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Ne4ueva [31]
3 years ago
7

Tony buys 4kg of potatoes at £1.60 per kilogram and 2kg of onions at £1.80 per kilogram.

Mathematics
2 answers:
natali 33 [55]3 years ago
3 0

Answer:

£10

Step-by-step explanation:

4 × 1.60 = 6.40

2× 1.80 = 3.60

6.40 + 3.60 = 10.00

zhuklara [117]3 years ago
3 0

Tony buys 4kg of potatoes at £1.60 per kilogram and 2kg of onions at £1.80 per kilogram.

She pays with a £20 note.

How much change should she receive? ​

4kg of potatoes at £1.60 = 4 x 1.60 = 6.40

2kg of onions at £1.80 = 2 x 1.80 = 3.60

4kg of potatoes + 2kg of onions = 6.40 + 3.60 = 10

£20 - £10 = £10

How much change should she receive? ​£10

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Answer:

B.) two

Step-by-step explanation:

These equations have two solutions, one for solving for each variable. There is a solution for solving for x, and one for y. For both of these equations,

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4 0
2 years ago
Using the frequency table given, find the mean, median, and mode.
aalyn [17]

Answer:

Mean = 69

Mode = 66

<em />Median = 69<em />

<em />

Step-by-step explanation:

Given

The frequency table

Required

Determine the mean, median and mode

Calculating Mean

Mean = \frac{\sum fx}{\sum f}

Where

fx = product of frequency and inches

f = frequency

So;

Mean = \frac{63 * 2 + 65 * 1 + 66 * 4 + 67 * 3 + 68 * 1 + 69 * 2 + 70 * 2 + 71 * 1 + 72 * 3 + 74 * 2 + 75 * 2}{2 + 1 + 4 + 3 + 1 + 2 + 2 + 1 + 3 + 2 + 2}

Mean = \frac{1587}{23}

Mean = 69

Calculating Mode

Mode = 66

Because it highest frequency of 4

Calculating Median

Median = \frac{\sum f}{2}th\ position

Median = \frac{2 + 1 + 4 + 3 + 1 + 2 + 2 + 1 + 3 + 2 + 2}{2}th\ position

Median = \frac{23}{2}th\ position

Median = 11.5th\ position

Approximate

Median = 12th\ position

At this point we, need to get the cumulative frequency (CF)

<em>Inches ---- Frequency ---- CF</em>

<em>63 -------------2------------------2</em>

<em>65 -------------1------------------3</em>

<em>66 -------------4------------------7</em>

<em>67 -------------3------------------10</em>

<em>68 -------------1------------------11</em>

<em>69 -------------2------------------13</em>

<em>70 -------------2------------------15</em>

<em>71 -------------1------------------16</em>

<em>72 -------------3------------------19</em>

<em>74 -------------2------------------21</em>

<em>75 -------------2------------------23</em>

From the above table

Since the median fall in the 12th position, then we consider the following data

<em>69 -------------2------------------13</em>

<em>because it has a CF greater than 12</em>

<em />

<em>Hence;</em>

<em />Median = 69<em />

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