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2 years ago

If a:b=5:7 and b:c=6:12, then a:b:c

Mathematics

1 answer:

uranmaximum [27]2 years ago

3 0

Step-by-step explanation:

A/B=5/7 , B/C=6/12

Taking B=12×7=84, we get

A=12×5=60, C=7×6=42

Hence,

60:84:42. Answer

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Determine if events A and B are mutually exclusive.

gladu [14]

Answer:

A) Not mutually exclusive

Step-by-step explanation:

Given

$P(A) = \frac{1}{4}$

$P(B) = \frac{3}{5}$

$P(A\ or\ B) = \frac{7}{10}$

Required

Determine if they are mutually exclusive or not

Mutually exclusive are defined by:

$P(A\ or\ B) = P(A) + P(B)$

So, we have:

$P(A\ or\ B) = \frac{1}{4} + \frac{3}{5}$

Take LCM

$P(A\ or\ B) = \frac{5+12}{20}$

$P(A\ or\ B) = \frac{17}{20}$

By comparing:

$P(A\ or\ B) = \frac{17}{20}$ ---- Calculated

$P(A\ or\ B) = \frac{7}{10}$ ---- Given

We can conclude that A and B are not mutually exclusive because:

$\frac{17}{20} \ne \frac{7}{10}$

6 0

3 years ago

Unit 6.6, Lesson 9: Practice Problems

7nadin3 [17]

Answer:

the answer is 798 + 789 mawdgghhji the same time as the work of the Maroons

8 0

1 year ago

Factor the following perfect square trinomial

yKpoI14uk [10]

25x^2+10x+1

25x^2+5x+5x+1

5x(5x+1)+1(5x+1)

(5x+1)(5x+1)

(5x+1)^2

6 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)$

The direction of flight Y to the nearest degree is 39 degrees.

7 0

2 years ago

Thirty cows. Twenty-eight chickens. How many didn't?

Anon25 [30]

20 didn't do it for this one.

3 0

2 years ago

If a:b=5:7 and b:c=6:12, then a:b:c​

If a:b=5:7 and b:c=6:12, then a:b:c