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elixir [45]
2 years ago
12

Sandy's orchard has only 4 apple trees for every 3 pear trees. She is wanting to plant enough avocado trees so that she has 2 av

ocado trees for every 5 trees.
Sandy is planting 18 avocado trees, making the ratio of avocado trees be _ to _.

PLEASE HELP!
Mathematics
1 answer:
LUCKY_DIMON [66]2 years ago
4 0
<span>current ratio: 4 apple trees for every 3 pear trees. wants: 2 avocado trees for every 5 pear trees in the orchard. Sandy is planting 18 avocado trees so 2 avocado trees for every 5 pear trees =1 avocado trees for every 2.5 pear trees therefore if we have 18 avocado trees 18 x 2.5 = 45 pear trees 4 apple trees for every 3 pear trees 4/3 = 1.333 apple trees for every 1 pear tree therefore if we have 45 avocado trees 45 x 1.333 = 60 apple trees therefore ratio of avocado to apple trees = 18 : 60 18 : 60 = 9 : 30 = 3 : 10</span>
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A particular fruit's weights are normally distributed, with a mean of 406 grams and a standard deviation of 27 grams.The heavies
Tom [10]

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}

Final Answer

Therefore , The heaviest 14% of fruits weigh more than 435.16 gms

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1 year ago
Josh went to a restaurant for dinner. The cost of the
Svetllana [295]

Answer:

17.41 Dollars

Step-by-step explanation:

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LuckyWell [14K]
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Please help solve this system of equations
stepan [7]

Make a substitution:

\begin{cases}u=2x+y\\v=2x-y\end{cases}

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}

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}

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}

\implies\sqrt[3]{\left(\dfrac uv\right)^4}=81

\implies\dfrac uv=\pm27

\implies u=\pm27v

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

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Then

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

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2 years ago
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