Q. 5: A student noted the number of cars passing through a spot on a road for 100 periods each of 3 minutes and summarised it in
1 answer:
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Hi there!
The question sounds really complex, but the mathematical operations behind it is figuratively simple. Here, we are told to find 19 (18.7 rounded up is 19) percent of 1,000. To do this, we can set up a proportion to model the whole scenario.
Now, just solve for x by cross multiplying and simplifying.
Therefore, 190 people out of 1,000 people leave out pertinent information on their IRS forms. Hope this helped and have a terrific day!
The question sounds really complex, but the mathematical operations behind it is figuratively simple. Here, we are told to find 19 (18.7 rounded up is 19) percent of 1,000. To do this, we can set up a proportion to model the whole scenario.
Now, just solve for x by cross multiplying and simplifying.
Therefore, 190 people out of 1,000 people leave out pertinent information on their IRS forms. Hope this helped and have a terrific day!
Answer:
14.63% probability that a student scores between 82 and 90
Step-by-step explanation:
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.
In this problem, we have that:
What is the probability that a student scores between 82 and 90?
This is the pvalue of Z when X = 90 subtracted by the pvalue of Z when X = 82. So
X = 90
has a pvalue of 0.9649
X = 82
has a pvalue of 0.8186
0.9649 - 0.8186 = 0.1463
14.63% probability that a student scores between 82 and 90