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2 years ago

8

Ms. brown's class will raise rabbits for their spring science fair. they have 24 feet of fencing with which to build a rectangul

ar rabbit pen in which to keep the rabbits. if ms. brown's students want their rabbits to have as much room as possible, how long would each of the sides of the pen be?how long would each of the sides of the pen be if they had only 16 feet of fencing?

1 answer:

telo118 [61]2 years ago

7 0

Each side of the fence should have 6 feet if they have 24 feet and if they have 16 feet then each side will be 4 feet long.

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Suppose a simple random sample of size nequals81 is obtained from a population with mu equals 79 and sigma equals 18. (a) Descr

Daniel [21]

Answer:

(a) The sampling distribution of $\overline{X}$ = Population mean = 79

(b) P ( $\overline{X}$ greater than 81.2 ) = 0.1357

(c) P ( $\overline{X}$ less than or equals 74.4 ) = .0107

(d) P (77.6 less than $\overline{X}$ less than 83.2 ) = .7401

Step-by-step explanation:

Given -

Sample size ( n ) = 81

Population mean $(\nu)$ = 79

Standard deviation $(\sigma )$ = 18

(a) Describe the sampling distribution of $\overline{X}$

For large sample using central limit theorem

the sampling distribution of $\overline{X}$ = Population mean = 79

(b) What is Upper P ( $\overline{X}$ greater than 81.2 ) =

$P(\overline{X}> 81.2)$ = $P(\frac{\overline{X} - \nu }{\frac{\sigma }{\sqrt{n}}}> \frac{81.2 - 79}{\frac{18}{\sqrt{81}}})$

= $P(Z> 1.1)$

= $1 - P(Z< 1.1)$

= 1 - .8643 =

= 0.1357

(c) What is Upper P ( $\overline{X}$ less than or equals 74.4 ) =

$P(\overline{X}\leq 74.4)$ = $P(\frac{\overline{X} - \nu }{\frac{\sigma }{\sqrt{n}}}\leq \frac{74.4- 79}{\frac{18}{\sqrt{81}}})$

= $P(Z\leq -2.3)$

= .0107

(d) What is Upper P (77.6 less than $\overline{X}$ less than 83.2 ) =

$P(77.6< \overline{X}< 83.2)$ = $P(\frac{77.6- 79}{\frac{18}{\sqrt{81}}})< P(\frac{\overline{X} - \nu }{\frac{\sigma }{\sqrt{n}}}\leq \frac{83.2- 79}{\frac{18}{\sqrt{81}}})$

= $P(- 0.7< Z< 2.1)$

= $(Z< 2.1) - (Z< -0.7)$

= 0.9821 - .2420

= 0.7401

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2 years ago

A rectangular storage container with an open top is to have a volume of 24 cubic meters. The length of its base is twice the wid

Mariana [72]

Answer:

419.25

Step-by-step explanation:

The calculation of the cost of materials for the cheapest such container is shown below:-

We assume

Width = x

Length = 2x

Height = h

where, length = $2 \times width$

Base area = lb

= $2x^2$

Side area = 2lh + 2bh

= 2(2x)h + 2(x)h

= 4xh + 2xh

Volume = 24 which is lbh = 24

$h = \frac{24}{2x^2} \\\\ h = \frac{12}{x^2}$

Now, cost is

$= 13(2x^2) + 9(4xh + 2xh)\\\\ = 13(2x^2) + 9(4x + 2x)\times \frac{12}{x^2} \\\\ = 26x^2 + \frac{648}{x}$

now we have to minimize C(x)

So, we need to compute the C'(x)

$= 52x - \frac{648}{x^2}$

C"(x) $= 52x - \frac{1,296}{x^3}$

now for the critical points, we will solve the equation C'(x) = 0

$= 52x - \frac{648}{x^2} = 0\\\\ x = \frac{648}{52}^{\frac{1}{3}}$

$C" = ((\frac{648}{52} ^{\frac{1}{3} } = 52 + \frac{1296}{(\frac{648}{52})^\frac{1}{3} )^3}\\\\ = 52 + \frac{1296}{\frac{648}{52} } >0$

So, x is a point of minima that is

= $(\frac{648}{52} )^\frac{1}{3}$

Now, Base material cost is

$= 13(2x^2)\\\\ = 26(\frac{648}{52} )^\frac{2}{3}$

= 139.75

Side material cost is

$= \frac{648}{x} \\\\ = \frac{648}{(\frac{648}{52})^\frac{1}{3} }$

= 279.50

and finally

Total cost is

= 139.75 + 279.50

= 419.25

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