Em um experimento foram colocados, em um mesmo recipiente, dois tipos de microrganismos, A

Determine the minimum value of f(x, y, z) = 2x 2 + y 2 + 2z 2 + 5 subject to the constraint 3x + 2y + 3z = 4. +

vlada-n [284]

We want to optimize f(x,y,z)=x^2 y^2 z^2, subject to g(x,y,z) = x^2 + y^2 + z^2 = 289.

Then, ∇f = λ∇g ==> <2xy^2 z^2, 2x^2 yz^2, 2x^2 y^2 z> = λ<2x, 2y, 2z>.

Equating like entries:
xy^2 z^2 = λx
x^2 yz^2 = λy
x^2 y^2 z = λz.

Hence, x^2 y^2 z^2 = λx^2 = λy^2 = λz^2.

(i) If λ = 0, then at least one of x, y, z is 0, and thus f(x,y,z) = 0 <---Minimum
(Note that there are infinitely many such points.)
(f being a perfect square implies that this has to be the minimum.)

(ii) Otherwise, we have x^2 = y^2 = z^2.
Substituting this into g yields 3x^2 = 289 ==> x = ±17/√3.

This yields eight critical points (all signage possibilities)
(x, y, z) = (±17/√3, ±17/√3, ±17/√3), and
f(±17/√3, ±17/√3, ±17/√3) = (289/3)^3 <----Maximum

I hope this helps!

8 0

3 years ago

Find the difference -1/10 - 8/10 -9/10 9/10 -7/10 7/10

TEA [102]

The answer to this would be A: -9/10. Hope this helps!

5 0

3 years ago

Read 2 more answers

Graph the circle (x+6)^2 + (y+5)^2 =9

Andrei [34K]

Answer:

This a circle centered at the point $(-6,-5)$ , and of radius "3" as it is shown in the attached image.

Step-by-step explanation:

Recall that the standard formula for a circle of radius "R", and centered at the point $(x_0,y_0)$ is given by:

$(x-x_0)^2+(y-y_0)^2=R^2$

Therefore, in our case, by looking at the standard equation they give us, we extract the following info:

1) $R^2 = 9\,\,\,then\,\,\,R=3$ since the radius must be a positive number and ( $R=-3$ ) is not a viable answer.

2) $x_0=-6$ for ( $x-x_0$ ) to equal $(x+6)$

3) $y_0=-5$ for ( $y-y_0$ ) to equal $(y+5)$

Therefore, we are in the presence of a circle centered at the point $(-6,-5)$ , and of radius "3". That is what we draw as seen in the attached image.

5 0

3 years ago

Daily-high temperature measurements for 40 consecutive days are recorded for a particular city. The mean daily-high temperature

Vesnalui [34]

Answer:

$t=\frac{21.5-22}{\frac{1.5}{\sqrt{40}}}=-2.108$

$p_v =P(t_{39}$

Conclusion

If we compare the p value and the significance level given $\alpha=0.01$ we see that $p_v>\alpha$ so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, so we can conclude that the mean temperature actually its NOT significant less then 22 at 1% of signficance.

(D) P-val=0.021, fail to reject the null hypothesis

Step-by-step explanation:

1) Data given and notation

$\bar X=21.5$ represent the mean for the temperatures

$s=1.5$ represent the sample standard deviation

$n=40$ sample size

$\mu_o =22$ represent the value that we want to test

$\alpha=0.01$ represent the significance level for the hypothesis test.

t would represent the statistic (variable of interest)

$p_v$ represent the p value for the test (variable of interest)

State the null and alternative hypotheses.

We need to conduct a hypothesis in order to check if the mean is less than 22C, the system of hypothesis would be:

Null hypothesis: $\mu \geq 22$

Alternative hypothesis: $\mu < 22$

If we analyze the size for the sample is > 30 and we don't know the population deviation so is better apply a t test to compare the actual mean to the reference value, and the statistic is given by:

$t=\frac{\bar X-\mu_o}{\frac{s}{\sqrt{n}}}$ (1)

t-test: "Is used to compare group means. Is one of the most common tests and is used to determine if the mean is (higher, less or not equal) to an specified value".

Calculate the statistic

We can replace in formula (1) the info given like this:

$t=\frac{21.5-22}{\frac{1.5}{\sqrt{40}}}=-2.108$

P-value

We can calculate the degrees of freedom like this:

$df=n-1=40-1=39$

Since is a one left tailed test the p value would be:

$p_v =P(t_{39}$

Conclusion

If we compare the p value and the significance level given $\alpha=0.01$ we see that $p_v>\alpha$ so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, so we can conclude that the mean temperature actually its NOT significant less then 22 at 1% of signficance.

The best option would be:

(D) P-val=0.021, fail to reject the null hypothesis

5 0

3 years ago

Question number 3 pls

marissa [1.9K]

Factoring f(x) would give a factor (x-1-2i), so you can divide f(x) by this, leaving (x-3)(x+4)(x-1+2i), so the other roots are:

x=3
x=-4
x=1-2i

4 0

3 years ago