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

such that

You have

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:
X=
Step-by-step explanation:
Hello!
All lines within a circle are the same length
We first have to find the length of one of the lines.
7 + 4 = 11
Now we find what we are missing from the second line
5 + z = 11
z = 6
The answer is C)6
Hope this helps!
Sure
Answer:
x = -2
If the steps aren't belong to the lesson that you want, just tell me the lesson title so I can get the answer for you !!!!
...
The steps
Multiply all terms by (x-3)(x+3) and cancel:
x(x+3)+2x(x−3)=18
3x2−3x=18(Simplify both sides of the equation)
3x2−3x−18=18−18(Subtract 18 from both sides)
3x2−3x−18=0
3(x+2)(x−3)=0(Factor left side of equation)
x+2=0 or x−3=0(Set factors equal to 0)
x=−2 or x=3
Check answers. (Plug them in to make sure they work.)
x=−2 (Works in original equation)
x=3 (Doesn't work in original equation)
Answer:
0.636
Step-by-step explanation:
First off, we'll interpret the fraction bar to mean "divided by." This means that 7/11 is the same as 7 divided by 11.
Second, you have requested to see 2 decimal places in your answer, so we'll write the 7 as 7.00 instead (a 7 with 2 zeros after the decimal point).
Now, we'll just do what the fraction bar says: divide 7.00 by 11:
And that's about it! 7/11 written as a decimal to 3 decimal places is 0.636.