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:
The non-equivalent ratio is 4:5.
Step-by-step explanation:
2/3 4/6 and 8/12 are equivalent because they are all multiples of each other.
Answer:
False solution; [1⅐, -3 3⁄7]
Step-by-step explanation:
{x - 2y = 8
{4x - y = 8
-¼[4x - y = 8]
{x - 2y = 8
{-x + ¼y = -2 >> New Equation
____________
-1¾y = 6
y = -3 3⁄7 [Plug this back into both equations to get the x-coordinate of 1⅐]; 1⅐ = x
I am joyous to assist you anytime.
3) 1/1 2/0 -2/0 good luck i hope this helped
for the formula is
Y = starting population x (1-rate of change)^number of years
2010-2001 = 9
Y= 567 x (1-0.015)^9 = 494.89, round up to 495 people