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timama [110]
3 years ago
15

7 3/4 + (12 - 3 6/11) =

Mathematics
2 answers:
nataly862011 [7]3 years ago
5 0

Answer:

\frac{713}{44}  = 16 \frac{9}{44}

Step-by-step explanation:

7 3/4+ (12 - 3 6/11)

31/4 + (12 - 3 6/11)

31/4 + (12 - 39/11)

31/4 + 93/11

= <u>7</u><u>1</u><u>3</u><u>/</u><u>4</u><u>4</u><u> </u><u>=</u><u> </u><u>1</u><u>6</u><u> </u><u>9</u><u>/</u><u>4</u><u>4</u>

ladessa [460]3 years ago
5 0

Answer:

{ \boxed { \sf{Answer = 16 \frac{9}{44} }}}

Step-by-step explanation:

\\  \sf {7} \frac{3}{4}  + (12 - 3 \frac{6}{11} )

\\  \sf =  \frac{31}{4}  + (12 -  \frac{39}{11} )

\\  \sf =  \frac{31}{4} ( \frac{132 - 39}{11} )

\\  \sf =  \frac{31}{4}  +  \frac{93}{11}

\\  \sf =  \frac{341 + 372}{44}

\\  \sf =  \frac{713}{44}

\\  \sf = 16\frac{9}{44}

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To compare this two sets of data, you need to use a t-student test:

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t= \frac{X1-X2}{ \sqrt{ \frac{(n1-1)* s1^{2}+(n2-1)* s2^{2} }{n1+n2-2}} * \sqrt{ \frac{1}{n1}+ \frac{1}{n2}} } =2,2510

To calculate the degrees of freedom you need to use the following equation:

df= \frac{ (\frac{ s1^{2}}{n1} + \frac{ s2^{2}}{n2})^{2}}{ \frac{(s1^{2}/n1)^{2}}{n1-1}+ \frac{(s2^{2}/n2)^{2}}{n2-1}}=33,89≈34

The tabulated value at 0,05 level (using two-tails, as the distribution is normal) is 2,03. https://www.danielsoper.com/statcalc/calculator.aspx?id=10

So, as the calculated value is higher than the critical tabulated one, we can conclude that the average speed for all vehicles was higher on Monday than on Wednesday.



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