Answer:
Step-by-step explanation:
Quadratic function is given as 
Let's find a, b and c:
Substituting (0, 6):
Now that we know the value of c, let's derive 2 system of equations we would use to solve for a and b simultaneously as follows.
Substituting (2, 16), and c = 6
=> (Equation 1)
Substituting (3, 33), and c = 6
=> (Equation 2)
Subtract equation 1 from equation 2 to solve simultaneously for a and b.
Replace a with 4 in equation 2.
The quadratic function that represents the given 3 points would be as follows:
Answer:
option 3
Step-by-step explanation:
not sure if right
so hmmm let's get the area of the whole hexagon, and then get the area of the circle inside it, then <u>subtract the area of the circle from that of the hexagon's</u>, what's leftover is what we didn't subtract, namely the shaded part.
![\textit{area of a regular polygon}\\\\ A=\cfrac{1}{4}ns^2\cot\stackrel{\stackrel{degrees}{\downarrow }}{\left( \frac{180}{n} \right)}~ \begin{cases} n=\textit{number of sides}\\ s=\textit{length of a side}\\[-0.5em] \hrulefill\\ n=\stackrel{hexagon}{6}\\ s=\frac{9}{2} \end{cases}\implies A=\cfrac{1}{4}(6)\left( \cfrac{9}{2} \right)^2 \cot\left( \cfrac{180}{6} \right)](https://tex.z-dn.net/?f=%5Ctextit%7Barea%20of%20a%20regular%20polygon%7D%5C%5C%5C%5C%20A%3D%5Ccfrac%7B1%7D%7B4%7Dns%5E2%5Ccot%5Cstackrel%7B%5Cstackrel%7Bdegrees%7D%7B%5Cdownarrow%20%7D%7D%7B%5Cleft%28%20%5Cfrac%7B180%7D%7Bn%7D%20%5Cright%29%7D~%20%5Cbegin%7Bcases%7D%20n%3D%5Ctextit%7Bnumber%20of%20sides%7D%5C%5C%20s%3D%5Ctextit%7Blength%20of%20a%20side%7D%5C%5C%5B-0.5em%5D%20%5Chrulefill%5C%5C%20n%3D%5Cstackrel%7Bhexagon%7D%7B6%7D%5C%5C%20s%3D%5Cfrac%7B9%7D%7B2%7D%20%5Cend%7Bcases%7D%5Cimplies%20A%3D%5Ccfrac%7B1%7D%7B4%7D%286%29%5Cleft%28%20%5Ccfrac%7B9%7D%7B2%7D%20%5Cright%29%5E2%20%5Ccot%5Cleft%28%20%5Ccfrac%7B180%7D%7B6%7D%20%5Cright%29)
![A=\cfrac{1}{4}(6)\cfrac{9^2}{2^2} \cot(30^o)\implies A=\cfrac{243}{8}\cot(30^o)\implies A=\cfrac{243\sqrt{3}}{8} \\\\[-0.35em] ~\dotfill\\\\ \textit{area of circle}\\\\ A=\pi r^2~~ \begin{cases} r=radius\\[-0.5em] \hrulefill\\ r=\frac{4}{5} \end{cases}\implies A=\pi \left( \cfrac{4}{5} \right)^2\implies A=\cfrac{16\pi }{25} \\\\[-0.35em] ~\dotfill](https://tex.z-dn.net/?f=A%3D%5Ccfrac%7B1%7D%7B4%7D%286%29%5Ccfrac%7B9%5E2%7D%7B2%5E2%7D%20%5Ccot%2830%5Eo%29%5Cimplies%20A%3D%5Ccfrac%7B243%7D%7B8%7D%5Ccot%2830%5Eo%29%5Cimplies%20A%3D%5Ccfrac%7B243%5Csqrt%7B3%7D%7D%7B8%7D%20%5C%5C%5C%5C%5B-0.35em%5D%20~%5Cdotfill%5C%5C%5C%5C%20%5Ctextit%7Barea%20of%20circle%7D%5C%5C%5C%5C%20A%3D%5Cpi%20r%5E2~~%20%5Cbegin%7Bcases%7D%20r%3Dradius%5C%5C%5B-0.5em%5D%20%5Chrulefill%5C%5C%20r%3D%5Cfrac%7B4%7D%7B5%7D%20%5Cend%7Bcases%7D%5Cimplies%20A%3D%5Cpi%20%5Cleft%28%20%5Ccfrac%7B4%7D%7B5%7D%20%5Cright%29%5E2%5Cimplies%20A%3D%5Ccfrac%7B16%5Cpi%20%7D%7B25%7D%20%5C%5C%5C%5C%5B-0.35em%5D%20~%5Cdotfill)
So if we think of a test tube, it looks sort of like a cylinder. This means that its cross-section would be a circle. To find out how many turns a piece of thread would make around the test tube, we need to find the circumference of the test tube, then divide the length of the string by the circumference.
Step 1) Find the circumference
C = pi x diameter
C = 3.14 x 3
C = 9.42
Step 2) Divide the length of the string by the circumference
90.42 / 9.42 = 9.5987
The string would make approximately 9.60 turns around the test tube.
Hope this helps!! :)
A. The point estimate of μ1 − μ2 is calculated using the value of x1 - x2, therefore:
μ1 − μ2 = x1 – x2 =
7.82 – 5.99
μ1 − μ2 = 1.83
B. The formula for
confidence interval is given as:
Confidence interval
= (x1 –x2) ± z σ
where z is a value
taken from the standard distribution tables at 99% confidence interval, z =
2.58
and σ is calculated
using the formula:
σ = sqrt [(σ1^2 /
n1) + (σ2^2 / n2)]
σ = sqrt [(2.35^2 /
18) + (3.17^2 / 15)]
σ = 0.988297
Going back to the
confidence interval:
Confidence interval
= 1.83 ± (2.58) (0.988297)
Confidence interval
= 1.83 ± 2.55
Confidence interval
= -0.72, 4.38
