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Bad White [126]
3 years ago
13

in a certain examination 52 candidate offers biology,60 offers history,96 offers maths, if 21 offered both biology and history,2

2 offered maths and biology, and 16 offered maths and history,if 7 candidates offered all the subject. how many candidates were there for the examination​
Mathematics
1 answer:
Naddik [55]3 years ago
3 0

Answer: There are 135 candidates.

Step-by-step explanation:

we know that:

52 offer biology

60 offers history

96 offers maths

21 offered both biology and history

22 offered maths and biology

16 offered maths and history

if 7 candidates offered all the subject.

We want to find the total number of candidates.

We start by adding the numbers for each particular exam:

(52 + 60 + 96)

Now, there are 21 that offered both biology and history, then these 21 are counted in the 52 for biology and in the 60 for history, then we need to subtract 21 (because we are counting it two times)

(52 + 60 + 96) - 21

The same happens for the 22 that offered maths and biology, and 16 that offered maths and history, then we get:

(52 + 60 + 96) - 21 - 22 - 16

And there are another 7 that offered for the 3 subjects, then we are counting these 7 ones 3 times, this means that we need to subtract 7 two times.

then we get:

(52 + 60 + 96) - 21 - 22 - 16 - 7 - 7

This is the total number of candidates, if we solve the equation we get:

candidates = (52 + 60 + 96) - 21 - 22 - 16 - 7 - 7 = 135

So we can conclude that there are 135 candidates.

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Yes the new method if sample size was less than 20 than that of old method or identical sample numbers of old and new the differences still prove the new operation is better. As 88 patients minus 1% still shows us 76.7475 significance of  old method being low point 67.5 = 30% of 225 and proved a 65.25 low point and 69.75 high point which is also a 20% jump to new methods low point significance.

You cna show this as workings to prove or follow any of the below statements.

Where new method of 88 patients -0.01 significance rate stands at 76.7475. This figure has reduced by 11.2525 from 88 patients to 76.7 we compare this to the old method if reversing significance we find  = 62.5 and it's 30% standing value of 67.5  as +1% increase shows us 31% = 69.7  ( 0.31 x 225 = 69.74)

Step-by-step explanation:

88/225  = 0.39111111111 = 39.11%%

P value 01 = 1%  =  225.225 or 5% range of alternative hypotheses.

To graph the P value we take the distance between the sample mean and the null hypothesis value (225 + 1% of sample - x nhv) = y ). We can graph the probability of obtaining a sample mean  (225 +/- ( x +1% of sample) where nhv has a decimal if needed to utilize the 1% added). we would replace nvp in this example with  Ha or H1 which means the alternative hypotheses as the data shows less than or equal to.

We can then show 225.225 -  Ha or H1  then graph the probability of obtaining a sample mean that is at least extreme in both tails  Ha or H1

However it would be the other way round where you take the first set of data and use the sample as the 30% significance of that sample indicates it may be a larger sample or a higher significance. Therefore this would be used in the graphing - 1%

We prove that 30-1 =29  where 29% of 225 = 225 x 0.29 = 65.25

this way we have proved that the new set of data being equal to 88 patients regaining their eyesight is <23 and can be written like this 65.25< x <88

This means that sample mean has taken the 1% to show on the graph we can show 225> 33.11 +1 .

We can prove that both indifference of significance would reduce when 1% is added and close based on being a higher percentage to begin with.

34.11 = 0.3411 x 225 = 76.7475 for second surgeon = 33.11% +1

Where as shown

30 = 0.3 x 225 = 67.5

76.5475 - 67.5 = 9.04 difference when comparing old method = +1%

where new method stands at 76.7475 has reduced by 11.2525 from 88 patients and where old method if reversing = 62.5 and has reduced from 67.5  as +1% and 31% = 69.7  ( 0.31 x 225 = 69.74)

You would therefore graph each higher methods first if comparing both by 0.01 or show 88 on graph and 76.7475 = +1%

NB/ if sample size was 20 more in the old data then 225+20 = 245 x 0.29 = 71.05 and would still be lower than new data. = 2.0 increase level of significance and not relevant unless you are looking for the decrease which means new is greater than 20% success than that of old method findings where 30% = 67.5.

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Here the roots are: $ +6 , +6 $.

The factored form would simply be:  $ (x - 6) (x - 6) $.

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