₦airaland Forum

I will be posting questions and after few attempts(if wrong), I will post the answer.....

You can as well solve on paper and send the picture....

*hope we have participants

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QUESTION 1

A plane P takes off from an airport A at 10:15am and flies on a bearing of 150° at a speed of 700km/h.

How far due south of the airport A is the plane at 1:45pm? Correct to the nearest kilometers..

Total time is 3 hrs 30 min.

Distance travelled is 3.5hrs × 700km/hr= 2450km.

Due south is 2450× Cos30 =2121.76km

Therefore, plane p is 2121.76km due south of airport A at 1:45 pm.

LionDeLeo:
Total time is 3 hrs 30 min.

Distance travelled is 2450km.

Due south, 2121.76km

That was correct.......Engineer, physicist or mathematician, which one are you sir?

AbdelKabir:
That was correct.......Engineer, physicist or mathematician, which one are you sir?

Engineer.

I have just modified to show the working.

LionDeLeo:
Engineer

Great! Will be posting more, hope you'll be around the corner to answer them ....


Honestly the intent of this thread is to actually help high school students....

AbdelKabir:
Great! Will be posting more, hope you'll be around the corner to answer them ....


Honestly the intent of this thread is to actually help high school students....

If I'm around, I will try.

Will post one more question on "bearing and distance" before moving to another aspect...

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QUESTION 2

Two ships A and B left a port P at the same time. A travels on a bearing of 159° and B travels on a bearing of 215°, after sometime, A is 9km from P and the bearing of B from A is 256°. Calculate the distance of B from A.

[s]

LionDeLeo:
Engineer.

I have just modified to show the working.

[/s]
lol those that do Yahoo Yahoo electrical engineering in India
lol

AbdelKabir:
Will post one more question on "bearing and distance" before moving to another aspect...

---

QUESTION 2

Two ships A and B left a port P at the same time. A travels on a bearing of 159° and B travels on a bearing of 215°, after sometime, A is 9km from P and the bearing of A from B is 256°. Calculate the distance of B from A.

Pls, check very well, is it bearing of A from B or B from A?

@ AbdelKabir.

LionDeLeo:
Total time is 3 hrs 30 min.

Distance travelled is 3.5hrs × 700km/hr= 2450km.

Due south is 2450× Cos30 =2121.76km

Therefore, plane p is 2121.76km due south of airport A at 1:45 pm.

awww i love this

gonna give it to you this time

resemble gaussian elimination and Laplace theory

you're really good in solving

cursedAbiola:
awww i love this

gonna give it to you this time

resemble gaussian elimination and Laplace theory

you're really good in solving

Thank you.

LionDeLeo:
Pls, check very well, is it bearing of A from B or B from A?

@ AbdelKabir.

Sorry its actually B from A.... Hope you can help with diagrams too?

AbdelKabir:
Sorry its actually B from A.... Hope you can help with diagrams too?

Distance of B from A is 11.37km

Rough work attached.

LionDeLeo:
Distance of B from A is 11.37km

Rough work attached.

That was also correct sir.......Thanks!

I should perhaps move to probability distribution.....

QUESTION 3

A die is tossed 3 times. What is the probability of

(a) No fives turning up?

(b) 1 five?

(c) 3 fives?

AbdelKabir:
That was also correct sir.......Thanks!

I should perhaps move to probability distribution.....

Sorry, I just noticed, I forgot to attach the working.

AbdelKabir:
QUESTION 3

A die is tossed 3 times. What is the probability of

(a) No fives turning up?

(b) 1 five?

(c) 3 fives?

(a) 125/216

(b) 25/216 (only 1 five) or 1/2 (at least 1 five)

(c) 1/216

LionDeLeo:
(a) 125/216

(b) 25/216 (only 1 five) or 1/2 (at least 1 five)

(c) 1/216

Emm, I think you need to check that "b" again......BTW its for only 1 five.

AbdelKabir:
Emm, I think you need to check that "b" again......BTW its for only 1 five.

75/216, this is what you want, I guess.

I don't really agree with this actually but since I'm not an authority, let's level it at that.

LionDeLeo:
75/216, this is what you want, I guess.

I don't really agree with this actually but since I'm not an authority, let's level it at that.

Can you explain why you don't agree with that?

*I might learn something new...

AbdelKabir:
Can you explain why you don't agree with that?

*I might learn something new...

It is based on this working:

1/6 ×5/6 ×5/6 or (+) 5/6×1/6×5/6 or (+) 5/6×5/6×1/6.

But in the actual sense, the dice was tossed just 3 times.

My argument is that since it was tossed just 3 times and the probability required is just the appearance of 5 only once, it is more appropriate to assume the probability of 5 appearing only once out of the three tosses which leads to 1/6 (for 5) × 5/6×5/6 (for non 5) regardless of the order in which it occurs.

I once argued this while undergoing one HSE course in my present job but the resource person out argued me.

