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ratelena [41]
3 years ago
6

① 5J+2S =12.20 ② 5J+ 10S =15.80 elimination method

Mathematics
1 answer:
hram777 [196]3 years ago
4 0

Answer:

<h2>S = 9/20</h2><h2>J = 113/50</h2>

Step-by-step explanation:

5J +2S =12.20 ........(1)\\5J+10S=15.80......(2)\\Multiply \:equation \:(1)\:by\:the\:coefficient\:of\:x\:in equation\:(2)\\\\Multiply\:equation(2)\: by\: the\: coefficient\:of \:x\: in \:equation (1)\\\\5J +2S =12.20 ........(1) \times 5\\5J+10S=15.80......(2)\times 5\\\\25J+10S=61......(3)\\25J+50S = 79......(4)\\Subtract\:equation\:(4)\:from\: equation\: (3)\\-40S =-18\\\frac{-40S}{-40}=\frac{-18}{-40}\\  S = 9/20\\

Substitute \:9/20\:for \:x \:in\:equation\:(1)\:or \:equation\:(2)\\5J+2S = 12.20\\5J +2(9/20) =12.20\\5J +9/10=12.20\\5J=12.20-9/10\\5J=113/10\\Cross-Multiply\\50J = 113\\J = 113/50

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Answer:

The true statements are:

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Step-by-step explanation:

The interquartile range is the difference between the first and third quartiles

Steps to find the interquartile range:

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The interquartile range is not sensitive to outliers

Now let us find the true statements

A. Subtract the lowest and highest values to find the interquartile range ⇒ NOT true (<em>because the interquartial range is the difference between the lower and upper quartiles</em>)

B. Interquartile ranges are not significantly impacted by outliers ⇒ True <em>(because it does not depends on the smallest and largest data)</em>

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C. Lower and upper quartiles are needed to find the interquartile range ⇒ True <em>(because IQR = Q3 - Q2)</em>

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D. A small interquartile range means the data is spread far away from the median ⇒ NOT true (<em>because a small interquartile means data is not spread far away from the median</em>)

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