Write an equation of the line that passes through (3,1) and (0,10) .
1 answer:
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Answer:
The answer is -3
Step-by-step explanation:
First of all we find the difference -4-2 = -(4+2) and than we add the numbers answer is -6 .
4÷2=
And than we simplify this fraction to 2 and find the answer -3. <em>I</em><em> </em><em>w</em><em>o</em><em>u</em><em>l</em><em>d</em><em> </em><em>b</em><em>e</em><em> </em><em>g</em><em>l</em><em>a</em><em>d</em><em> </em><em>i</em><em>f</em><em> </em><em>y</em><em>o</em><em>u</em><em> </em><em>c</em><em>h</em><em>o</em><em>o</em><em>s</em><em>e</em><em> </em><em>m</em><em>y</em><em> </em><em>a</em><em>n</em><em>s</em><em>w</em><em>e</em><em>r</em><em> </em><em>t</em><em>h</em><em>e</em><em> </em><em>b</em><em>e</em><em>s</em><em>t</em><em>.</em><em> </em><em>G</em><em>o</em><em>o</em><em>d</em><em> </em><em>l</em><em>e</em><em>s</em><em>s</em><em>o</em><em>n</em><em>s</em><em> </em><em>♡</em>
Answer:
0.1% probability that the average score of a random sample of 20 students exceeds 59.5.
Step-by-step explanation:
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
and standard deviation
Normal probability distribution
Problems of normally distributed samples can be 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.
In this problem, we have that:
What is the probability that the average score of a random sample of 20 students exceeds 59.5?
This is 1 subtracted by the pvalue of Z when . So
has a pvalue of 0.9990.
So there is a 1-0.9990 = 0.001 = 0.1% probability that the average score of a random sample of 20 students exceeds 59.5.