The playing time X of jazz CDs has the normal distribution with mean 52 and standard deviation 7; N(52, 7). According to the 68-95-99.7 rule, what percentage of jazz CDs play between 38 and 59 minutes?

68% of jazz CDs play between 45 and 59 minutes.

Step-by-step explanation:

The correct question is: The playing time X of jazz CDs has the normal distribution with mean 52 and standard deviation 7; N(52, 7).

According to the 68-95-99.7 rule, what percentage of jazz CDs play between 45 and 59 minutes?

Let X = playing time of jazz CDs

SO, X ~ Normal()

The z-score probability distribution for the normal distribution is given by;

                                    Z  =    ~ N(0,1)

Now, according to the 68-95-99.7 rule, it is stated that;

Here, we have to find the percentage of jazz CDs play between 45 and 59 minutes;

   For 45 minutes, z-score is =    = -1

   For 59 minutes, z-score is =    = 1

This means that our data values lie within 1 standard deviation points, so it is stated that 68% of jazz CDs play between 45 and 59 minutes.

i believe the answer is 68cm

step-by-step explanation:

i just took 34 x 2 because the space between v and y is the same as y to k.

the answer is b: 18

step-by-step explanation:

use the pythagorean theorem to .

a^2+b^2=c^2

plug 18 for "r"

the 2 legs are 18 and 24

18^2+24^2=c^2

324+576=c^2

900=c^2

c=30

30-12=18

18 is the radius

: )