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Eva8 [605]
3 years ago
15

A gardener has 10 pounds of soil. He used 5/8 of the soil for his garden. How many pounds of soil did he use in the Garden? How

many pounds did he have left?
Mathematics
2 answers:
olchik [2.2K]3 years ago
8 0
A gardener has 10 pounds of soil. Now he 5/8 of that soil for his garden. Let's solve how many pounds of soil did he used and how many pounds is left.
=> 5/8
=> 10 lbs / 8 = 1.25 lbs
=> 1.25 lbs * 5 = 6.25 lbs.
He used 6.25 lbs of soil for his garden and has 3.75 lbs of soil left.
aleksandrvk [35]3 years ago
3 0
  • 6.25 pounds he used.
  • 3.75 pounds he has left.
<h3>Further explanation</h3>

<u>Given:</u>

  • A gardener has 10 pounds of soil.
  • He used \frac{5}{8} of the soil for his garden.

<u>Question:</u>

  • How many pounds of soil did he use in the garden?
  • How many pounds did he have left?

<u>The Process:</u>

From the information above, let us create a suitable diagram as follows:

\boxed{\frac{10}{8}}\boxed{\frac{10}{8}}\boxed{\frac{10}{8}}\boxed{\frac{10}{8}}\boxed{\frac{10}{8}}\boxed{\frac{10}{8}}\boxed{\frac{10}{8}}\boxed{\frac{10}{8}} = 10 \ pounds

\boxed{\frac{10 \div 2}{8 \div 2} \rightarrow \frac{5}{4}}. The diagram has rewritten as follows:

\boxed{\frac{5}{4}}\boxed{\frac{5}{4}}\boxed{\frac{5}{4}}\boxed{\frac{5}{4}}\boxed{\frac{5}{4}}\boxed{\frac{5}{4}}\boxed{\frac{5}{4}}\boxed{\frac{5}{4}} = 10 \ pounds

\boxed{ \ Hence, \ 1 \ unit = \frac{5}{4} \ pounds \ }

He used \frac{5}{8} of the soil for his garden, or 5 of 8 units of 10 pounds. It means he has left 3 of 8 units.

Let us calculate how many pounds of soil did he use in the garden.

\boxed{ \ = 5 \times \frac{5}{4} \ pounds \ }

\boxed{\boxed{ \ \frac{25}{4} \ pounds \ }}

In mixed fraction:

\boxed{\boxed{ \ \frac{25}{4} = \frac{24}{4} + \frac{1}{4} = 6\frac{1}{4} \ pounds \ }}

In decimal:

\boxed{\boxed{ \ \frac{25}{4} \times \frac{25}{25} = \frac{625}{100} = 6.25 \ pounds \ }}

It could also be like this:

\boxed{\boxed{ \ 6\frac{1}{4} = 6\frac{25}{100} = 6.25 \ pounds \ }}

Let us calculate how many pounds of soil did he have left.

\boxed{ \ = 10 - \frac{25}{4} \ pounds \ }

\boxed{ \ = \frac{40}{4} - \frac{25}{4} \ pounds \ }}

\boxed{\boxed{ \ \frac{15}{4} \ pounds \ }}

In mixed fraction:

\boxed{\boxed{ \ \frac{15}{4} = \frac{12}{4} + \frac{3}{4} = 3\frac{3}{4} \ pounds \ }}

In decimal:

\boxed{\boxed{ \ \frac{15}{4} \times \frac{25}{25} = \frac{375}{100} = 3.75 \ pounds \ }}

It could also be like this:

\boxed{\boxed{ \ 3\frac{3}{4} = 3\frac{75}{100} = 3.75 \ pounds \ }}

- - - - - - - - - -

<u>Quick Steps:</u>

He used \frac{5}{8} of the soil for his garden.

\boxed{ \ = \frac{5}{8} \times 10 \ pounds \ }

\boxed{ \ = \frac{50}{8} \ pounds \ }

\boxed{ \ = \frac{50 \div 2}{8 \div 2} = \frac{25}{4} \ pounds \ }

\boxed{\boxed{ \ \frac{25}{4} = 6\frac{1}{4} \ pounds \ }}

\boxed{\boxed{ \ \frac{25}{4} \times \frac{25}{25} = \frac{625}{100} = 6.25 \ pounds \ }}

He has left \frac{3}{8} of the soil.

