Answer:
You did the same on both exams.
Step-by-step explanation:
To compare both the scores, we need to compute the z scores of both the exams and then compare the values. The formula for z-score is:
<u>Z = (X - μ)/σ</u>
Where X = score obtained
μ = mean score
σ = standard deviation
For Exam 1:
Z = (95 - 79)/8
= 16/8
<u>Z = 2</u>
For Exam 2:
Z = (90 - 60)/15
= 30/15
<u>Z = 2</u>
<u>The z-scores for both the tests are same hence the third option is correct i.e. </u><u>you did the same on both exams.</u>
Percent increase from 2000 to 2001 is 12% and from 2001 to 2002 is 12.6%. Thus period of 2001 to 2002 has higher increase percentage.
<u>Solution:</u>
Given that,
Total circulation of local newspaper in 2000 = 3250
Total circulation of local newspaper in 2001 = 3640
Total circulation of local newspaper in 2002 = 4100
<em><u>Finding percent of increase in the newspaper’s circulation from 2000 to 2001:</u></em>


<em><u>Finding percent of increase in the newspaper’s circulation from 2001 to 2002:</u></em>

As we can see 12% < 12.6%
So period of 2001 to 2002 has higher increase percentage.
Answer:
89°
Step-by-step explanation:
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