Answer:
Frankly, substitution is almost always a better method. Graphing is a very precise and time consuming process where you have to calculate several..
ANSWER:
The surface area of the star is 3.2700 x
square kilometres.
EXPLANATIONS:
Diameter of the star = 1.8083 x
Km.
Surface area of the star = 4
Where n is the radius of the star.
So that;
n =
= 0.90415 x 
n = 0.90415 x
Km
Thus,
Surface area = 4 x 
= 326994889
Surface area = 3.2700 x

Therefore, the surface area of the star is 3.2700 x
square kilometres.
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Answer:
y = 11.
Step-by-step explanation:
Set both of the equations equal:
5x - 9 = x² - 3x + 7
Rearrange the equation:
0 = x² - 3x - 5x + 9 + 7
Combine like terms:
0 = x² - 8x + 16
Factor:
0 = (x -4)²
Solve for x:
0 = x - 4
x = 4.
Plug this into an equation to solve for 'y':
y = 5(4) - 9
y = 20 - 9
y = 11.
Answer:
y=5
m<A= 25
m<E= 35
Step-by-step explanation:
since <E equals <B due to CPCTC the equation:
3y+20=10+5y
is the equation
We have a sample of 28 data points. The sample mean is 30.0 and the sample standard deviation is 2.40. The confidence level required is 98%. Then, we calculate α by:

The confidence interval for the population mean, given the sample mean μ and the sample standard deviation σ, can be calculated as:
![CI(\mu)=\lbrack x-Z_{1-\frac{\alpha}{2}}\cdot\frac{\sigma}{\sqrt[]{n}},x+Z_{1-\frac{\alpha}{2}}\cdot\frac{\sigma}{\sqrt[]{n}}\rbrack](https://tex.z-dn.net/?f=CI%28%5Cmu%29%3D%5Clbrack%20x-Z_%7B1-%5Cfrac%7B%5Calpha%7D%7B2%7D%7D%5Ccdot%5Cfrac%7B%5Csigma%7D%7B%5Csqrt%5B%5D%7Bn%7D%7D%2Cx%2BZ_%7B1-%5Cfrac%7B%5Calpha%7D%7B2%7D%7D%5Ccdot%5Cfrac%7B%5Csigma%7D%7B%5Csqrt%5B%5D%7Bn%7D%7D%5Crbrack)
Where n is the sample size, and Z is the z-score for 1 - α/2. Using the known values:
![CI(\mu)=\lbrack30.0-Z_{0.99}\cdot\frac{2.40}{\sqrt[]{28}},30.0+Z_{0.99}\cdot\frac{2.40}{\sqrt[]{28}}\rbrack](https://tex.z-dn.net/?f=CI%28%5Cmu%29%3D%5Clbrack30.0-Z_%7B0.99%7D%5Ccdot%5Cfrac%7B2.40%7D%7B%5Csqrt%5B%5D%7B28%7D%7D%2C30.0%2BZ_%7B0.99%7D%5Ccdot%5Cfrac%7B2.40%7D%7B%5Csqrt%5B%5D%7B28%7D%7D%5Crbrack)
Where (from tables):

Finally, the interval at 98% confidence level is: