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].
Answer:
$46,053
Step-by-step explanation:
- Convert 29% into a decimal, which is 0.29.
- Multiply $35,700 by 29% as a decimal (0.29), which is 10,353.
- Since the keyword is more, we want to add $35,700 and 10,353 (29% of $35,700). The answer will be $46,053.
Answer:
20 believe me i got it right if not srry
Step-by-step explanation:
Answer:
y=4
Step-by-step explanation:
hello :
Y-3 (4-2Y) = 16
y-12+6y = 16
7y-12=16
so :7y =12+16
7y=28
y=28/7 =4
Answer:
2.02/2%
Step-by-step explanation:
Divide