I think that it is b sorry if I’m wrong
Answer:
y= -6
Step-by-step explanation:
4y=2(y−5)−2
4y = 2y -10 -2
4y = 2y -12
2y = -12
y = -6
By definition, the area of a circle is given by:
![A = \pi * r ^ 2 ](https://tex.z-dn.net/?f=A%20%3D%20%5Cpi%20%2A%20r%20%5E%202%0A)
Where,
r: radius of the circle.
Substituting the values in the given expression, we have:
![A = 3.14 * (3) ^ 2 A = 28.26 feet ^ 2](https://tex.z-dn.net/?f=A%20%3D%203.14%20%2A%20%283%29%20%5E%202%0A%0A%20A%20%3D%2028.26%20feet%20%5E%202)
Rounding to the nearest tenth we have:
Answer:
Approximately needed:
Answer:
A yes ; it can be rotated 360° or less and match the original figure
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!