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navik [9.2K]
2 years ago
13

What are the advantages and disadvantages to representing a function as an equation instead of a graph

Mathematics
1 answer:
Serjik [45]2 years ago
3 0

Answer:

Advantages of representing a function as an equation instead of graph :

  • No chance of misinterpretation of the data as making graph involves more data as compared to data needed to represent function as an equation.
  • Drawing graphs may be inconvenient in case you need to plot fraction and decimal functions.

Disadvantages of representing a function as an equation instead of graph :

  • Visualization is easy in case of graph. Many factors such as maximum, minimum can be easily seen with the help of graph as compared to that of equation of the function.
  • With the help of curve of the graph, we can identify the nature of the function much easily than by the equation of the function.
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The law of exponent used in is _____
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A normally distributed population has mean 57,800 and standard deviation 750. Find the probability that a single randomly select
Stels [109]

Answer:

(a) Probability that a single randomly selected element X of the population is between 57,000 and 58,000 = 0.46411

(b) Probability that the mean of a sample of size 100 drawn from this population is between 57,000 and 58,000 = 0.99621

Step-by-step explanation:

We are given that a normally distributed population has mean 57,800 and standard deviation 75, i.e.; \mu = 57,800  and  \sigma = 750.

Let X = randomly selected element of the population

The z probability is given by;

           Z = \frac{X-\mu}{\sigma} ~ N(0,1)  

(a) So, P(57,000 <= X <= 58,000) = P(X <= 58,000) - P(X < 57,000)

P(X <= 58,000) = P( \frac{X-\mu}{\sigma} <= \frac{58000-57800}{750} ) = P(Z <= 0.27) = 0.60642

P(X < 57000) = P( \frac{X-\mu}{\sigma} < \frac{57000-57800}{750} ) = P(Z < -1.07) = 1 - P(Z <= 1.07)

                                                          = 1 - 0.85769 = 0.14231

Therefore, P(31 < X < 40) = 0.60642 - 0.14231 = 0.46411 .

(b) Now, we are given sample of size, n = 100

So, Mean of X, X bar = 57,800 same as before

But standard deviation of X, s = \frac{\sigma}{\sqrt{n} } = \frac{750}{\sqrt{100} } = 75

The z probability is given by;

           Z = \frac{Xbar-\mu}{\frac{\sigma}{\sqrt{n} } } ~ N(0,1)  

Now, probability that the mean of a sample of size 100 drawn from this population is between 57,000 and 58,000 = P(57,000 < X bar < 58,000)

P(57,000 <= X bar <= 58,000) = P(X bar <= 58,000) - P(X bar < 57,000)

P(X bar <= 58,000) = P( \frac{Xbar-\mu}{\frac{\sigma}{\sqrt{n} } } <= \frac{58000-57800}{\frac{750}{\sqrt{100} } } ) = P(Z <= 2.67) = 0.99621

P(X < 57000) = P( \frac{Xbar-\mu}{\frac{\sigma}{\sqrt{n} } } < \frac{57000-57800}{\frac{750}{\sqrt{100} } } ) = P(Z < -10.67) = P(Z > 10.67)

This probability is that much small that it is very close to 0

Therefore, P(57,000 < X bar < 58,000) = 0.99621 - 0 = 0.99621 .

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3 years ago
Water weighs 62.4 pounds per cubic foot. How much does the water in a tank that measures 7 ft × 4 ft × 9 in, weigh?
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Answer:

  • <u>1,310 lb</u>

Explanation:

<u>1) Calculate the volume of the water in the tank.</u>

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  • Height of the tank: H = 9 in = 9 in × 1 ft / 12 in = 0.75 ft
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<u>2) Calculate the weight of 21 ft³ of water.</u>

Since this is not a chemistry question but a math question, I will not use the fomula of density but set a proportion with one unknown:

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Solve for x:

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So, rounding to the next integer, the water in the tank weighs 1,310 pounds, when it is full.

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Answer:1=70 2=65 3=95

Step-by-step explanation:

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Angles on a straight line add up to 180°

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