Rip van Winkle fell asleep for a very long time. When he fell asleep, his beard was 8 millimeters long, and each passing week it
grew 2 additional millimeters.
Graph the length of Rip van Winkle's beard (in millimeters) as a function of time (in weeks).
Please help me to understand how to graph this problem.
Graph the length of Rip van Winkle's beard (in millimeters) as a function of time (in weeks).
Please help me to understand how to graph this problem.
2 answers:
Answer:
Given,
The original length of the beard = 8 mm
Each week additional length of beard = 2 mm
So, the addition length after x weeks = 2x mm,
Thus, the total length of beard after x weeks ( say f(x) ) = original length + additional length
⇒ f(x) = 8 + 2x
Which is the required function,
f(x) = 8 + 2x is a line,
If f(x) = 0, x = - 4,
So, the line passes through (-4,0),
if x = 0, f(x) = 8
So, the line passes through (0,8),
Hence, we can graph the above function by joining the points (-4,0) and (0,8) in the graph. ( shown below )
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Answer:
100% probability that the sample mean scores will be between 87 and 124 points
Step-by-step explanation:
To solve this question, we have to understand the normal probability distribution and the central limit theorem.
Normal probability distribution:
Problems of normally distributed samples are solved using the z-score formula.
In a set with mean and standard deviation
, the zscore of a measure X is given by:
The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.
Central limit theorem:
The Central Limit Theorem estabilishes that, for a random variable X, with mean and standard deviation
, a large sample size can be approximated to a normal distribution with mean \mu and standard deviation, which is also called standard error
In this problem, we have that:
What is the probability that the sample mean scores will be between 87 and 124 points
This is the pvalue of Z when X = 124 subtracted by the pvalue of Z when X = 87. So
X = 124
By the Central Limit Theorem
has a pvalue of 1
X = 87
has a pvalue of 0
1 - 0 = 1
100% probability that the sample mean scores will be between 87 and 124 points