The actual mass of a packet of cheese is [215 g, 225 g), hence the total mass of 250 packets is within the interval [53750 g, 56250 g) or [53.75 kg, 56.25 kg). 53.6 kg correct to the nearest 0.1 kg lies outside the interval of [53.75 kg, 56.25 kg). Therefore the claim is false.
(a)
and
take the intersection of these two intervals,
(b)
-4.5 < x < 3 implies negative integer x could be -4, -3, … , -1, a total of 4 negative integers.
Let F be the number of female passengers and M be the number of male passengers. The first sentence implies
Now, 24 female passengers leave.
Substitute the first equation into the second yields,
Since , then
There are 35 male passengers.
(a) 7a = 6b implies a:b = 6:7
Since 6b=7a, so 18b=21a.
Therefore a:b:c = 6:7:17.
(b)
By inscribed angle theorem,
By exterior angle theorem,
hence,
By inscribed angle of diameter,
(a)
Given that AC // DB. By alternate angles are equal, angle C and angle D are equal. And angle A and angle B are equal. Then by third angles of the triangle are equal, angle AEC and angle BED are equal. Therefore,
(b)
Since the corresponding sides of the two similar triangles are proportional,
Therefore triangle ACE is not a right triangle and hence triangle BDE is not a right triangle as well because they are similar.
(a) Range = 27 hence
(b) 14/24 = 7/12
(a) Line Γ is the perpendicular bisector of line segment AB.
(b) Γ: y = -3x + 12, hence
(i)
hence the perpendicular slope is
or
(ii) Center of the circle is the intersection of the two lines.
Substitute x = 3y + 14 into 3x + y – 12 = 0,
a) mean = 2, hence
So, total number of data = 8 + 5 + 6 + 1 = 20.
Q1 = 1 and Q3 = 3, and hence IQR = 3 – 1 = 2.
(b) The range would change if the student who has 4 calculators is withdrawn from the class. But this is impossible because in order for the mean remains unchanged, the other withdrawn student must have zero calculator but no one has zero calculator.
Given that for some real numbers a and b such that f(10) = 62 and f(15) = 122.
(a)
Subtract the second equation by the first yields,
(b)
Since angle UWV = 90 degrees, UV is diameter of the circle. Hence the circumference is
(a) deg(h) = 4, deg(g) = 3, hence the degree of the quotient function = deg(q) = 1. Since remainder is the same as the quotient, r(x) = q(x). Hence,
Let q(x) = 3x+k, for some constant k.
hence by comparing the coefficients of the equation,
Therefore the quotient function is 3k + 4.
(b)
For , and 73 isn’t a perfect square.
So has no rational root.
Therefore, h(x)=0 has 2 rational roots.
(a) Since the curved surface area of the cone is πrs where s is the slant height of the cone, then
Therefore the height is 48 cm.
(b) (i)
Since the two triangles in the figure are similar, the ratio of the curved surface area of X and Y is
Therefore,
(ii)
hence,
Therefore, the diameter of each sphere is 21 cm.
(a)
(b)
Case 1: Drawing 3 R from 10R in the bag,
Case 2: Drawing 3R from 9R1B in the bag.
Case 3: Drawing 3R from 8R2B in the bag.
Therefore, the probability that the three chosen balls are of the same color is
Since p and 5p are the roots of the given equation, by sum of roots and product of roots,
then,
Substitute this into , then
(b) Since OQ : QR = 1 : 4, by similar triangles,
hence and
Plug into , then
Since and are roots of (*), hence by part (a),
(a) By sine law,
(b) Project Z onto triangle WXY and label it as O. Since slant side = s = WZ = XZ = YZ, we have OX = OY = OW = r = circumradius of triangle WXY. By part(a),
Since O is center of the circumcircle,
Obviously , so
Triangle WOZ is a 30-60-90 special triangle, so
In triangle XOM,
Let θ be the angle between triangles WXY and XYZ.
Therefore the angle between triangles WXY and XYZ does not exceed 45 degrees.
(a) Suppose α, 7, β form a geometric sequence, then
(b) Let , then the given arithmetic sequence becomes
Let d be the common difference,
Case ,
This contradicts with . So reject.
Case ,
Case ,
hence, contradict with the fact that , so we reject.
Therefore the common difference is 1.
P(50, 0), Q(32, t), where t > 0. Let R(x, y). Since Q is midpoint of PR, hence
(a) Suppose that G is circumcenter. Let M be midpoint of OP. M(25, 0), so G(25, y).
Let H(14, y).
(b) S lies on OP such that and .
(i) then
(ii) Now,
Therefore, OQ // OG // GQ and hence O, G, and Q are collinear.
(iii)
hence, â–³OPR is isosceles such that OP = OR.
Since Q is midpoint of PR, by SSS,
hence, H and I lie on OQ.
Since RQ = PQ and they are the altitudes of â–³GHR and â–³IPQ respectively, then
So, I(20, y),
Therefore the ratio of the area of â–³GHR to the area of â–³IPQ is 11:12.
