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ira [324]
4 years ago
12

Scores on an exam follow an approximately normal distribution with a mean of 76.4 and a standard deviation of 6.1 points. what i

s the minimum score you would need to be in the top 5%?
Mathematics
1 answer:
Helga [31]4 years ago
5 0
The top 5% of scores belong to the 95th percentile, which means the cutoff score k is such that

\mathbb P(X

Transforming to the standard normal distribution, you have

\mathbb P(X

A left-tail probability of 95% corresponds to a z-score of about k^*=1.6449, which means the cutoff score must be around

1.6449=\dfrac{k-76.4}{6.1}\implies k\approx86.4
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Solve for m 1/C+1/m=1/z
9966 [12]
Answer:  " m = zC / (C − z) " .
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Explanation:
_________________________
Given:  1/C + 1/m = 1/z ;  Solve for "m".

Subtract  "1/C" from each side of the equation:
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1/C + 1/m − 1/C = 1/z − 1/C  ;

to get:  1/m = 1/z − 1/C ;
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Now, multiply the ENTIRE EQUATION (both sides); by "(mzC"); to get ride of the fractions:
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mzC {1/m = 1/z − 1/C} ;

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Factor out an "m" on the "right-hand side" of the equation:

zC = m(C − z) ;  Divide EACH side of the equation by "(C − z)" ; to isolate "m" on one side of the equation;

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zC/ (C − z) = m ;   ↔   m = zC/ (C − z) .
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3 years ago
Jada has 60% of her goal of $70 saved for her trip.
Amiraneli [1.4K]

Answer:

60% of $70 is $42, so i think its $42

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3 years ago
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