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3 years ago

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Komodo dragons are the heaviest lizards on earth. A baby Komodo dragon is 1 3/4 feet long when it hatches. It’s mother is 8 3/4

feet long. An adult male Komodo dragon is 6 times as long as the baby Komodo dragon.
How long is the male?

How much longer is the male than the mother?

1 answer:

sukhopar [10]3 years ago

4 0

Answer:

An adult male Komodo dragon is $10\frac12$ feet.

Therefore the male is $1\frac34$ feet longer than the mother.

Step-by-step explanation:

Given that,

A baby dragon is $1\frac34$ feet long.

It's mother is $8\frac34$ feet.

An adult male Komodo dragon is 6 times as long as the baby Komodo dragon.

So, an adult male Komodo dragon is =(6× $1\frac34$ ) feet

$=6\times \frac74$ feet

$=\frac61\times \frac74$ feet

= $\frac {21}2$ feet.

$=10\frac12$ feet.

$\therefore 10\frac 12- 8\frac34$

$=\frac{21}{2}-\frac{35}4$

$=\frac{21\times 2-35\times 1}{4}$

$=\frac{42-35}{4}$

$=\frac74$

$=1\frac34$

Therefore the male is $1\frac34$ feet longer than the mother.

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Answer:

a) 0.27% probability that the mean is less than 1995 milliliters.

b) 2002.3 milliliters or more will occur only 10% of the time for the sample of 100 bottles.

Step-by-step explanation:

To solve this problem, it is important 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 random variable X, with mean $\mu$ and standard deviation $\sigma$ , a large sample size can be approximated to a normal distribution with mean $\mu$ and standard deviation $s = \frac{\sigma}{\sqrt{n}}$

In this problem, we have that:

$\mu = 2000, \sigma = 18$

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$Z = -2.78$ has a pvalue of 0.0027

So 0.27% probability that the mean is less than 1995 milliliters.

B) What mean overfill or more will occur only 10% of the time for the sample of 100 bottles?

This is the value of X when Z has a pvalue of 1-0.1 = 0.9.

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