Which of the following are techniques you have learned so far for solving a

In triangle ABC, AB measures 25 cm and AC measures 35 cm. The inequality < s < represents the possible third side length o

Flura [38]

we know that

The Triangle Inequality Theorem states that the sum of any 2 sides of a triangle must be greater than the measure of the third side

so

Let

s------> the length of the third side

$25+s > 35 \\ s > 35-25 \\ s> 10\ cm$

$25+35 > s \\ 60 > s \\ s< 60\ cm$

$10\ cm < s < 60\ cm$

therefore

The answer part a) is

$10\ cm < s < 60\ cm$

we know that

the perimeter of a triangle is the sum of the length sides

In this problem

$P=25+35+s\\P=60+s$

For $s> 10\ cm$

the perimeter is equal to

$P > 60+10\\ P>70\ cm$

For $s< 60\ cm$

the perimeter is equal to

$P < 60+60\\ P$

so

$70\ cm < P < 120\ cm$

therefore

the answer part b) is

$70\ cm < P < 120\ cm$

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3 years ago

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Suppose that the walking step lengths of adult males are normally distributed with a mean of 2.5 feet and a standard deviation o

Mumz [18]

Answer:

0% probability that the mean of the sample taken is less than 2.2 feet.

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 normal distributions can be solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$ , the z-score 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 p-value, we get the probability that the value of the measure is greater than X.

Central Limit Theorem

The Central Limit Theorem establishes 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 $s = \frac{\sigma}{\sqrt{n}}$ .