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eduard
3 years ago
15

A total of 150 new recruits join an army training course. After one month, 30 have dropped out. what is the percentage decrease

in the number of trainees after one month
Mathematics
2 answers:
Pachacha [2.7K]3 years ago
4 0
So what will it be a percentage of? it will be the percentage of the original recruits, of the original 150.

for this, we need to divide the number of recruits that left over all recruits, and multiply by 100%:\frac{30}{150} *100%= \frac{1}{5} *100%=20%

so the number of trainees decreased by 20 %
eimsori [14]3 years ago
3 0
Originally there were 150 recruits 
30 dropped out 
the percentage of decrease would be 30/150
which can be simplified to 1/5 
1/5 as a percentage is 20%
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Answer:

(0.1875-0.274) - 1.96 \sqrt{\frac{0.1875(1-0.1875)}{400} +\frac{0.274(1-0.274)}{500}}=-0.1412  

(0.1875-0.274) + 1.96 \sqrt{\frac{0.1875(1-0.1875)}{400} +\frac{0.274(1-0.274)}{500}}=-0.0318  

And the 95% confidence interval would be given (-0.1412;-0.0318).  

We are confident at 95% that the difference between the two proportions is between -0.1412 \leq p_A -p_B \leq -0.0318

Step-by-step explanation:

Previous concepts

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".  

The margin of error is the range of values below and above the sample statistic in a confidence interval.  

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".  

p_A represent the real population proportion for business  

\hat p_A =\frac{75}{400}=0.1875 represent the estimated proportion for Business

n_A=400 is the sample size required for Business

p_B represent the real population proportion for non Business

\hat p_B =\frac{137}{500}=0.274 represent the estimated proportion for non Business

n_B=500 is the sample size required for non Business

z represent the critical value for the margin of error  

The population proportion have the following distribution  

p \sim N(p,\sqrt{\frac{p(1-p)}{n}})  

Solution to the problem

The confidence interval for the difference of two proportions would be given by this formula  

(\hat p_A -\hat p_B) \pm z_{\alpha/2} \sqrt{\frac{\hat p_A(1-\hat p_A)}{n_A} +\frac{\hat p_B (1-\hat p_B)}{n_B}}  

For the 95% confidence interval the value of \alpha=1-0.95=0.05 and \alpha/2=0.025, with that value we can find the quantile required for the interval in the normal standard distribution.  

z_{\alpha/2}=1.96  

And replacing into the confidence interval formula we got:  

(0.1875-0.274) - 1.96 \sqrt{\frac{0.1875(1-0.1875)}{400} +\frac{0.274(1-0.274)}{500}}=-0.1412  

(0.1875-0.274) + 1.96 \sqrt{\frac{0.1875(1-0.1875)}{400} +\frac{0.274(1-0.274)}{500}}=-0.0318  

And the 95% confidence interval would be given (-0.1412;-0.0318).  

We are confident at 95% that the difference between the two proportions is between -0.1412 \leq p_A -p_B \leq -0.0318

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3 years ago
Using f(x)=2x-3 and g(x)=4x, what is f(2) - g(3)
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First fine f(2).

f(2) = 2*2 - 3
f(2) = 4 - 3
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Now g(3).

g(3) = 4*3
g(3) = 12

Now subtract 1 - 12 = -11

-11 is your answer.
6 0
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