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2 answers:
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So... let's say the amounts invested were "a" at 6% and "b" at 7.5%.
ok.. hmm what's 6% of a? well, (6/100) * a or 0.06a.
what's 7.5% of b? well, (7.5/100) * b or 0.075b.
now... we know, whatever "a" and "b" are, they total the investment of 5000 bucks, thus a + b = 5000
and the interest yielded was 337.50, thus 0.06a + 0.075b = 337.50
thus
solve for "a", to see how much was invested at 6%.
what about "b"? well, b = 5000 - a.
ok.. hmm what's 6% of a? well, (6/100) * a or 0.06a.
what's 7.5% of b? well, (7.5/100) * b or 0.075b.
now... we know, whatever "a" and "b" are, they total the investment of 5000 bucks, thus a + b = 5000
and the interest yielded was 337.50, thus 0.06a + 0.075b = 337.50
thus
solve for "a", to see how much was invested at 6%.
what about "b"? well, b = 5000 - a.
Answer:
Modification has not reduced the number of accidents.
Step-by-step explanation:
A B C D E F G H
Before modification 5 7 6 4 8 9 8 10
After modification 3 7 7 0 4 6 8 2
Difference -2 0 1 -4 -4 -3 0 -8
Mean of differences
Standard deviation of differences =
Formula of paired t test :
Df = n-1 = 8-1 =7
t critical> t calculated
So, We failed to reject null hypothesis
Hence modification has not reduced the number of accidents.