Even numbers can’t be simplified
Step-by-step explanation:
A better way to word this is how much more is 4/9 than 2/9 which is 2/9 more because 2/9 plus 2/9 is 4/9 and the denominator stays the same
Answer: The required ratio will be
![84:1034](https://tex.z-dn.net/?f=84%3A1034)
Step-by-step explanation:
Since we have given that
Ratio of AD to AB is 3:2
Length of AB = 30 inches
So, it becomes
![2x=30\\\\x=\frac{30}{2}=15\ inches](https://tex.z-dn.net/?f=2x%3D30%5C%5C%5C%5Cx%3D%5Cfrac%7B30%7D%7B2%7D%3D15%5C%20inches)
So, Length of AD becomes
![3x=3\times 15=45\ inches](https://tex.z-dn.net/?f=3x%3D3%5Ctimes%2015%3D45%5C%20inches)
Now, at either end , there is a semicircle.
Radius of semicircle along AB is given by
![\frac{30}{2}=15\ inches](https://tex.z-dn.net/?f=%5Cfrac%7B30%7D%7B2%7D%3D15%5C%20inches)
So, Area of semicircle along AB and CD is given by
![2\times \frac{\pi r^2}{2}\\\\=\frac{22}{7}\times 15\times 15\\\\=\frac{4950}{7}\ in^2](https://tex.z-dn.net/?f=2%5Ctimes%20%5Cfrac%7B%5Cpi%20r%5E2%7D%7B2%7D%5C%5C%5C%5C%3D%5Cfrac%7B22%7D%7B7%7D%5Ctimes%2015%5Ctimes%2015%5C%5C%5C%5C%3D%5Cfrac%7B4950%7D%7B7%7D%5C%20in%5E2)
Radius of semicircle along AD is given by
![\frac{45}{2}=22.5\ inches](https://tex.z-dn.net/?f=%5Cfrac%7B45%7D%7B2%7D%3D22.5%5C%20inches)
Area of semicircle along AD and BC is given by
![2\times \frac{1}{2}\pi r^2\\\\=\frac{22}{7}\times \frac{45}{2}\times \frac{45}{2}\\\\=\frac{445500}{28}\ in^2](https://tex.z-dn.net/?f=2%5Ctimes%20%5Cfrac%7B1%7D%7B2%7D%5Cpi%20r%5E2%5C%5C%5C%5C%3D%5Cfrac%7B22%7D%7B7%7D%5Ctimes%20%5Cfrac%7B45%7D%7B2%7D%5Ctimes%20%5Cfrac%7B45%7D%7B2%7D%5C%5C%5C%5C%3D%5Cfrac%7B445500%7D%7B28%7D%5C%20in%5E2)
And the combined area of the semicircles is given by
![\frac{4950}{7}+\frac{445500}{28}\\\\=\frac{465300}{28}\ in^2](https://tex.z-dn.net/?f=%5Cfrac%7B4950%7D%7B7%7D%2B%5Cfrac%7B445500%7D%7B28%7D%5C%5C%5C%5C%3D%5Cfrac%7B465300%7D%7B28%7D%5C%20in%5E2)
Area of rectangle is given by
![Length\times width\\\\=AD\times AB\\\\=45\times 30\\\\=1350\ in^2](https://tex.z-dn.net/?f=Length%5Ctimes%20width%5C%5C%5C%5C%3DAD%5Ctimes%20AB%5C%5C%5C%5C%3D45%5Ctimes%2030%5C%5C%5C%5C%3D1350%5C%20in%5E2)
Hence, Ratio of the area of the rectangle to the combined area of the semicircles is given by
![1350:\frac{465300}{28}\\\\=1350\times 28:465300\\\\=37800:465300\\\\=84:1034](https://tex.z-dn.net/?f=1350%3A%5Cfrac%7B465300%7D%7B28%7D%5C%5C%5C%5C%3D1350%5Ctimes%2028%3A465300%5C%5C%5C%5C%3D37800%3A465300%5C%5C%5C%5C%3D84%3A1034)
Hence, the required ratio will be
![84:1034](https://tex.z-dn.net/?f=84%3A1034)
All outcomes of three children (using G for girl and B for boy)
GGG, GGB,GBG,GBB,BGG,BGB,BBG,BBB
1) exactly two is 3 out of 8 total outcomes, or 3/8
2) all sons is one outcome, or 1/8
Answer:
hmmmmmmm
Step-by-step explanation:
nah I am good thx for the free points