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Mathematics

1 answer:

ivanzaharov [21]3 years ago

7 0

14. 32 out of the 125 people surveyed preferred Company B. If the same proportion holds for the larger sample, we expect

... (32/125)×750 = 192 . . . . people would prefer Company B

15. There are a total of 45 marbles in the bag, of which 28 are NOT blue. The probability of drawing a NOT blue marble is 28/45.

16. We can desginate the point where the lines meet as O. Angles NOK and MOL are vertical angles, so are equal. The diagram tells you the measure of NOK is 90° + 15° = 105°, so that is the measure of (5x)°. The value of x is

... 105/5 = x = 21

17. Division by a fraction is the same as multiplication by its inverse.

$-\dfrac{10}{19}\times -\dfrac{7}{5}=\dfrac{10}{5}\cdot\dfrac{7}{19}=2\cdot\dfrac{7}{19}\\\\=\dfrac{14}{19}$

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The measures of the angles of a triangle are in the extended ratio 4 : 5 : 6. What is the measure of the smallest angle?

anyanavicka [17]

The sum of the interior angles of a triangle is 180 degrees

4x + 5x + 6x = 180
15x = 180
x = 180/15
x = 12

The measure of the smallest angle = 4x = 4*12 = 48°

8 0

3 years ago

An aeroplane X whose average speed is 50°km/hr leaves kano airport at 7.00am and travels for 2 hours on a bearing 050°. It then

Zigmanuir [339]

Answer:

(a)123 km/hr

(b)39 degrees

Step-by-step explanation:

Plane X with an average speed of 50km/hr travels for 2 hours from P (Kano Airport) to point Q in the diagram.

Distance = Speed X Time

Therefore: PQ =50km/hr X 2 hr =100 km

It moves from Point Q at 9.00 am and arrives at the airstrip A by 11.30am.

Distance, QA=50km/hr X 2.5 hr =125 km

Using alternate angles in the diagram:

$\angle Q=110^\circ$

(a)First, we calculate the distance traveled, PA by plane Y.

Using Cosine rule

$q^2=p^2+a^2-2pa\cos Q\\q^2=100^2+125^2-2(100)(125)\cos 110^\circ\\q^2=34175.50\\q=184.87$ km$

SInce aeroplane Y leaves kano airport at 10.00am and arrives at 11.30am

Time taken =1.5 hour

Therefore:

Average Speed of Y

$=184.87 \div 1.5\\=123.25$ km/hr\\\approx 123$ km/hr (correct to three significant figures)$

(b)Flight Direction of Y

Using Law of Sines

$\dfrac{p}{\sin P} =\dfrac{q}{\sin Q}\\\dfrac{125}{\sin P} =\dfrac{184.87}{\sin 110}\\123 \times \sin P=125 \times \sin 110\\\sin P=(125 \times \sin 110) \div 184.87\\P=\arcsin [(125 \times \sin 110) \div 184.87]\\P=39^\circ $ (to the nearest degree)$