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Arte-miy333 [17]
3 years ago
12

Pls check the attachment :) f(x)=x^5-9x^3​

Mathematics
1 answer:
Galina-37 [17]3 years ago
5 0

Answer:

As x —> negative infinity, f(x) —> negative infinity

As x —> positive infinity, f(x) —> positive infinity.

Step-by-step explanation:

y =  {x}^{5}  - 9 {x}^{3}

An odd-degree function, meaning that the graph starts from negative infinity at x —> negative infinity and positive infinity at x —> positive infinity.

As x —> negative infinity, f(x) —> negative infinity

As x —> positive infinity, f(x) —> positive infinity.

An odd-degree function is an one-to-one function so whenever x approaches positive, f(x) will also approach positive.

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$12 per hour

Step-by-step explanation:

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Solve the Quadratic Equation . Show all of your work for full credit. 4x^2− 2x − 5= 0
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Out of the thirty
labwork [276]

Answer:

She answered 76 percent of the question correct.

Step-by-step explanation:

Convert fraction (ratio) 23 / 30 Answer: 76.6%

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3 years ago
A recent study found that the average length of caterpillars was 2.8 centimeters with a
pogonyaev

Using the normal distribution, it is found that there is a 0.0436 = 4.36% probability that a randomly selected caterpillar will have a length longer than (greater than) 4.0 centimeters.

<h3>Normal Probability Distribution</h3>

The z-score of a measure X of a normally distributed variable with mean \mu and standard deviation \sigma is given by:

Z = \frac{X - \mu}{\sigma}

  • The z-score measures how many standard deviations the measure is above or below the mean.
  • Looking at the z-score table, the p-value associated with this z-score is found, which is the percentile of X.

In this problem, the mean and the standard deviation are given, respectively, by:

\mu = 2.8, \sigma = 0.7.

The probability that a randomly selected caterpillar will have a length longer than (greater than) 4.0 centimeters is <u>one subtracted by the p-value of Z when X = 4</u>, hence:

Z = \frac{X - \mu}{\sigma}

Z = \frac{4 - 2.8}{0.7}

Z = 1.71

Z = 1.71 has a p-value of 0.9564.

1 - 0.9564 = 0.0436.

0.0436 = 4.36% probability that a randomly selected caterpillar will have a length longer than (greater than) 4.0 centimeters.

More can be learned about the normal distribution at brainly.com/question/24663213

#SPJ1

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