Answer:
Different traits
Step-by-step explanation:
A mutation in one's gene means something is going to be altered in the genetic code of an organism's DNA, so traits will be different because of that. They can be good or bad, but it depends really...
Y - 12 = -10
The answer is y = 2
![\bf Step-by-step~explanation:](https://tex.z-dn.net/?f=%5Cbf%20Step-by-step~explanation%3A)
![\bf Step~1:](https://tex.z-dn.net/?f=%5Cbf%20Step~1%3A)
To solve this, we have to simplify these numbers by adding like terms.
![42x^2+34-8-12\\\\(34-8=26, 26-12=14)\\\\42x^2+14\\](https://tex.z-dn.net/?f=42x%5E2%2B34-8-12%5C%5C%5C%5C%2834-8%3D26%2C%2026-12%3D14%29%5C%5C%5C%5C42x%5E2%2B14%5C%5C)
![\bf Step~2:](https://tex.z-dn.net/?f=%5Cbf%20Step~2%3A)
Next, we factor our expression. To do so, we look for the greatest common factor of 42 and 14, which is 14. We know that because 42 divided by 14 is 3.
![\frac{42}{14} =3](https://tex.z-dn.net/?f=%5Cfrac%7B42%7D%7B14%7D%20%3D3)
![\boxed{ \bf Our~final~answer: 14(3x^2 + 1)}](https://tex.z-dn.net/?f=%5Cboxed%7B%20%5Cbf%20Our~final~answer%3A%2014%283x%5E2%20%2B%201%29%7D)
From the question we are told that:
Sample size ![n=40](https://tex.z-dn.net/?f=n%3D40)
Generally Number of class (K) is Mathematically given as
![2^k \geq 40](https://tex.z-dn.net/?f=2%5Ek%20%5Cgeq%2040)
Therefore
![2^k \geq 2^6](https://tex.z-dn.net/?f=2%5Ek%20%5Cgeq%202%5E6)
![k=6](https://tex.z-dn.net/?f=k%3D6)
Generally with class interval of k
Class Width is mathematically given as
![W=\frac{highest\ score-lowest\ score}{k}](https://tex.z-dn.net/?f=W%3D%5Cfrac%7Bhighest%5C%20score-lowest%5C%20score%7D%7Bk%7D)
![W=\frac{97-42}{6}](https://tex.z-dn.net/?f=W%3D%5Cfrac%7B97-42%7D%7B6%7D)
![W=10](https://tex.z-dn.net/?f=W%3D10)
The complete Table is attached below
2)
The complete Image of the Histogram is attached below
Recall that the area of an equilateral triangle with side length
is
.
In the
plane, the base is given by two equations:
![x^2+y^2=9\implies y=\pm\sqrt{9-x^2}](https://tex.z-dn.net/?f=x%5E2%2By%5E2%3D9%5Cimplies%20y%3D%5Cpm%5Csqrt%7B9-x%5E2%7D)
so that for any given
, the vertical distance between the two sides of the circle is
![\sqrt{9-x^2}-\left(-\sqrt{9-x^2}\right)=2\sqrt{9-x^2}](https://tex.z-dn.net/?f=%5Csqrt%7B9-x%5E2%7D-%5Cleft%28-%5Csqrt%7B9-x%5E2%7D%5Cright%29%3D2%5Csqrt%7B9-x%5E2%7D)
and this is the side of length of each triangular cross-section for each
. Then the area of each cross-section is
![\dfrac{\sqrt3}4(2\sqrt{9-x^2})^2=\sqrt3(9-x^2)](https://tex.z-dn.net/?f=%5Cdfrac%7B%5Csqrt3%7D4%282%5Csqrt%7B9-x%5E2%7D%29%5E2%3D%5Csqrt3%289-x%5E2%29)
and the volume of the solid is
![\displaystyle\int_{-3}^3\sqrt3(9-x^2)\,\mathrm dx=\boxed{36\sqrt3}](https://tex.z-dn.net/?f=%5Cdisplaystyle%5Cint_%7B-3%7D%5E3%5Csqrt3%289-x%5E2%29%5C%2C%5Cmathrm%20dx%3D%5Cboxed%7B36%5Csqrt3%7D)