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3 years ago

The function f(x)=x+π is a polynomial. A. True B. False

2 answers:

7 0

A polynomial function is a function with multiple terms of a or multiple variables separated by addition. In this function you have a term of the variable x with a coefficient of one being added to a constant term of pi.

baherus [9]3 years ago

3 0

It is true that t<span>he function f(x)=x+π is a polynomial. Pi is considered to be a constant of about 3.14. </span>

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the fee for the parking garage at the airport is $16 for the first day plus $4.50 for each additional day. which equation repres

11111nata11111 [884]

It will be a. f=4.5d+16. Unlike the other equations. F represents the fee for the parking garage, and d represents the number of days.

6 0

3 years ago

What is the logarithmic function modeled by the following table? x f(x) 9 2 27 3 81 4

Nataly_w [17]

Answer:

Required logarithmic function is :

$f\left(x\right)=\log_3\left(x\right)$

Step-by-step explanation:

We have been fiven that table for the logarithmic function is:

x f(x)

9 2

27 3

81 4

which can be rewritten as:

x f(x)

3^2 2

3^3 3

3^4 4

Which are basically powers of 3

So we can use logarithmic function as

$\log_3\left(9\right)=\log_3\left(3^2\right)=2\log_3\left(3\right)=2(1)=2$

Hence required logarithmic function is :

$f\left(x\right)=\log_3\left(x\right)$

8 0

3 years ago

Simplify the expression. 4√ 6+ 5√6

pshichka [43]

Answer:

9√6

Step-by-step explanation:

4√6+5√6

=(4+5)√6

=9√6

5 0

3 years ago

Rhianna and Myra, both teachers, are adding books to their class libraries. Rhianna's classroom started out with a collection of

Hunter-Best [27]

Answer:

21 books; 5 weeks

Step-by-step explanation:

3 0

2 years ago

The average length of a field goal in the National Football League is 38.4 yards, and the s. d. is 5.4 yards. Suppose a typical

Tasya [4]

Answer:

a) The sample is larger than 30, so, by the Central Limit Theorem, the distribution of the sample means will be normally distributed with mean 38.4 and standard deviation 0.8538.

b) 0.25% probability that his average kicks is less than 36 yards

c) 0.11% probability that his average kicks is more than 41 yards

d-a) The sample is larger than 30, so, by the Central Limit Theorem, the distribution of the sample means will be normally distributed with mean 38.4 and standard deviation 1.08

d-b) 1.32% probability that his average kicks is less than 36 yards

d-c) 0.80% probability that his average kicks is more than 41 yards

Step-by-step explanation:

To solve this question, we need to understand the normal probability distribution and the central limit theorem.

Normal probability distribution

Problems of normally distributed samples are solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$ , the zscore of a measure X is given by:

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

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

Central Limit Theorem

The Central Limit Theorem estabilishes that, for a normally distributed random variable X, with mean $\mu$ and standard deviation $\sigma$ , the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean $\mu$ and standard deviation $s = \frac{\sigma}{\sqrt{n}}$ .

For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.

In this problem, we have that:

$\mu = 38.4, \sigma = 5.4, n = 40, s = \frac{5.4}{\sqrt{40}} = 0.8538$

a. What is the distribution of the sample mean? Why?

The sample is larger than 30, so, by the Central Limit Theorem, the distribution of the sample means will be normally distributed with mean 38.4 and standard deviation 0.8538.

b. What is the probability that his average kicks is less than 36 yards?

This is the pvalue of Z when X = 36. So

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

By the Central Limit Theorem

$Z = \frac{X - \mu}{s}$

$Z = \frac{36 - 38.4}{0.8538}$

$Z = -2.81$

$Z = -2.81$ has a pvalue of 0.0025

0.25% probability that his average kicks is less than 36 yards

c. What is the probability that his average kicks is more than 41 yards?

This is 1 subtracted by the pvalue of Z when X = 41. So

$Z = \frac{X - \mu}{s}$

$Z = \frac{41 - 38.4}{0.8538}$

$Z = 3.05$

$Z = 3.05$ has a pvalue of 0.9989

1 - 0.9989 = 0.0011

0.11% probability that his average kicks is more than 41 yards

d. If the sample size is 25 in the above problem, what will be your answer to part (a) , (b)and (c)?

Now $n = 25, s = \frac{5.4}{\sqrt{25}} = 1.08$

The sample is larger than 30, so, by the Central Limit Theorem, the distribution of the sample means will be normally distributed with mean 38.4 and standard deviation 1.08

$Z = \frac{X - \mu}{s}$

$Z = \frac{36 - 38.4}{1.08}$