Answer:
Step-by-step explanation:
Hello!
X: number of absences per tutorial per student over the past 5 years(percentage)
X≈N(μ;σ²)
You have to construct a 90% to estimate the population mean of the percentage of absences per tutorial of the students over the past 5 years.
The formula for the CI is:
X[bar] ±
* 
⇒ The population standard deviation is unknown and since the distribution is approximate, I'll use the estimation of the standard deviation in place of the population parameter.
Number of Absences 13.9 16.4 12.3 13.2 8.4 4.4 10.3 8.8 4.8 10.9 15.9 9.7 4.5 11.5 5.7 10.8 9.7 8.2 10.3 12.2 10.6 16.2 15.2 1.7 11.7 11.9 10.0 12.4
X[bar]= 10.41
S= 3.71

[10.41±1.645*
]
[9.26; 11.56]
Using a confidence level of 90% you'd expect that the interval [9.26; 11.56]% contains the value of the population mean of the percentage of absences per tutorial of the students over the past 5 years.
I hope this helps!
Answer:
mmm but picture of the triangle
Step-by-step explanation:
Answer:
The statement about Rosa's calculations that is true is (0.914m)/(1yd).
Step-by-step explanation:
The ratio of meter (m) to yard (yd) is 0.914 : 1.
Therefore, the conversion factor is 0.914 m/1 yd
Well I know that c. goes with I. just because I saw so many x^3 graphs.
For a. the parent function is x^2, so 0.5x^2 should look about the same as x^2, so the answer is VI.