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Vitek1552 [10]
3 years ago
9

Jane is planting a mixture of flowers. For every ounce of Marigold seeds she uses 2/5 of an ounce of lupine seeds how many ounce

s of Marigold seeds did Jane used if she used one ounce of lupine seeds
Mathematics
1 answer:
dusya [7]3 years ago
6 0

Answer:

\frac{5}{2} ounces of marigold seeds.

Step-by-step explanation:

We have been given that Jane uses 2/5 of an ounce of lupine seeds for every ounce of Marigold seeds. We are asked to find the ounces of Marigold seeds, which Jane will use, if she used one ounce of lupine seeds.  

We will use proportions to solve our given problem.  

\frac{\text{Marigold seeds}}{\text{Lupine seeds}}=\frac{1}{\frac{2}{5}}

Let us simplify our proportion as shown below:

\frac{\text{Marigold seeds}}{\text{Lupine seeds}}=\frac{1\cdot 5}{2}

\frac{\text{Marigold seeds}}{\text{Lupine seeds}}=\frac{5}{2}

Since Jane used 1 pounce of lupine seeds, so we will substitute this value in our proportion as:

\frac{\text{Marigold seeds}}{1}=\frac{5}{2}

\text{Marigold seeds}}=\frac{5}{2}

Therefore, Jane will use \frac{5}{2} ounces of marigold seeds, if she used one ounce of lupine seeds.

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A delivery truck is transporting boxes of two sizes: large and small. The combined weight of a large box and a small box is 60 p
boyakko [2]

Answer:

  • large: 40 lbs
  • small: 20 lbs

Step-by-step explanation:

A system of equations can be written for the weights of the boxes based on the relationships given in the problem statement. One equation will be for the total weight of 1 large and 1 small box; the other will be for the total weight of 70 large and 60 small boxes.

Let L and S represent the weights of Large and Small boxes, respectively. The system of equations is ...

  L + S = 60 . . . . . . combined weight is 60 lbs

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We can solve this by substituting for s in the second equation.

  70L +60(60 -L) = 4000

  10L = 400 . . . . . . . . . subtract 3600, simplify

  L = 40

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A large box weighs 40 pounds; a small box weighs 20 pounds.

7 0
2 years ago
A man invests a total of $9,493in two savings accounts. One account yields 9% simple interest and the other 10% simple interest.
Tju [1.3M]

$ 1850 was invested in 9% account

<u>Solution:</u>

Given that  

Total amount invested by man in two saving accounts = $9493

Simple interest on one account =9%

Simple interest on second account = 10%

Total interest earned = $930.80

Need to determine amount invested in 9 % account.

Let assume amount invested in account where Simple Interest is 9% = x

And assume amount invested in account where Simple Interest is 10% = y

As total amount invested in two accounts is $9493

=> x + y = 9493      

=> y = 9493 - x                              ------(1)

\text { Simple Interest }=\frac{\text { Amount Invested } \times \text {rate of interest } \times \text {time}}{100}

\begin{array}{l}{\text { Simple interest when rate of interest is } 9 \%=\frac{x \times 9 \times 1}{100}=\frac{9 x}{100}} \\\\ {\text { Simple interest when rate of interest is } 10 \%=\frac{y \times 10 \times 1}{100}=\frac{10 y}{100}}\end{array}

As total interest earned = $930.80

\begin{array}{l}{\Rightarrow \frac{9 x}{100}+\frac{10 y}{100}=930.80} \\\\ {\Rightarrow 9 x+10 y=930.80 \times 100} \\\\ {\Rightarrow 9 x+10 y=93080}\end{array}

On substituting value of y from equation(1) in above equation , we get

9x + 10 (9493 – x) = 93080

=> 9x  + 94930 – 10x = 93080

=> -x = 93080 – 94930

=> -x = -1850

=> x = 1850  

Amount invested in account where Simple Interest is 9% = x  = $1850

Hence $1850 was invested in 9% account.

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