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.
Hello :
<span>7x-2y=1 ...(1)
2y=x-1...(2)
</span><span>substitution 2y in (1) :
</span>7x-(x-1) = 1
7x-x+1 = 1
7x = 0
x=0
in (2) : 2y = 0-1
2y = -1 y= -1/2
the <span>system have one solution : (0 ;-1/2)</span>
Okay 2
Really
4
During
32
Square
67
Is means =
less than is subtract
1/2a -8=2
if you need to know how to do it let me know
the answer is 5 pi/6.........