Answer:
So it wants you to first find the mean (average) of the data set, then find the distance everything is from the mean (average).
Step-by-step explanation:
3+5+9+14+16+18= 65
65 divided by the amount of numbers you put in (6) = 10.83 rounded to the nearest hundreth
So now we know the mean (average) now we need to see how far away each number is from the mean (average).10.83-3 is 7.83. 10.83-5 is 5.83. 10.83-9 is 1.83. 10.83-14 is -14.17, but -14.17 would be a normal 14.17 because it wants the absolute value. 10.3-16 is -5.17 (of course keep it as a positive), and finally 10.83-18 is -7.17 as a positive. Now for the mean absolute value of the data they likely mean from 0, so it would be 10.83.
Hope it helped and wasn't too confusing :)
<h3>
Answer: 12,201.90 dollars</h3>
You may need to delete the comma if you are entering this result into a computer system.
=====================================================
Work Shown:
P = 10,000 = deposit
r = 0.04 = decimal form of 4% interest rate
n = 4 = we're compounding 4 times a year
t = 5 years
-------------
A = P*(1+r/n)^(n*t) .... compound interest formula
A = 10,000*(1+0.04/4)^(4*5)
A = 12,201.9003994797
A = 12,201.90
After 5 years, there is $12,201.90 in the account.
This is assuming you do not deposit any more money, and it also assumes that you don't take any money out during the 5 year timespan either.
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!
