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.
The domain is all x-values.
The domain is -3, 2, 7
The range is all y-values
The range is 5, 0, -5
Best of Luck!
Answer:
A
B
C
Step-by-step explanation:
A
11=4(7)-17
11=28-17
11=11
B
-13=4(1)-17
-13=4-17
-13=-13
C
-1=4(4)-17
-1=16-17
-1=-1
It's easy to show that is strictly increasing on . This means
and
Then the integral is bounded by