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3 years ago

F(x)=x^2-x^3 what's the value of f(-2)

Mathematics

2 answers:

Anvisha [2.4K]3 years ago

4 0

(-2)^2-(-2)^3

=4-(-8)

=4+8

=12

Vlada [557]3 years ago

3 0

Answer:

Step-by-step explanation:

f(-2) = (-2)^2 - (-2)^3

= 4 - (-8)

= 4 + 12

= 16

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Researchers are interested in the effect of a certain nutrient on the growth rate of plant seeding. Use a hydroponics grow proce

Rashid [163]

Answer:

The 95% confidence interval for the mean would be given by (56.604;68.596)

The 99% confidence interval for the mean would be given by (53.196;72.004)

Step-by-step explanation:

1) Previous concepts

When we compute a confidence interval for the mean, we are interested on the parameter population mean, and we use the info from the sample to estimate this parameter.

2) Basic operations

The sample mean can be calculated with the following formula

$\sum_{i=1}^n \frac{x_i}{n}$

Using excel we can use this function to calculate the mean:

=AVERAGE(54.2,59.8,61.8,63.3,65.1,71.4)

The value obtained is $\bar X=62.6$

In order to find the sample deviation we can use this formula

$s=\sqrt{\sum_{i=1}^n \frac{(x_i-\bar X)^2}{n-1}}$

And using excel we can use this function to calculate the sample standard deviation:

=STDEV.S(54.2,59.8,61.8,63.3,65.1,71.4)

The value obtained is $s=5.713$

The sample size for this case is n=6, n<30 so then is better use the t distribution to calculate the margin of error. First we need to calculate the degrees of freedom, on this case $df=n-1=6-1=5$

The formula for the confidence interval would be given by:

$\bar X \pm t_{\alpha/2}\frac{s}{\sqrt{n}}$ (1)

3) Part a

If we want a 95% or 0.95 of confidence, then the value for the signficance is $\alpha=1-0.95=0.05$ , and $\alpha/2=0.025$ , and $1-\frac{\alpha}{2}=0.975$ so we can find the critical t value with the following formula in excel:

=T.INV(0.975,5)

And we got $t_{\alpha/2}= 2.571$

And we can replace into equation (1) and we got:

$62.6 \pm 2.571\frac{5.713}{\sqrt{6}}$

And using excel with the following formulas we got:

=62.6-2.571*(5.713/SQRT(6)) = 56.604

=62.6+2.571*(5.713/SQRT(6)) = 68.596

So the 95% confidence interval for the mean would be given by (56.604;68.596)

4) Part b

If we want a 99% or 0.99 of confidence, then the value for the signficance is $\alpha=1-0.99=0.01$ , and $\alpha/2=0.005$ , and $1-\frac{\alpha}{2}=0.995$ so we can find the critical t value with the following formula in excel:

=T.INV(0.995,5)

And we got $t_{\alpha/2}= 4.032$

And we can replace into equation (1) and we got:

$62.6 \pm 4.032\frac{5.713}{\sqrt{6}}$

And using excel with the following formulas we got:

=62.6-4.032*(5.713/SQRT(6)) = 53.196

=62.6+4.032*(5.713/SQRT(6)) = 72.004

So the 99% confidence interval for the mean would be given by (53.196;72.004)

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section covers somplifying algebraic expressions

Step-by-step explanation:

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Step-by-step explanation:

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3 years ago

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An elementary school teacher wants to know if the words in a book are too difficult for her students. She reads the first senten

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Answer:

d.every word in the first sentence of the first page of every chapter.

Step-by-step explanation

the sample space means the set of total possible outcomes or ALL possible outcomes.

so here, the situation is about the teacher reading a sentence.

the favorable outcome is the set which has all the words that the students would not know from all the words read by the teacher or simply from the sample space.

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