Answer:
Step-by-step explanation:
Hello!
The objective is to estimate the average time a student studies per week.
A sample of 8 students was taken and the time they spent studying in one week was recorded.
4.4, 5.2, 6.4, 6.8, 7.1, 7.3, 8.3, 8.4
n= 8
X[bar]= ∑X/n= 53.9/8= 6.7375 ≅ 6.74
S²= 1/(n-1)*[∑X²-(∑X)²/n]= 1/7*[376.75-(53.9²)/8]= 1.94
S= 1.39
Assuming that the variable "weekly time a student spends studying" has a normal distribution, since the sample is small, the statistic to use to perform the estimation is the student's t, the formula for the interval is:
X[bar] ±
* (S/√n)
![t_{n-1;1-\alpha /2}= t_{7;0.975}= 2.365](https://tex.z-dn.net/?f=t_%7Bn-1%3B1-%5Calpha%20%2F2%7D%3D%20t_%7B7%3B0.975%7D%3D%202.365)
6.74 ± 2.365 * (1.36/√8)
[5.6;7.88]
Using a confidence level of 95% you'd expect that the average time a student spends studying per week is contained by the interval [5.6;7.88]
I hope this helps!
Answer:
![\frac{5}{2}](https://tex.z-dn.net/?f=%5Cfrac%7B5%7D%7B2%7D)
I hope I helped you! If my answer is correct, please mark me brainiest. Thanks!
Answer:
a
Step-by-step explanation:
To be honest I don't know❤️
Answer:
x=7
Step-by-step explanation:
4(5x-2)=2(9x+3) do everything in the parentheses
20x-8=18x+6
+8 +8
20x=18x+14
get X by itself by adding same terms
20x=18x+14
-18x -18x
2x=14 divide
/2 /2
X=7