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Fed [463]
3 years ago
11

The cost per ounce of a drink, c, varies inversely as the number of ounces, n. Six ounces of the drink costs 60 cents per ounce.

Which equation represents this relationship?

Mathematics
1 answer:
kaheart [24]3 years ago
3 0

Answer:

c = 360/n

Step-by-step explanation:

From the question, we are told that:

The cost per ounce of a drink, c, varies inversely as the number of ounces, n.

Mathematically this is expressed as:

c ∝ 1/n

Using the constant of proportionality k

c = k/n

Six ounces of the drink costs 60 cents per ounce.

c =60 cents per ounce

n = 6 ounces

Mathematically we have

c = k/n

60 = k/6

Cross multiply to find the constant k

k = 60 × 6

k = constant of proportionality = 360

Therefore, using the mathematically expression "c = k/n" the equation that represents this relationship is

c = 360/n

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x^2+x+(\frac{1}{4})= 2

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x+\frac{1}{2}=\pm \sqrt{2}
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x+\frac{1}{2}=- \sqrt{2}
x=-\frac{1}{2}- \sqrt{2} <=

AND

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6 0
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A shoe manufacturer was investigating the weights of men's soccer cleats. He felt that the weight of these cleats was less than
aleksklad [387]

Answer:

 The conclusion is that the researcher was correct

Step-by-step explanation:

From the question we are told that

     The sample size is  n =  13

    The sample mean is  \= x =  9.63

     The standard deviation is  s =  0.585

      The significance level is  \alpha  =  0.05

The Null Hypothesis is  H_o :  \mu = 0

The Alternative  Hypothesis  is  H_a =  \mu < 10

The test statistic is  mathematically represented as

          t =  \frac{\= x - \mu }{\frac{s}{\sqrt{n} } }

Substituting values

          t =  \frac{9.63  - 10 }{\frac{0.585}{\sqrt{13} } }

         t =  -  2.280

Now the critical value for \alpha is  

     t_{\alpha } = 1.645

This obtained from the critical value table

  So comparing the critical value of alpha and the test value we see that the test value is less than the critical value so the Null Hypothesis is rejected

 The conclusion is that the researcher was correct

 

 

6 0
4 years ago
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