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:
20 cm
Step-by-step explanation:
We are given a trapezoid, where the length of shorter base or on of the parllel line is 16 cm and the length of other parallel side is 24 cm.
Let the two parallel sides be x and y that is x = 16 cm and y = 24 cm.
A median of a trapezoid is a line segment that divides the non parallel sides of a trapezoid equally or a line segment that passes through the mid points of non-parallel sides of a trapezoid.
The length of median of a trapezoid = 
= 
= 20 cm.
Thus, the length of median of trapezoid is 20 cm.
Answer:
Daniel can read his data and refer to line as best line of fit and estimate an average per set of hours.
Step-by-step explanation:
A line of fit draws a solid conclusion to the average for the hours spent during the amount of indicated hours. We draw a line of fit central fit and aim similar centrality as that similar results of the mean (without working out the mean we can draw a line perpendicular to the number of mean, but in line of fit we go central to all the descending or cascading results to include all results but just using one line), with one further consideration and that is balance if anything sticks out from the norm ie) weather conditions including data, we suggest if there is nothing to weigh the line of fit to a balancing outcome that shows the opposite of kilometres walked (eg. extreme higher mileage within the hour/s) then it may just alter the line a fraction of how many treks he did, but not in data less than 30 entries. Have attached an example where they classify in economics something outside the norm is called a misfit. Daniel can read his data and refer to line as best line of fit and estimate an average per set of hours. Here on the attachment you can read any misfit info and use the line coordination perpendicular to guide the indifference, the attachment shows it is not really included in the best line of fit as other dominating balances have occurred and therefore we have a misfit, all whilst using best line of fit to balance everything fairly.
Answer:
The Median should be 89.5
And the mean should be 69.75
I hope that helped
6(x+5)² + 5(x+5)-4 = 0
(x+5)(x+5) = x(x+5)+5(x+5) = x² + 5x + 5x + 25 = x² + 10x + 25
6(x²+10x+25) + 5(x+5) - 4 = 0
6x² + 60x + 150 + 5x + 25 - 4 = 0
6x² + 60x + 5x + 150 + 25 - 4 = 0
6x² + 65x + 171 = 0
<span>u = (x + 5)</span>
