<h2>
Answer: 1/5 or 0.2
_____________________________________</h2><h3>
Isolate the variable by dividing each side by factors that don't contain the variable.
Exact Form: x = 1/5
Decimal Form:
x = 0.2</h3><h3>
______________________________________________</h3>
Hope this helps!
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<em>Only if I'm right of course...</em>
Answer:
Step-by-step explanation:
answer to the question:
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Answer:
21.759
Step-by-step explanation:
Given that :
Mean (m) = 25
Standard deviation (s) = 12.5
Sample size (n) = 40
α = 90%
The confidence interval is obtained using the relation:
Mean ± Zcritical * s/sqrt(n)
Zcritical at 90% confidence interval = 1.64
25 ± 1.64 * (12.5/sqrt(40))
Lower boundary : 25 - 1.64(1.9764235) = 21.75866546
Upper boundary : 25 + 1.64(1.9764235) = 28.24133454
(21.759, 28.241)
Hence, lower bound of confidence interval is : 21.759
Answer:
the numerical value of the correlation between percent of classes attended and grade index is r = 0.4
Step-by-step explanation:
Given the data in the question;
we know that;
the coefficient of determination is r²
while the correlation coefficient is defined as r = √(r²)
The coefficient of determination tells us the percentage of the variation in y by the corresponding variation in x.
Now, given that class attendance explained 16% of the variation in grade index among the students.
so
coefficient of determination is r² = 16%
The correlation coefficient between percent of classes attended and grade index will be;
r = √(r²)
r = √( 16% )
r = √( 0.16 )
r = 0.4
Therefore, the numerical value of the correlation between percent of classes attended and grade index is r = 0.4
