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Alborosie
3 years ago
7

Take away 1/5 of 50% =

Mathematics
1 answer:
zaharov [31]3 years ago
5 0
30%,
1-5=20%
so 50%-20%= 30%
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For 5 through 8, write a statistical question that could be used to gather data on each topic. 5. Distances members of the track
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Step-by-step explanation:

We must ask a question that your answer corresponds to each of the topics in the statement, therefore we will review each one:

5. Distances members of the track team jogged last week

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The equation C = 5 9 F − 160 9 gives the relation between temperature readings in Celsius and Fahrenheit.
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Step-by-step explanation:

The question is wrong. The correct equation is :

C=\frac{5}{9}F-\frac{160}{9}

We know that the equation gives the relation  between temperature readings in Celsius and Fahrenheit.

Therefore, giving that we know the value in Fahrenheit ''F'' we can find the reading in Celsius ''C''. This define a function C(F) that depends of the variable ''F''.

So for the incise (a) we answer Yes, C is a function of F.

For (b) we need to find the mathematical domain of this function. Giving that we haven't got any mathematical restriction, the mathematical domain of the function are all real numbers.

Dom (C) = ( - ∞ , + ∞)

For (c) we know that the water in liquid state and at normal atmospheric pressure exists between 0 and 100 Celsius.

Therefore the range will be

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Now, we need to find the domain for this range. We do this by equaliting and finding the value for the variable ''F'' :

For C = 0 :

0=\frac{5}{9}F-\frac{160}{9} ⇒ F=32

And for C = 100 :

100=\frac{5}{9}F-\frac{160}{9} ⇒ F=212

Therefore, the domain as relating temperatures of water in its liquid state is

Dom (C) = (32,212)

For (d) we only need to replace in the equation by F=71 and find the value of C ⇒

C=\frac{5}{9}F-\frac{160}{9} ⇒

C=(\frac{5}{9})(71)-\frac{160}{9}

C=\frac{65}{3} ≅ 21.67

C(71) = 21.67 °C

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