Answer:
The range in which we can expect to find the middle 68% of most pregnancies is [245 days , 279 days].
Step-by-step explanation:
We are given that the lengths of pregnancies in a small rural village are normally distributed with a mean of 262 days and a standard deviation of 17 days.
Let X = <u><em>lengths of pregnancies in a small rural village</em></u>
SO, X ~ Normal(
)
Here,
= population mean = 262 days
= standard deviation = 17 days
<u>Now, the 68-95-99.7 rule states that;</u>
- 68% of the data values lies within one standard deviation points.
- 95% of the data values lies within two standard deviation points.
- 99.7% of the data values lies within three standard deviation points.
So, middle 68% of most pregnancies is represented through the range of within one standard deviation points, that is;
[
,
] = [262 - 17 , 262 + 17]
= [245 days , 279 days]
Hence, the range in which we can expect to find the middle 68% of most pregnancies is [245 days , 279 days].
(how many more needed) divided by (how many innings left)
17-6=11
4 left
11/4=2.75
needs to score 2.75 per inning (even though you can't score 0.75 point)
Answers:
- cos(2theta) = -119/169
- cos(theta) = -5/13
=====================================================
The given info is:
- sin(theta) = 12/13
- theta is in quadrant II
From that, we can use the pythagorean trig identity to find that cos(theta) = -5/13. Keep in mind that cosine is negative in quadrant II.
Now use the trig identity below to compute cos(2theta)

Other options you could use are these identities

or

Answer:
3(3*3)
Step-by-step explanation:
yes
Answer:
32 days is my final answer
Step-by-step explanation: