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Mathematics: Post your doubts here!

areeba240

areeba240

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upload_2016-4-13_9-29-21-png.60286

how to solve the last two parts i.e. (b) and (c) of this qtn...plzzzzz helppppp
thanks in advance:)
 
techgeek

techgeek

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osama ahmed ibrahim said:
View attachment 60302 View attachment 60301
help in III AND IV
Click to expand...
For (iii) we know the range of sin x is -1 to 1
-1 < sin x < 1
-3 <3 sin x <3
3 > -3sinx> -3
3+4 > 4 - 3 sinx > -3 +4
7 > 4-3sinx > 1

For (iv), you might know that only 1-1 function has inverse, i.e. the values of y at one certain point maps only to one value of x
When you sketch the graph, it would be apparent that 3/2 pi is the greatest value where the function could exist as one to one
 
Saad the Paki

Saad the Paki

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20160413_224214.png In the graph should I use the boundaries as it is or do I have to make it 40.5. And when do we do this adding and subtracting of 0.5?
This what the er says "A large number of candidates plotted on an adjusted upper boundary, such as 39·5 or 40·5. "
 
techgeek

techgeek

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Saad the Paki said:
View attachment 60307 In the graph should I use the boundaries as it is or do I have to make it 40.5. And when do we do this adding and subtracting of 0.5?
This what the er says "A large number of candidates plotted on an adjusted upper boundary, such as 39·5 or 40·5. "
Click to expand...
We only adjust the boundaries when there are gaps between boundaries. If the boundaries given were 1-40 then 41-50, you might see that data between 40-41 is not included. When we plot the graph, our line is continuous, (it isn't broken from 40-41 where data doesn't exist!), in order to make the data continuous, we create mid-boundaries, so that the lower range is up to 40.5 and the next range starts from 40.5. This kind of makes bridges between the data . In this scenario, x<50 includes ALL those which are less than 50, there is no restriction, so we just take the upper boundaries of each range only and plot accordingly.
For instance, see this graph:
cumulative frequency graph.jpg

They adjusted the boundaries here because the data given was broken in between. It was 21-24 for the first range and then 25-28, so they extended the lower range up to 24.5 and started the upper one from 24.5.
 
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techgeek

techgeek

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Sarosh Jameel said:
kindly help with part 1 of this question.
Click to expand...
Collect the information given:

probability that children jump higher than 127 cm is 8/10 = 0.8

probability that children jump higher than 135 cm is 1/3= 0.333

Applying normal distribution as usual:

for 127 cm:

P (X> 127) = 1 - ϕ (127-μ) /σ

0.8 = 1 - ϕ (127-μ) /σ


for 135 cm:

P (X> 135) = 1 - ϕ (135-μ) /σ

0.333 = 1 - ϕ (135-μ) /σ


When you solve these two you end up with two equations:

μ=127 + 0.842σ

and μ=135 - 0.431 σ


Solving these two simultaneously:
you'll get
σ= 6.28

and

μ=132
 
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