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mixas84 [53]
3 years ago
11

1. The amount paid for different weights of oranges at a farmer’s market are shown below. Oranges (lb) Price ($) 2 3 3 4.50 5 7.

50 ? 12 (a) Is the relationship between the price and the weight of the oranges proportional? Use the first three rows of data to answer this question. Show your work and explain. How many pounds of oranges cost $12? Show your work.
due today.
Mathematics
1 answer:
torisob [31]3 years ago
8 0

Answer:

Relationship is proportional ;

8 oranges

Step-by-step explanation:

Oranges (lb) 2 ___ 3 ____5 ___?

Price ($) __ 3 ___4.50___7.50_12

For proportionality :

Oranges = k * price

Where, k = constant of proportionality

2 = 3k

k = 2 /3

k = 0.666666

Using k :

Number of oranges at price = 4.50 should be:

0.66666 * 4.50 = 3

Number of oranges at price = 7.50 should be:

0.66666 * 7.50 = 5

Hence, relationship is proportional

Number of oranges at $12

0.66666 * 12 = 8

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The Candela brothers own two pizza restaurants, one on Park Street and one on Bridge Road.
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The mean, median and mode are measures of central tendency, that is they tend to indicate the location middle of the data

Required values;

(a) The performance for the week for Park Street

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The performance for the week for Bridge Road

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The standard deviation is $<u>3250</u>

The Interquartile range is $<u>6075</u>

Reason:

The table of values that maybe used to find a solution to the question is given as follows;

\begin{array}{|l|l|l|}\mathbf{Variable} &\mathbf{Park}&\mathbf{Bridge}\\N&36&40\\Mean&6611&5989\\SE \ Mean&597&299\\StDev&3580&1794\\Minimum&800&1800\\Q_1&3600&5225\\Median&6600&6000\\Q_3&9675&7625\\Maximum&14100&8600\end{array}\right]

(a) Park Street revenue = $7,500

Bridge Road's revenue = $7,100

The two stores sold close to but below the 75th percentile

Bridge Road revenue;

The z-score is given as follows;

Z = \dfrac{x - \mu }{\sigma }

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From the Z-Table, we have;

The percentile= 0.7291

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Park Street revenue;

The z-score is given as follows;

  • Z = \dfrac{7500 - 6611}{3580} \approx 0.25

From the Z-Table, we have;

The percentile = <u>0.5987</u>

  • Therefore, the sale for the week is better than <u>59.87 %</u> of all the sales

(b) Given that the operating cost is $3,000, frim which we have;

The subtracted value is subtracted from the mean and median to find the new value

Profit = The revenue - Cost

New mean = 6611 - 3000 = 3611

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The new median = 6600 - 3000 = 3600

  • The new median = <u>$3,600</u>

The standard deviation and the interquartile range remain the same, therefore, we have;

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The interquartile range = 9675 - 3600 = 6075

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