Given two equations containing x and y you are able to find two solutions x and y if they exist.
Hope this helps.
Let one integer be x, the consecutive odd integer is x+2
reciprocals are 1/x and 1/(x+2)
1/x +1/(x+2)=24/143
make the denominators the same:
(x+2)/[x(x+2)] +x/[x(x+2)]=24/143
143(2x+2)=24(x)(x+2)
286x+286=24x²+48x
24x²-238x-286=0
use the quadratic formula: x=22 this doesn't work because 22 is not an odd number.
or x=-2.1666666666 (this is not an integer)
weird.
X^2 = 73 - y^2
<span>x^2 - y^2 = 55 </span>
<span>substituting in to it </span>
<span>73 - 55 = 2y^2 </span>
<span>18 = 2y^2 </span>
<span>y = 3 </span>
<span>x^2 - 73 - 9 </span>
<span>x = 8 </span>
<span>answer is A</span>
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].