Answer:

Step-by-step explanation:
Assuming this complete question:
"Suppose a certain species of fawns between 1 and 5 months old have a body weight that is approximately normally distributed with mean
kilograms and standard deviation
kilograms. Let x be the weight of a fawn in kilograms. Convert the following z interval to a x interval.
"
Previous concepts
Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".
The Z-score is "a numerical measurement used in statistics of a value's relationship to the mean (average) of a group of values, measured in terms of standard deviations from the mean".
Solution to the problem
Let X the random variable that represent the weights of a population, and for this case we know the distribution for X is given by:
Where
and 
And the best way to solve this problem is using the normal standard distribution and the z score given by:

We know that the Z scale and the normal distribution are equivalent since the Z scales is a linear transformation of the normal distribution.
We can convert the corresponding z score for x=42.6 like this:

So then the corresponding z scale would be:

Answer:
the answer would be 1/15, or approximately 0.067.
Step-by-step explanation:
Answer:
$8.33
Step-by-step explanation:
24.99 ÷ 3 = 8.33
Answer:
See below
Step-by-step explanation:
It could be a positive square root l like √10 ( the number not being a perfect square).
He would have obtained this value from the application of the Pythagoras theorem. For example the length and width of the rectangle might have been 3 and 1 foot respectively, so the diagonal would have length √(3^2 + 1^2) = √10.
He could give an estimate of the length to nearest hundredth using his calculator. This would be 3.16 feet.
Answer: C. essays about work experience
Step-by-step explanation: I would chose C because when you apply for a job you are asked to write and an essay on your work experience. But in college, you are no applying for a job.