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:
g(h(x)) = [x + 3]^2 or x^2 + 6x + 9
Step-by-step explanation:
g[h(x)] signifies that h(x) is the input to g(x).
Writing out g(x) = x^2 and replacing "x" with [x + 3], we get:
g(h(x)) = [x + 3]^2 or x^2 + 6x + 9
we know that
The measure of the external angle is the semidifference of the arcs that it covers
so
∠MLN=(1/2)*(mayor arc MN-minor arc MN)
Let
x------> minor arc MN
major arc MN=360-x
substitute in the formula above
∠MLN=(1/2)*(mayor arc MN-minor arc MN)
75.86=(1/2)*(360-x-x)------> 151.72=(360-2x)------>360-151.72=2x
x=104.14°
therefore
the answer is
the measure of arc MN (minor arc) is 104.14°
Answer:
X= 5.52
Step-by-step explanation:
1. Solve the equation:
e×2x + 12= 42
x=15/e
2. Round x=15/e to the required place
x=5.52
4 *3 = 12
12 * 20 = 240
240 / 2 = 120 units^3
answer is C