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Debora [2.8K]
4 years ago
6

Which type of graph would be most appropriate to help you organize a set of data in numerical order?

Mathematics
2 answers:
s344n2d4d5 [400]4 years ago
6 0

Answer:

The answer is - bar graph

Step-by-step explanation:

Which type of graph would be most appropriate to help you organize a set of data in numerical order?

Numerical data means measuring or counting a numerical value. So, graphs like bar graph and pie charts show quantitative data. Numerical data is quantitative data.

Here, bar graph is the option.

mixer [17]4 years ago
5 0
A bar graph or line graph. Hope I helped!!
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10. Three kinds of teas are worth $4.60 per pound, $5.75 per pound, and $6.50 per pound. They are to be
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Answer:

The mass of the $4.60/lb tea that should be used in the mixture is 10 lb

The mass of the $5.75/lb tea that should be used in the mixture is 8 lb

The mass of the $6.50/lb tea that should be used in the mixture is 2 lb

Step-by-step explanation:

The parameters of the question are;

The worth of the three teas are

Tea A = $4.60/lb

Tea B = $5.75/lb

Tea C = $6.50/lb

The mass of the mixture of the three teas = 20 lb

The worth of the mixture of the three teas = $5.25 per pound = $5.25/lb

The amount of the $4.60 in the mixture = The sum of the amount of the other two teas

Therefore, given that the mass of the mixture = 20 lb, we have in the mixture;

The mass of tea A + The mass of Tea B + The mass of Tea C = 20 lb

The mass of tea A = The mass of Tea B + The mass of Tea C

Therefore;

The mass of tea A + The mass of tea A = 20 lb

2 × The mass of tea A in the mixture = 20 lb

The mass of tea A in the mixture = 20 lb/2 = 10 lb

The mass of tea A in the mixture = 10 lb

The mass of Tea B + The mass of Tea C = The mass of tea A = 10 lb

The mass of Tea B + The mass of Tea C = 10 lb

The mass of Tea B  = 10 lb - The mass of Tea C

Where the mass of Tea C in the mixture = x, we have;

The mass of Tea B in the mixture = 10 lb - x

The cost of the 10 lb of tea A = 10 × $4.60 = $46.0

The worth of the tea mixture = 20 × $5.25 = $105

The worth of the remaining 10 lb of the mixture comprising of tea A and tea B is given as follows;

The worth of Tea B + The worth of Tea C in the mixture = $105.00 - $46.00 = $59.00

Therefore, we have;

x lb × $6.50/lb + (10 - x) lb × $5.75/lb = $59.00

x × $6.50 - x × $5.75 + $57.50 = $59.00

x × $0.75 = $59.00 -  $57.50 = $1.50

x =  $1.50/$0.75 = 2 lb

∴ The mass of Tea C in the mixture = 2 lb

The mass of Tea B in the mixture = 10 lb - x = 10 lb - 2 lb = 8 lb

The mass of Tea B in the mixture = 8 lb

Therefore, since we have;

Tea A = $4.60/lb

Tea B = $5.75/lb

Tea C = $6.50/lb

The mass of tea A in the mixture = 10 lb

The mass of tea B in the mixture = 8 lb

The mass of tea C in the mixture = 2 lb, we find;

The mass of the $4.60/lb tea that should be used in the mixture = 10 lb

The mass of the $5.75/lb tea that should be used in the mixture = 8 lb

The mass of the $6.50/lb tea that should be used in the mixture = 2 lb.

6 0
3 years ago
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