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padilas [110]
3 years ago
8

NEED HELP ASAP. I don not understand why I'm wrong. What is the right answer???

Mathematics
2 answers:
galina1969 [7]3 years ago
8 0

\huge \boxed{\mathfrak{Question} \downarrow}

  • Simplify \huge \sf\frac { \frac { 3 } { 4 } - \frac { 5 } { 2 } } { \frac { 3 } { 5 } - \frac { 3 } { 2 } } \\

\large \boxed{\mathfrak{Answer \: with \: Explanation} \downarrow}

\huge \sf\frac { \frac { 3 } { 4 } - \frac { 5 } { 2 } } { \frac { 3 } { 5 } - \frac { 3 } { 2 } } \\

  • The least common multiple of 4 and 2 is 4. Convert \sf\frac{3}{4}and \sf\frac{5}{2}to fractions with denominator 4.

\huge \sf \frac{\frac{3}{4}-\frac{10}{4}}{\frac{3}{5}-\frac{3}{2}}  \\

  • Because \sf\frac{3}{4}and \sf \frac{10}{4} have the same denominator, subtract them by subtracting their numerators.

\huge \sf\frac{\frac{3-10}{4}}{\frac{3}{5}-\frac{3}{2}}  \\

  • Subtract 10 from 3 to get -7.

\huge \sf\frac{-\frac{7}{4}}{\frac{3}{5}-\frac{3}{2}}  \\

  • The least common multiple of 5 and 2 is 10. Convert \sf\frac{3}{5} and \sf \frac{3}{2} to fractions with denominator 10.

\huge \sf\frac{-\frac{7}{4}}{\frac{6}{10}-\frac{15}{10}}  \\

  • Because \sf \frac{6}{10} and \sf \frac{15}{10} have the same denominator, subtract them by subtracting their numerators.

\huge \sf\frac{-\frac{7}{4}}{\frac{6-15}{10}}  \\

  • Subtract 15 from 6 to get -9.

\huge \sf\frac{-\frac{7}{4}}{-\frac{9}{10}}  \\

  • Divide \sf-\frac{7}{4} by \sf-\frac{9}{10} by multiplying \sf-\frac{7}{4} by the reciprocal of \sf-\frac{9}{10}.

\huge \sf-\frac{7}{4}\left(-\frac{10}{9}\right)

  • Multiply \sf-\frac{7}{4} by \sf-\frac{10}{9} by multiplying the numerator by the numerator and the denominator by the denominator.

\huge \sf\frac{-7\left(-10\right)}{4\times 9}

  • Carry out the multiplications in the fraction \sf\frac{-7\left(-10\right)}{4\times 9}.

\huge \sf\frac{70}{36}

  • Reduce the fraction \sf\frac{70}{36} to its lowest terms by extracting and cancelling out 2.

\huge \boxed{ \bf\frac{35}{18}\approx 1.944..}

hammer [34]3 years ago
6 0

Answer:

<u>Answer</u><u>:</u><u> </u><u>3</u><u>5</u><u>/</u><u>1</u><u>8</u>

Step-by-step explanation:

\frac{ \{ \frac{3}{4} -  \frac{5}{2}  \} }{ \{ \frac{3}{5}  -  \frac{3}{2}  \}}   =  \frac{( -  \frac{7}{4} )}{ (-  \frac{9}{10} )}  \\  \\   =  \frac{7}{4}  \div  \frac{9}{10}

use the reciprocal of 9/10 :

=  \frac{7}{4}  \times  \frac{10}{9}  \\  \\  =  \frac{35}{18}

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a) p_A represent the real population proportion for women

\hat p_A =\frac{123}{150}=0.82 represent the estimated proportion for women

n_A=150 is the sample size selected for women

b) p_B represent the real population proportion for male

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c) (0.82-0.6) - 1.96 \sqrt{\frac{0.82(1-0.82)}{150} +\frac{0.6(1-0.6)}{170}}=0.124  

(0.82-0.6) + 1.96 \sqrt{\frac{0.82(1-0.82)}{150} +\frac{0.6(1-0.6)}{170}}=0.316  

And the 95% confidence interval would be given (0.124;0.316).  

We are confident at 95% that the difference between the two proportions is between 0.124 \leq p_A -p_B \leq 0.316

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".  

Part a

p_A represent the real population proportion for women

\hat p_A =\frac{123}{150}=0.82 represent the estimated proportion for women

n_A=150 is the sample size selected for women

Part b

p_B represent the real population proportion for male

\hat p_B =\frac{102}{170}=0.6 represent the estimated proportion for male

n_B=170 is the sample size elected for male

z represent the critical value for the margin of error  

Part c

The population proportion have the following distribution  

p \sim N(p,\sqrt{\frac{p(1-p)}{n}})  

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:  

(0.82-0.6) - 1.96 \sqrt{\frac{0.82(1-0.82)}{150} +\frac{0.6(1-0.6)}{170}}=0.124  

(0.82-0.6) + 1.96 \sqrt{\frac{0.82(1-0.82)}{150} +\frac{0.6(1-0.6)}{170}}=0.316  

And the 95% confidence interval would be given (0.124;0.316).  

We are confident at 95% that the difference between the two proportions is between 0.124 \leq p_A -p_B \leq 0.316

8 0
3 years ago
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