Given
mean of 406 grams and a standard deviation of 27 grams.
Find
The heaviest 14% of fruits weigh more than how many grams?
Explanation
given
mean = 406 gms
standard deviation = 27 gms
using standard normal table ,
![\begin{gathered} P(Z>z)=14\% \\ 1-P(Zso , [tex]\begin{gathered} x=z\times\sigma+\mu \\ x=1.08\times27+406 \\ x=435.16 \end{gathered}](https://tex.z-dn.net/?f=%5Cbegin%7Bgathered%7D%20P%28Z%3Ez%29%3D14%5C%25%20%5C%5C%201-P%28Zso%20%2C%20%5Btex%5D%5Cbegin%7Bgathered%7D%20x%3Dz%5Ctimes%5Csigma%2B%5Cmu%20%5C%5C%20x%3D1.08%5Ctimes27%2B406%20%5C%5C%20x%3D435.16%20%5Cend%7Bgathered%7D)
Final Answer
Therefore , The heaviest 14% of fruits weigh more than 435.16 gms
Answer:
17.41 Dollars
Step-by-step explanation:
First, you take the discount, which is 2 dollars, so now the cost without any additional costs is 22 dollars. Next, find 8 percent of 22 by multipling 22*0.08 and then you get 1.76. Subtract that from 22 and 20.24. Then take 14 percent of 20.24 by multiplying it by 0.14, which you get 2.8336. Round that down and you get 2.83. Subtract that from that and you get 17.41.
$24.54 I hope this is the right answer
The answer is 63.3
9.5 * 3 + 8.7 * 4 = 63.3
Make a substitution:

Then the system becomes
![\begin{cases}\dfrac{2\sqrt[3]{u}}{u-v}+\dfrac{2\sqrt[3]{u}}{u+v}=\dfrac{81}{182}\\\\\dfrac{2\sqrt[3]{v}}{u-v}-\dfrac{2\sqrt[3]{v}}{u+v}=\dfrac1{182}\end{cases}](https://tex.z-dn.net/?f=%5Cbegin%7Bcases%7D%5Cdfrac%7B2%5Csqrt%5B3%5D%7Bu%7D%7D%7Bu-v%7D%2B%5Cdfrac%7B2%5Csqrt%5B3%5D%7Bu%7D%7D%7Bu%2Bv%7D%3D%5Cdfrac%7B81%7D%7B182%7D%5C%5C%5C%5C%5Cdfrac%7B2%5Csqrt%5B3%5D%7Bv%7D%7D%7Bu-v%7D-%5Cdfrac%7B2%5Csqrt%5B3%5D%7Bv%7D%7D%7Bu%2Bv%7D%3D%5Cdfrac1%7B182%7D%5Cend%7Bcases%7D)
Simplifying the equations gives
![\begin{cases}\dfrac{4\sqrt[3]{u^4}}{u^2-v^2}=\dfrac{81}{182}\\\\\dfrac{4\sqrt[3]{v^4}}{u^2-v^2}=\dfrac1{182}\end{cases}](https://tex.z-dn.net/?f=%5Cbegin%7Bcases%7D%5Cdfrac%7B4%5Csqrt%5B3%5D%7Bu%5E4%7D%7D%7Bu%5E2-v%5E2%7D%3D%5Cdfrac%7B81%7D%7B182%7D%5C%5C%5C%5C%5Cdfrac%7B4%5Csqrt%5B3%5D%7Bv%5E4%7D%7D%7Bu%5E2-v%5E2%7D%3D%5Cdfrac1%7B182%7D%5Cend%7Bcases%7D)
which is to say,
![\dfrac{4\sqrt[3]{u^4}}{u^2-v^2}=\dfrac{81\times4\sqrt[3]{v^4}}{u^2-v^2}](https://tex.z-dn.net/?f=%5Cdfrac%7B4%5Csqrt%5B3%5D%7Bu%5E4%7D%7D%7Bu%5E2-v%5E2%7D%3D%5Cdfrac%7B81%5Ctimes4%5Csqrt%5B3%5D%7Bv%5E4%7D%7D%7Bu%5E2-v%5E2%7D)
![\implies\sqrt[3]{\left(\dfrac uv\right)^4}=81](https://tex.z-dn.net/?f=%5Cimplies%5Csqrt%5B3%5D%7B%5Cleft%28%5Cdfrac%20uv%5Cright%29%5E4%7D%3D81)


Substituting this into the new system gives
![\dfrac{4\sqrt[3]{v^4}}{(\pm27v)^2-v^2}=\dfrac1{182}\implies\dfrac1{v^2}=1\implies v=\pm1](https://tex.z-dn.net/?f=%5Cdfrac%7B4%5Csqrt%5B3%5D%7Bv%5E4%7D%7D%7B%28%5Cpm27v%29%5E2-v%5E2%7D%3D%5Cdfrac1%7B182%7D%5Cimplies%5Cdfrac1%7Bv%5E2%7D%3D1%5Cimplies%20v%3D%5Cpm1)

Then

(meaning two solutions are (7, 13) and (-7, -13))