Answer:
A = 1/625 or 0.0016
Step-by-step explanation:
Answer:
336x + 624
Step-by-step explanation:
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!
D)y=6(x+1)²-16= 6(x²+2x+1)-16= 6x²+12x+6-16= 6x²+12x-10
<span>answer: D</span>
9w+45= amount of money she’ll have after any amount of weeks.
So, since W is weeks... and you have 9 weeks, you do (9x9)+45.
Then you get 81+45, which is 126. She will have $126 after 9 weeks