Answer:
a) and d)
Step-by-step explanation:
Given in the question, 4 equations.
We will find the solution of all one by one to see which equation have a different solution.
a)
y - 5 = -7
y = -7 + 5
y = -2
b)
-y + 4 = 5
-y = 5 - 4
-y = 1
y = -1
C)
(y/-1) – 6 = -5
-y/1 =-5 + 6
-y = 1
y = -1
D)
(y/3) – 5 = -5
y/3 = -5 +5
y/3 = 0
Answer:
The volume of the sphere is ![V=288\pi\ m^3](https://tex.z-dn.net/?f=V%3D288%5Cpi%5C%20m%5E3)
Step-by-step explanation:
<u><em>The question in English is</em></u>
Calculate the volume in m^3 of the sphere in which the area of one of its maximum circles is 36pi m^2
we know that
The radius of the maximum circle in the sphere is equal to the radius of the sphere
Step 1
Find the radius of the maximum circle
The area of the circle is
![A=\pi r^{2}](https://tex.z-dn.net/?f=A%3D%5Cpi%20r%5E%7B2%7D)
we have
![A=36\pi\ m^2](https://tex.z-dn.net/?f=A%3D36%5Cpi%5C%20m%5E2)
substitute and solve for r
![36\pi=\pi r^{2}](https://tex.z-dn.net/?f=36%5Cpi%3D%5Cpi%20r%5E%7B2%7D)
Simplify
![36=r^{2}](https://tex.z-dn.net/?f=36%3Dr%5E%7B2%7D)
take the square root both sides
![r=6\ m](https://tex.z-dn.net/?f=r%3D6%5C%20m)
Step 2
Find the volume of the sphere
The volume of the sphere is
![V=\frac{4}{3}\pi r^{3}](https://tex.z-dn.net/?f=V%3D%5Cfrac%7B4%7D%7B3%7D%5Cpi%20r%5E%7B3%7D)
substitute the value of r
![V=\frac{4}{3}\pi (6)^{3}](https://tex.z-dn.net/?f=V%3D%5Cfrac%7B4%7D%7B3%7D%5Cpi%20%286%29%5E%7B3%7D)
![V=288\pi\ m^3](https://tex.z-dn.net/?f=V%3D288%5Cpi%5C%20m%5E3)
The greatest common factor of 14x-24 is 2. So it would be 7x - 12
Answer:
The required confidence inteval = 94.9%.
Step-by-step explanation:
Confidence interval: Mean ± Margin of error
Given: A confidence interval for the true mean diameter of all oak trees in the neighbourhood is calculated to be (36.191, 42.969).
i.e. Mean + Margin of error = 42.969 (i)
Mean - Margin of error = 36.191 (ii)
Adding (i) and (ii), we get
![2Mean =79.16\\\\\Rightarrow\ Mean= 39.58](https://tex.z-dn.net/?f=2Mean%20%3D79.16%5C%5C%5C%5C%5CRightarrow%5C%20Mean%3D%2039.58)
Margin of error = 42.969-39.58 [from (i)]
= 3.389
Margin of error = ![t^* \dfrac{\sigma}{\sqrt{n}}](https://tex.z-dn.net/?f=t%5E%2A%20%5Cdfrac%7B%5Csigma%7D%7B%5Csqrt%7Bn%7D%7D)
here n= 25 ![, \ \sigma=8.25](https://tex.z-dn.net/?f=%2C%20%5C%20%5Csigma%3D8.25)
i.e.
![3.389=t^*\dfrac{8.25}{5}\\\\\Rightarrow\ t^* = \dfrac{3.389}{1.65}\\\\\Rightarrow\ t^* =2.0539 \](https://tex.z-dn.net/?f=3.389%3Dt%5E%2A%5Cdfrac%7B8.25%7D%7B5%7D%5C%5C%5C%5C%5CRightarrow%5C%20t%5E%2A%20%3D%20%5Cdfrac%7B3.389%7D%7B1.65%7D%5C%5C%5C%5C%5CRightarrow%5C%20t%5E%2A%20%3D2.0539%20%5C)
Using excel function 1-TDIST.2T(2.054,24)
The required confidence inteval = 94.9%.