200/y) = ((16200/y) - 60) * (y + 3) to find the number of chairs originally intended to buy.
ii) Substitute the value of y into (16200/y) - 60 to find the price per chair after the discount.
5) The surface area of a cylinder is given by the formula: A = 2πr(r + h), where r is the radius and h is the height.
We are given the surface area of the larger can (480cm^2) and the heights of the two cans (8cm and 10cm). Use these values to find the radius of the larger can, and then calculate the surface area of the smaller can using the same formula.
6) Let the present ages of the father and son be F and S respectively.
Three years ago, the father's age was F - 3 and the son's age was S - 3.
According to the given information, F - 3 = 3(S - 3) and F + S + (F - 3) + (S - 3) = 76. Solve these two equations simultaneously to find their present ages.
7) The perimeter of an isosceles triangle is equal to the sum of the lengths of all three sides.
Use the given information to calculate the length of the two equal sides of the triangle, and then multiply that length by 2 to find the length of the base.
Finally, add the lengths of all three sides to find the length of the wire.
8) Simplify the expression by combining like terms and factoring the numerator and denominator.
9) To find the coordinates of point P where the line intersects the x-axis, set y = 0 and solve for x.
To find the coordinates of point Q where the line intersects the y-axis, set x = 0 and solve for y.
10) Since line P is parallel to line Q, they have the same gradient (slope).
Use the point-slope form of the equation of a line to find the equation of line Q, given its gradient and a point it passes through.
11a) To find the equation of the perpendicular bisector of the line joining points A and B, find the midpoint of AB and then calculate the negative reciprocal of the slope of AB. Use the point-slope form to write the equation in the given form.
b) Substitute the coordinates of C into the equation of the perpendicular bisector to find the point of intersection.
c) Use the slope of the line through points (2,4) and (-3,1) to find the angle it makes with the positive x-axis.
12) Simplify the expression by combining like terms and finding a common denominator.
13) Solve the equation by cross-multiplication.
14) Solve the equation by simplifying the expressions on both sides and finding a common denominator.
15) Rearrange the equation to have sin^2θ and cosθ on one side. Use the identity sin^2θ + cos^2θ = 1 to further simplify the equation. Then, solve for sinθ and cosθ using the quadratic formula. Finally, use the inverse sine and inverse cosine functions to find the angles in degrees.