Erm I would say it comes from the same money pot (account) but not 100% sure what your tryin to ask
The median is 84.5. Feel free to ask me any other questions.
Answer:
1. 2x
2. 22
Step-by-step explanation:
question 1
f(x)-g(x)
5x-3-(3x-3)
5x-3-3x+3
5x-3x-3+3
2x+0
2x
question 2
f(x)=y
f(-4)= -5(-4)+ (-4/-2)
= 20 + 2
=22
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!