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topjm [15]
2 years ago
12

2. John and Jen went to the farmers market to buy fruit. John bought 3 boxes of oranges and 9 boxes of cherries for $78. Jen bou

ght 8 boxes of oranges and 4 boxes cherries for $58. The system of equations ( shown in the pic i uploaded) can be used to represent this situation. A. Find how much does a box of oranges and a box of cherries cost ?

Mathematics
1 answer:
jenyasd209 [6]2 years ago
7 0

Answer:

Let the cost of orange be X and cost of cherry be y

3x + 9y = 78

8x + 4y = 58

Solving equation using elimination method

multiplying eq 1 with 8 and eq 2 with 3

8(3x + 9y ) = 78(8)

3( 8x + 4y) = 3(58)

24x + 72 y = 624

24x + 12y = 174

subtracting,

60 y = 450

y = 7.5

3x + 9(7.5) = 78

3x + 67.5 = 78

3x = 78 - 67.5

3x = 10.5

x = 3.5

<h2>Box of orange = $3.5 </h2><h2>Box of cherry = $7.5</h2>
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7238 rounded to the nearest whole number.

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DerKrebs [107]

Answer:

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Step-by-step explanation:

To solve this question, we need to understand the normal probability distribution and the central limit theorem.

Normal probability distribution:

Problems of normally distributed samples can be solved using the z-score formula.

In a set with mean \mu and standard deviation \sigma, the zscore of a measure X is given by:

Z = \frac{X - \mu}{\sigma}

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

Central limit theorem:

The Central Limit Theorem estabilishes that, for a random variable X, with mean \mu and standard deviation \sigma, the sample means with size n of at least 30 can be approximated to a normal distribution with mean \mu and standard deviation s = \frac{\sigma}{\sqrt{n}}

In this problem, we have that:

\mu = 48564, \sigma = 3293, n = 281, s = \frac{3293}{\sqrt{281}} = 196.44

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This is the pvalue of Z when X = 48101. So

Z = \frac{X - \mu}{\sigma}

By the Central Limit Theorem

Z = \frac{X - \mu}{s}

Z = \frac{48101 - 48564}{196.44}

Z = -2.36

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Solve by factoring <br> 2x^2=19x+33
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</span>
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