document.write( "Question 1131491: Break-downs occur on an old car with a rate of 6 break-downs every four weeks. The owner of the car is planning to have a trip on his car for 4 days. What is the probability that he will return home without experiencing a car break-down? \r
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Algebra.Com's Answer #748231 by ikleyn(52832)\"\" \"About 
You can put this solution on YOUR website!
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document.write( "6 break-downs every four weeks = 6 break-downs every 28 days = \"6%2F7\" break-downs every 4 days  ====>  \r\n" );
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document.write( "the probability do not have a break-down in 4 days is the COMPLEMENT to \"6%2F7\" = 1 - \"6%2F7\" = \"1%2F7\".      ANSWER\r\n" );
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