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3 years ago

Which expression is equivalent to

Mathematics

2 answers:

disa [49]3 years ago

5 0

Answer:

Expression 1

Step-by-step explanation:

14x + 35 - 21y

$14x + 35 - 21y \\ \frac{14x}{7} + \frac{35}{7} - \frac{21y}{7} \\ 2x + 5 - 3y$

Rus_ich [418]3 years ago

5 0

Answer:

Its Expression 2: 7(2x + 5 - 3y)

Step-by-step explanation:

just got it right

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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)

8 0

3 years ago

the change in water vapor in a cloud is modeled by a polynomial function, c (x). describe how to find the x - intercepts of c (x

lukranit [14]

The change in the water vapors is modeled by the polynomial function c(x). In order to find the x-intercepts of a polynomial we set it equal to zero and solve for the values of x. The resulting values of x are the x-intercepts of the polynomial.

Once we have the x-intercepts we know the points where the graph crosses the x-axes. From the degree of the polynomial we can visualize the end behavior of the graph and using the values of maxima and minima a rough sketch can be plotted.

Let the polynomial function be c(x) = x ² -7x + 10

To find the x-intercepts we set the polynomial equal to zero and solve for x as shown below:

x ² -7x + 10 = 0

Factorizing the middle term, we get:

x ² - 2x - 5x + 10 = 0

x(x - 2) - 5(x - 2) =0

(x - 2)(x - 5)=0

x - 2 = 0 ⇒ x=2

x - 5 = 0 ⇒ x=5

Thus the x-intercept of our polynomial are 2 and 5. Since the polynomial is of degree 2 and has positive leading coefficient, its shape will be a parabola opening in upward direction. The graph will have a minimum point but no maximum if the domain is not specified. The minimum points occurs at the midpoint of the two x-intercepts. So the minimum point will occur at x=3.5. Using x=3.5 the value of the minimum point can be found. Using all this data a rough sketch of the polynomial can be constructed. The figure attached below shows the graph of our polynomial.