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.
Answer:
Step-by-step explanation:
Probability of success =
(number of possibilities that would be considered successful)
divided by
(total number of different possible outcomes) .
Total number of different possible outcomes =
total number of different possible values of the last 2 digits
= the quantity of numbers from 00 to 99
= 100 of them .
The number of possibilities that will match your age = only 1 .
Probability that the last 2 digits match your age = 1/100 = 0.01 = 1 % .
When you have an equation that has variables and you know the value of those variables, just plug it it. So plugging the vaules we get
Just evaluating we get 4
2040 = n/2(20+(n-1)4)
4080 = n(20+4n-4)
4080= 20n +4n^2 -4n
1020 = 4n + n^2
n^2 +4n -1020 =0
use common formula (can't write out so just look at answers. sorry)
which gives answers of n=-34 and n=30. since n can only be positive, n=30 so there are 30 rows. I liked that challenge