Evaluate the following function :<br> f(-3) if f(x)=-8x-9

Given that a person's normal body temperature is 98.6 ° F, and according to physicians, a person's body temperature should not be more than 0.5 ° F from the normal temperature, to determine how you could use an absolute value inequality to represent the temperatures that fall outside of normal range, the following logical-mathematical reasoning must be carried out:

As long as the normal temperature is 98.6 ° F, and its variation should not be greater than 0.5 ° F in its increase or decrease, it is correct to say that the range of normal body temperatures is equal to 98.6 - 0.5 to 98.6 + 0.5, that is, it has a variability that goes from 98.1 ° F to 99.1 ° F.

Thus, the absolute value inequality of 0.5 (both subtracting and adding) determines the limits of the temperature parameter considered normal.

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6 0

3 years ago

In the past week, Joe has recorded the number of tickets that a movie theater has sold during each day from Monday to Saturday.

fiasKO [112]

Based on the information shown through the bar graph, a misrepresentation of the data is that: B. the categories along the horizontal axis are missing one of the days.

<h3>What is a bar graph?</h3>

A bar graph can be defined as a type of chart that is used for the graphical represent of data (information), especially through rectangular bars or vertical columns.

Based on the information shown through the bar graph in the image attached below, we can logically deduce that a misrepresentation of the data is that the categories along the horizontal axis are missing one of the days (Friday).

In conclusion, this bar graph misrepresented the data because Friday is missing from the categories along the horizontal axis.

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5 0

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

So

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

b)

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

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