Up till now I am not satisfied with his analysis and that is why I said, since I'm not an authority, let's leave it at that, perhaps I may see practical reasons to agree in future, can't say for now.

LionDeLeo:
It is based on this working:

1/6 ×5/6 ×5/6 or (+) 5/6×1/6×5/6 or (+) 5/6×5/6×1/6.

But in the actual sense, the dice was tossed just 3 times.

My argument is that since it was tossed just 3 times and the probability required is just the appearance of 5 only once, it is more appropriate to assume the probability of 5 appearing only once out of the three tosses which leads to 1/6 (for 5) × 5/6×5/6 (for non 5) regardless of the order in which it occurs.

I once argued this while undergoing one HSE course in my present job but the resource person out argued me.

Up till now I am not satisfied with his analysis and that is why I said, since I'm not an authority, let's leave it at that, perhaps I may see practical reasons to agree in future, can't say for now.

Don't you think the order in which it appears is also important?

Anyway since the thread is targeting "high school" scope, then your suggestion of leaving it at that is the best...

QUESTION 4

Hospital records show that of patients suffering from a certain disease, 75% die of it. What is the probability that of 6 randomly selected patients, 4 will recover?

AbdelKabir:
QUESTION 4

Hospital records show that of patients suffering from a certain disease, 75% die of it. What is the probability that of 6 randomly selected patients, 4 will recover?

9/4096

AbdelKabir:
Can you explain why you don't agree with that?

*I might learn something new...

LionDeLeo:
It is based on this working:

1/6 ×5/6 ×5/6 or (+) 5/6×1/6×5/6 or (+) 5/6×5/6×1/6.

But in the actual sense, the dice was tossed just 3 times.

My argument is that since it was tossed just 3 times and the probability required is just the appearance of 5 only once, it is more appropriate to assume the probability of 5 appearing only once out of the three tosses which leads to 1/6 (for 5) × 5/6×5/6 (for non 5) regardless of the order in which it occurs.

I once argued this while undergoing one HSE course in my present job but the resource person out argued me.

Up till now I am not satisfied with his analysis and that is why I said, since I'm not an authority, let's leave it at that, perhaps I may see practical reasons to agree in future, can't say for now.

Shut up. Who are you deceiving? Continue asking questions and be quoting yourself, na today?

LionDeLeo:
It is based on this working:

1/6 ×5/6 ×5/6 or (+) 5/6×1/6×5/6 or (+) 5/6×5/6×1/6.

But in the actual sense, the dice was tossed just 3 times.

My argument is that since it was tossed just 3 times and the probability required is just the appearance of 5 only once, it is more appropriate to assume the probability of 5 appearing only once out of the three tosses which leads to 1/6 (for 5) × 5/6×5/6 (for non 5) regardless of the order in which it occurs.

I once argued this while undergoing one HSE course in my present job but the resource person out argued me.

Up till now I am not satisfied with his analysis and that is why I said, since I'm not an authority, let's leave it at that, perhaps I may see practical reasons to agree in future, can't say for now.

I agree with you on that but also remember the number of possible ways this can occur. You are tossing the die three times either the first, second or third turns up a five while the rest don't. This will now give you a combination problem so you have
3c1 × 25/216
= 3 × 25/216
= 75/216

LionDeLeo:
9/4096

That doesn't go well too, perhaps you followed your theory?

AbdelKabir:
QUESTION 4

Hospital records show that of patients suffering from a certain disease, 75% die of it. What is the probability that of 6 randomly selected patients, 4 will recover?

135/4096

blazingkings:
I agree with you on that but also remember the number of possible ways this can occur. You are tossing the die three times either the first, second or third turns up a five while the rest don't. This will now give you a combination problem so you have
3c1 × 25/216
= 3 × 25/216
= 75/216

Yes, we dealing with probability(uncertainty) so we are not sure how it will come, hence the order it follows is very important lest it is no longer "probability".

blazingkings:
135/4096

That's correct....

QUESTION 5

A representative from the National Football League's Marketing Division randomly selects people on a random street in Kansas City, Kansas until he finds a person who attended the last home football game. Let p, the probability that he succeeds in finding such a person, equal 0.20. And, let X denote the number of people he selects until he finds his first success.

a) What is the probability that the marketing representative must select 4 people before he finds one who attended the last home football game?

b) What is the probability that the marketing representative must select more than 6 people before he finds one who attended the last home football game?

AbdelKabir:
QUESTION 5

A representative from the National Football League's Marketing Division randomly selects people on a random street in Kansas City, Kansas until he finds a person who attended the last home football game. Let p, the probability that he succeeds in finding such a person, equal 0.20. And, let X denote the number of people he selects until he finds his first success.

a) What is the probability that the marketing representative must select 4 people before he finds one who attended the last home football game?

b) What is the probability that the marketing representative must select more than 6 people before he finds one who attended the last home football game?

A) 0.1024
B) 0.262