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 area of the entire circle is pi*(7)^2, or 49pi.
Due to the 90 degree angle, the area of the larger blue sector is
90
------- * 49 pi = 49/4 units^2
360
The area of the smaller blue sector is
48
------- 49 pi = (2/15)(49 pi) = 98 pi/15 units^2
360
The total area of the blue sectors is the sum of 98 pi/15 and 49 pi/4 (units^2):
59 units^2
Volume = 45
Bc
3 rows wide
5 rows long
3 rows tall
5x3=15
15x3=45
-9/15y+3/21=5/15y-14/21
Move 5/15y to the other side. Sign changes from +5/15y to -5/15y
-9/15y-5/15y+3/21=5/15y-5/15y-14/21
-14/15y+3/21=-14/21
Move 3/21 to the other side. Sign changes from +3/21 to -3/21.
-14/15y+3/21-3/21=-14/21-3/21
-14/15y=-14/21-3/21
-14/21-3/21=-17/21
-14/15y=-17/21
Multiply both sides by -15/14
-14/15y(-15/14)
Cross out 15 and 15, divide by 15 then becomes 1
Cross out 14 and 14, divide by 14 then becomes 1
1*1*y=y
-17/21*-15/14
Cross out 15 and 21 , divide by 3. 15/3=2, 21/3=7
17/7*5/14=85/98
Answer: c. y=85/98