In a normal distribution, the median is the same as the mean (25.3). The first quartile is the value of

such that
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You have
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For the standard normal distribution, the first quartile is about

, and by symmetry the third quartile would be

. In terms of the MCAT score distribution, these values are


The interquartile range (IQR) is just the difference between the two quartiles, so the IQR is about 8.8.
The central 80% of the scores have z-scores

such that

That leaves 10% on either side of this range, which means

You have

Converting to MCAT scores,


So the interval that contains the central 80% is

(give or take).
Answer:
5.15
Step-by-step explanation:
well, let's keep in mind that the SAS postulate, so if one Side and the Angle next to it and the following Side after the angle are equal on both triangles, both triangles are congruent. Now, we have the angle 30° with sides and 9 and 2x and sides 9 and x + 4, well, the 9's are equal, dohh, you know, if only the 2x = x + 4, we'd be golden

You would first do 2 multiplied by 2x and that would give you 4x, then you would do 2 multiplied by 3y which would give you 6y, you would put 4x under 2x and put 6y under 3y and bring down your addition symbol
The distance around a rectangle or a square is as you might remember called the perimeter. The distance around a circle on the other hand is called the circumference. It is calculated as follows:
C = 2pi(r)
From this, we can solve for r and evetually solve for the diameter.
44 = 2pi(r)
r = 7.00
d = 2r
d = 14
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