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3 years ago

A box has a volume given by the trinomial x^3 5x^2 -24x What are the possible dimensions of the box?

2 answers:

vladimir2022 [97]3 years ago

8 0

Volume is legnth times width times height so 3 dimentions
thee teacher probabl y wants you to factor the trinomial into 3 factors so
x^3+5x^2-24x
first factor out x
x(x^2+5x-24)
then factor
find what 2 numbers multiply to get -24 and add to get 5
the numbers are -3 and 8
x^2+5x-24 factors to
(x-3)(x+8)
the factored form is
(x)(x-3)(x+8)=volume
no dimention can be ≤0 so therefor
x≤0

x-3≤0
add 3
x≤3

x+8≤0
subtract 8
x≤-8

so the values for x is x<3

so if you had the volume, let's say 42 then
x^3+5x^2-24x=42
subtract 42 from both sides
x^3+5x^2-24x-42=0
factor and set each to zero to find the dimentions

hoa [83]3 years ago

6 0

Answer:

116

Step-by-step explanation:

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2 years ago

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3 years ago

Read 2 more answers

Make r the subject of the formula <br>v = \pi \: h {}^{2}(r - \frac{h}{3})v=πh2(r−3h) <br>

mel-nik [20]

Answer:

$\boxed{r = \frac{h}{3} + \frac{v}{\pi {h}^{2} } }$

Step-by-step explanation:

$Solve \: for \: r: \\ = > v= \pi {h}^{2}(r - \frac{h}{3} ) \\ \\ v=\pi {h}^{2}(r - \frac{h}{3} )is \: equivalent \: to \: {h}^{2}\pi(r - \frac{h}{3} ) = v: \\ = > {h}^{2}\pi(r - \frac{h}{3} ) = v \\ \\ Divide \: both \: sides \: by \: \pi {h}^{2} : \\ = > r - \frac{h}{3} = \frac{v}{\pi {h}^{2} } \\ \\ Add \: \frac{h}{3} \: to \: both \: sides: \\ = > r = \frac{h}{3} + \frac{v}{\pi {h}^{2} }$

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3 years ago

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3 years ago

6.Suppose the Gallup Organization wants to estimate the population proportion of those who think there should be a law that woul

drek231 [11]

Answer:

A sample of $n = (\frac{1.96\sqrt{0.3696*0.6304}}{E})^2$ is needed, in which E is the desired margin of error, as a proportion. If we find a decimal value, we round up to the next whole number.

Step-by-step explanation:

In a sample with a number n of people surveyed with a probability of a success of $\pi$ , and a confidence level of $1-\alpha$ , we have the following confidence interval of proportions.

$\pi \pm z\sqrt{\frac{\pi(1-\pi)}{n}}$

In which

z is the zscore that has a pvalue of $1 - \frac{\alpha}{2}$ .

The margin of error is of:

$M = z\sqrt{\frac{\pi(1-\pi)}{n}}$

In a previous study of 1012 randomly chosen respondents, 374 said that there should be such a law.

This means that $n = 1012, \pi = \frac{374}{1012} = 0.3696$

95% confidence level

So $\alpha = 0.05$ , z is the value of Z that has a pvalue of $1 - \frac{0.05}{2} = 0.975$ , so $Z = 1.96$ .

How large a sample size is needed to be 95% confident with a margin of error of E?

A sample size of n is needed, and n is found when M = E.

$M = z\sqrt{\frac{\pi(1-\pi)}{n}}$

$E = 1.96\sqrt{\frac{0.3696*0.6304}{n}}$

$E\sqrt{n} = 1.96\sqrt{0.3696*0.6304}$

$\sqrt{n} = \frac{1.96\sqrt{0.3696*0.6304}}{E}$

$(\sqrt{n})^2 = (\frac{1.96\sqrt{0.3696*0.6304}}{E})^2$

$n = (\frac{1.96\sqrt{0.3696*0.6304}}{E})^2$

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