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Mathematics

1 answer:

NARA [144]2 years ago

5 0

Looks like you got the answer allready selected in the photo

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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

PLEASE HELP!!!

tatuchka [14]

Asked and answered elsewhere.
brainly.com/question/10629871

How do you calculate it? You start by expressing the density in the units you have. Then you multiply by appropriate conversion factors to get to the units you want. Treat units as though they were any other variable. A value cancels that appears in both the numerator and denominator of a fraction.

$\dfrac{56.15\,kg}{0.222\,hh}\cdot \dfrac{1000\,g}{1\,kg}\cdot \dfrac{1\,pw}{1.55\,g}\cdot \dfrac{1\,hh}{238.5\,L}\cdot \dfrac{9.091\,L}{1\,pk}\\\\=\dfrac{56.15\cdot 1000\cdot 9.091\,pw}{0.222\cdot 1.55\cdot 238.5\,pk}\\\\ \approx6220\,\frac{pw}{pk}$