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Tems11 [23]
2 years ago
10

Albert wants to buy two pomegranates and some peaches

Mathematics
1 answer:
m_a_m_a [10]2 years ago
4 0

1.19p+3.18≤10.00 can be used to determine the pounds of peaches Albert can buy.

Step-by-step explanation:

Given,

Cost of one pomegranate = $1.59

Cost of two pomegranate = 1.59*2 = $3.18

Cost of peaches per pound = $1.19

Cost of p pounds of peaches = 1.19p

Total amount of spent = $10.00

Cost of p pounds of peaches + Cost of two pomegranates ≤ Total to spend

1.19p+3.18\leq 10.00

1.19p+3.18≤10.00 can be used to determine the pounds of peaches Albert can buy.

Keywords: inequality, addition

Learn more about addition at:

  • brainly.com/question/11007026
  • brainly.com/question/11207748

#LearnwithBrainly

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There are 5 red balls, 4 blue balls, 6 yellow balls and 10 green balls in a box,
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<h2 /><h2>Here we go ~ </h2>

According to given information there are :

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<h3>1. what is the probability that the ball chosen is red ?</h3>

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\qquad \sf  \dashrightarrow \: p(red) =  \dfrac{total \: red \: balls}{total \: balls}

\qquad \sf  \dashrightarrow \: p(red) =  \dfrac{5}{5 + 4 + 6 + 10}

\qquad \sf  \dashrightarrow \: p(red) =  \dfrac{5}{25}

\qquad \sf  \dashrightarrow \: p(red) =  \dfrac{1}{5}

<h3>2. what is the probability that the ball chosen is blue ?</h3>

\qquad \sf  \dashrightarrow \: p(blue) =  \dfrac{total \: blue \: balls}{total \: balls}

\qquad \sf  \dashrightarrow \: p(blue) =  \dfrac{4}{5 + 4 + 6 + 10}

\qquad \sf  \dashrightarrow \: p(blue) =  \dfrac{4}{25}

<h3>3. what is the probability that the ball chosen is yellow ?</h3>

\qquad \sf  \dashrightarrow \: p(yellow) =  \dfrac{total \: yellow\: balls}{total \: balls}

\qquad \sf  \dashrightarrow \: p(yellow) =  \dfrac{6}{5 + 4 + 6 + 10}

\qquad \sf  \dashrightarrow \: p(yellow) =  \dfrac{6}{25}

<h3>4. what is the probability that the ball chosen is green ?</h3>

\qquad \sf  \dashrightarrow \: p(green) =  \dfrac{total \: green\: balls}{total \: balls}

\qquad \sf  \dashrightarrow \: p(green) =  \dfrac{10}{5 + 4 + 6 + 10}

\qquad \sf  \dashrightarrow \: p(green) =  \dfrac{10}{25}

\qquad \sf  \dashrightarrow \: p(green) =  \dfrac{2}{5}

<h3>5. what is the probability that the ball chosen is not green ?</h3>

\qquad \sf  \dashrightarrow \: p(not \: green) =  \dfrac{total \: non \: green\: balls}{total \: balls}

\qquad \sf  \dashrightarrow \: p(not \: green) =  \dfrac{5 + 4 + 6}{5 + 4 + 6 + 10}

\qquad \sf  \dashrightarrow \: p(not \: green) =  \dfrac{15}{25}

\qquad \sf  \dashrightarrow \: p(not \: green) =  \dfrac{3}{5}

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