Answer:
68% of jazz CDs play between 45 and 59 minutes.
Step-by-step explanation:
<u>The correct question is:</u> 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 = <u>playing time of jazz CDs</u>
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;
- 68% of the data values lie within one standard deviation points from the mean.
- 95% of the data values lie within two standard deviation points from the mean.
- 99.7% of the data values lie within three standard deviation points from the mean.
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.
log 4 ( a^5 ) - log 4 ( b^6 ) =
log 4 ( a^5 / b^6 )
T = -4.5
Answer is One solution.
Answer:
-50x+-5
Step-by-step explanation:
-5.10x=-50x
-5.1=-5
Answer:
ans :240 cm²
Step-by-step explanation:
see the pic
....ik its messy but hope u understand
and I hope the ans is correct