\boxed{ \ = \frac{3}{8} \times 10 \ pounds \ }

\boxed{ \ = \frac{30}{8} \ pounds \ }

\boxed{ \ = \frac{30 \div 2}{8 \div 2} = \frac{15}{4} \ pounds \ }

\boxed{\boxed{ \ \frac{15}{4} = 3\frac{3}{4} \ pounds \ }}

\boxed{\boxed{ \ \frac{15}{4} \times \frac{25}{25} = \frac{375}{100} = 3.75 \ pounds \ }}

<h3>Learn more</h3>
  1. How many inches of gold chain did she have left? brainly.com/question/8937330
  2. How much money Elise have after paying for the stamp? brainly.com/question/3207704
  3. How many students are wearing a shirt other than blue or white? brainly.com/question/895261

Keywords: a gardener, has, 10 pounds of soil, he used, 5/8, the soil for his garden, how many, in the garden, have left, 3/8, 6.25, 3.75

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This question was not written completely

Complete Question

At one point the average price of regular unleaded gasoline was ​$3.39 per gallon. Assume that the standard deviation price per gallon is ​$0.07 per gallon and use​ Chebyshev's inequality to answer the following.

​(a) What percentage of gasoline stations had prices within 3 standard deviations of the​ mean?

​(b) What percentage of gasoline stations had prices within 2.5 standard deviations of the​ mean? What are the gasoline prices that are within 2.5 standard deviations of the​ mean?

​(c) What is the minimum percentage of gasoline stations that had prices between ​$3.11 and ​$3.67​?

Answer:

a) 88.89% lies with 3 standard deviations of the mean

b) i) 84% lies within 2.5 standard deviations of the mean

ii) the gasoline prices that are within 2.5 standard deviations of the​ mean is $3.215 and $3.565

c) 93.75%

Step-by-step explanation:

Chebyshev's theorem is shown below.

1) Chebyshev's theorem states for any k > 1, at least 1-1/k² of the data lies within k standard deviations of the mean.

As stated, the value of k must be greater than 1.

2) At least 75% or 3/4 of the data for a set of numbers lies within 2 standard deviations of the mean. The number could be greater.μ - 2σ and μ + 2σ.

3) At least 88.89% or 8/9 of a data set lies within 3 standard deviations of the mean.μ - 3σ and μ + 3σ.

4) At least 93.75% of a data set lies within 4 standard deviations of the mean.μ - 4σ and μ + 4σ.

​

(a) What percentage of gasoline stations had prices within 3 standard deviations of the​ mean?

We solve using the first rule of the theorem

1) Chebyshev's theorem states for any k > 1, at least 1-1/k² of the data lies within k standard deviations of the mean.

As stated, the value of k must be greater than 1.

Hence, k = 3

1 - 1/k²

= 1 - 1/3²

= 1 - 1/9

= 9 - 1/ 9

= 8/9

Therefore, the percentage of gasoline stations had prices within 3 standard deviations of the​ mean is 88.89%

​(b) What percentage of gasoline stations had prices within 2.5 standard deviations of the​ mean?

We solve using the first rule of the theorem

1) Chebyshev's theorem states for any k > 1, at least 1-1/k² of the data lies within k standard deviations of the mean.

As stated, the value of k must be greater than 1.

Hence, k = 3

1 - 1/k²

= 1 - 1/2.5²

= 1 - 1/6.25

= 6.25 - 1/ 6.25

= 5.25/6.25

We convert to percentage

= 5.25/6.25 × 100%

= 0.84 × 100%

= 84 %

Therefore, the percentage of gasoline stations had prices within 2.5 standard deviations of the​ mean is 84%

What are the gasoline prices that are within 2.5 standard deviations of the​ mean?

We have from the question, the mean =$3.39

Standard deviation = 0.07

μ - 2.5σ

$3.39 - 2.5 × 0.07

= $3.215

μ + 2.5σ

$3.39 + 2.5 × 0.07

= $3.565

Therefore, the gasoline prices that are within 2.5 standard deviations of the​ mean is $3.215 and $3.565

​(c) What is the minimum percentage of gasoline stations that had prices between ​$3.11 and ​$3.67​?

the mean =$3.39

Standard deviation = 0.07

Applying the 2nd rule

2) At least 75% or 3/4 of the data for a set of numbers lies within 2 standard deviations of the mean. The number could be greater.μ - 2σ and μ + 2σ.

the mean =$3.39

Standard deviation = 0.07

μ - 2σ and μ + 2σ.

$3.39 - 2 × 0.07 = $3.25

$3.39 + 2× 0.07 = $3.53

Applying the third rule

3) At least 88.89% or 8/9 of a data set lies within 3 standard deviations of the mean.μ - 3σ and μ + 3σ.

$3.39 - 3 × 0.07 = $3.18

$3.39 + 3 × 0.07 = $3.6

Applying the 4th rule

4) At least 93.75% of a data set lies within 4 standard deviations of the mean.μ - 4σ and μ + 4σ.

$3.39 - 4 × 0.07 = $3.11

$3.39 + 4 × 0.07 = $3.67

Therefore, from the above calculation we can see that the minimum percentage of gasoline stations that had prices between ​$3.11 and ​$3.67​ corresponds to at least 93.75% of a data set because it lies within 4 standard deviations of the mean.

4 0
4 years ago
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