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Crank
3 years ago
9

tmara is making a medium length necklace. write an expression that shows how nuch it will cost tamara for the chain,pendant,and

b beads that cost $0.25 each.Then find the total cost of the necklace if tamara uses 30 beads.
Mathematics
1 answer:
erastova [34]3 years ago
8 0

Answer:

1/4(n), $7.50

Step-by-step explanation:

Expanded: cost = 0.25 * chains + 0.25 * pendants + 0.25 * beads

Simplified: c = (1/4)(total number of items)

Expression only: 1/4n where n = # chains + # pendants + # beads purchased

Total cost of 30 beads = 1/4(30) = 7.5 = $7.50

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following frequency distribution shows the daily expenditure on milk of 30 households in a locality:Daily expenditure on milk(in
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<em>Note:</em> <em>As you missed to identify what we have to find in this question. But, after a little research, I am able to find that we had to find the Mode for the data given in your question. So, I am assuming we have to calculate the the Mode. Hopefully, it would clear your concept regarding this topic.</em>

Answer:

The mode of the data = 75

Step-by-step explanation:

Lets visualize the given data in a table to show the frequency distribution:

<em>Daily expenditure on milk (in Rs)           Number of households</em>

<em>                  0-30                                                           5</em>

<em>                 30-60                                                          6        </em>

<em>                 60-90                                                          9</em>

<em>                 90-120                                                         6</em>

<em>                 120-150                                                        4</em>

Here the maximum frequency is 9.

So, modal class is <em>60-90.</em>

<em />

As the formula to calculate the mode:

Mode = l_{1} + h (\frac{f_{1}-f_{0}}{2f_{1}-f_{0}-f_{2}} )

Here, the maximum

l_{1} =60, f_{1} =9, f_{0}=6, f_{0}=6, h=30

l= is the lower limit of the class

f_{1} = is the frequency of the modal class

f_{0} = is the frequency of the previous modal class

f_{2} = is the frequency of the next previous modal class

l= is the class size

So,

Mode = l_{1} + h (\frac{f_{1}-f_{0}}{2f_{1}-f_{0}-f_{2}} )

Mode = 60 + 30 (\frac{9-6}{2(9)-6-6} )

Mode = 60 + \frac{(30)(3)}{6}

Mode = 60 + 15=75

∴ The mode of the data = 75

<em>Keywords: mode, frequency distribution</em>

<em> Learn more about mode and frequency distribution from brainly.com/question/14354368</em>

<em> #learnwithBrainly </em>

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Dennis_Churaev [7]
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We simplify the equation to the form, which is simple to understand
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Simplifying:
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Simplifying:
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We move all terms containing m to the left and all other terms to the right.
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We simplify left and right side of the equation.
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We divide both sides of the equation by 0.5 to get m.
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