Answer:
Step-by-step explanation:
We want to determine a 95% confidence interval for the mean mean test score of students.
Number of sample, n = 25
Mean, u = 81.5
Standard deviation, s = 10.2
For a confidence level of 95%, the corresponding z value is 1.96. This is determined from the normal distribution table.
We will apply the formula
Confidence interval
= mean ± z score ×standard deviation/√n
It becomes
81.5 ± 1.96 × 10.2/√25
= 81.5 ± 1.96/× 2.04
= 81.5 ± 3.9984
The lower end of the confidence interval is 81.5 - 3.9984 =77.5
The upper end of the confidence interval is 81.5 + 3.9984 = 85.5
Therefore, with 95% confidence interval, the mean test score of students is between 77.5 and 85.5
Answer:
a) y = 4x + 7 (b) y = 5x + 1 (c) y = 6x + 2 (d) y = 9x + 13
As n goes onto infinity, the expression will approach . Note the 3 and 4 come from the leading coefficients of the numerator and denominator respectively. This only works because the degrees for the numerator and denominator are both the same (both are 5).
Since is not approaching 0 when n gets really large, adding on successive terms of these values will have the infinite sum diverge. After some very large n, we are effectively adding on 3/4 each time and that just makes the overall sum keep growing forever.
<h3>Answer: The sum diverges</h3>
Answer:
The answer is A.
Step-by-step explanation:
Segment BC is congruent to segment FE, not DE. This makes A incorrect.
Answer:
Step-by-step explanation:
<h3>Given</h3>
<u>The equation for simple interest:</u>
<u>And values</u>
<h3>Solution</h3>
<u>Interest amount is</u>
<u>Solving for annual interest rate</u>
- 160 = 2000*2r
- 160 = 4000r
- r = 160/4000
- r = 0.04
- 0.04*100% = 4%
Annual interest rate is 4%