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3 years ago

What is the range of f(x)? {x| -2

Mathematics

1 answer:

Lisa [10]3 years ago

5 0

Answer:

I thing it should be x-4 because I don't get the question

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The temperature dropped from 6 Celsius to -5 Celsius.How many degrees did the temperature drop?

kakasveta [241]

-11 degrees. 6+5=11, and it went down, so negative 11 degrees

7 0

3 years ago

Read 2 more answers

The heights of a certain type of tree are approximately normally distributed with a mean height p = 5 ft and a standard

arsen [322]

Answer:

A tree with a height of 6.2 ft is 3 standard deviations above the mean

Step-by-step explanation:

⇒ $1^s^t$ statement: A tree with a height of 5.4 ft is 1 standard deviation below the mean(FALSE)

an X value is found Z standard deviations from the mean mu if:

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

In this case we have: $\mu=5\ ft$ $\sigma=0.4\ ft$

We have four different values of X and we must calculate the Z-score for each

For X =5.4\ ft

$Z=\frac{X-\mu}{\sigma}\\Z=\frac{5.4-5}{0.4}=1$

Therefore, A tree with a height of 5.4 ft is 1 standard deviation above the mean.

⇒ $2^n^d$ statement:A tree with a height of 4.6 ft is 1 standard deviation above the mean. (FALSE)

For X =4.6 ft

$Z=\frac{X-\mu}{\sigma}\\Z=\frac{4.6-5}{0.4}=-1$

Therefore, a tree with a height of 4.6 ft is 1 standard deviation below the mean .

⇒ $3^r^d$ statement:A tree with a height of 5.8 ft is 2.5 standard deviations above the mean (FALSE)

For X =5.8 ft

$Z=\frac{X-\mu}{\sigma}\\Z=\frac{5.8-5}{0.4}=2$

Therefore, a tree with a height of 5.8 ft is 2 standard deviation above the mean.

⇒ $4^t^h$ statement:A tree with a height of 6.2 ft is 3 standard deviations above the mean. (TRUE)

For X =6.2\ ft

$Z=\frac{X-\mu}{\sigma}\\Z=\frac{6.2-5}{0.4}=3$

Therefore, a tree with a height of 6.2 ft is 3 standard deviations above the mean.

6 0

3 years ago

In 2011, the Institute of Medicine (IOM), a non-profit group affiliated with the Select one US National Academy of Sciences, rev

Sergio [31]

Answer:

B. Yes, there are at least 10 people with weak bones and 10 people with strong bones in each group.

Step-by-step explanation:

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 $\frac{\sigma}{\sqrt{n}}$