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ratelena [41]
3 years ago
10

kinyua spent 1/4 of his net january salary on school fees .he spent 1/4 of the remainder on electricity,he spent 1/9 of what rem

ained on transport,if he finally had sh 3400,calculate his net january salary ​
Mathematics
1 answer:
oee [108]3 years ago
4 0

Answer:

1-1/4 = 3/4.

3/4-1/4 = 2/4 = 1/2.

1/2-1/9 = 9/18-2/18 = 7/18.

7x/18 = 8400

X = $21,600 = Net Jan. salary.

Hope it helped u if yes mark me BRAINLIEST!

Tysm!

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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  

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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".  

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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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vampirchik [111]
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