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3 years ago

Which ordered pair is a solution to the system of equations?

Mathematics

2 answers:

Kazeer [188]3 years ago

7 0

Answer:

B) (-2,1)

Step-by-step explanation:

In order to find a solution to a system of equations, substitute the point's x and y values into both of the equations, solve, and see if it makes the equation true.

1) Let's try the point (-2,1). First, substitute its x and y values into the first equation, 3x - 3y = -9. Thus, substitute -2 for x and 1 for y in the equation and solve:

$3x-3y = -9\\3(-2)-3(1) = -9\\-6 - 3 = -9 \\- 9 = -9$

-9 does equal -9, thus (-2, 1) makes the first equation true.

2) Now, do the same thing but with the second equation, 2x + y = -3. Again, substitute -2 for x and 1 for y and solve:

$2x + y = -3\\2(-2) + (1) = -3\\-4 + 1 = -3\\-3 = -3$

-3 does equal -3, thus (-2, 1) makes the second equation true.

By substituting its values into both equations, we saw that it made both equations true. Thus, (-2,1) is the answer.

elixir [45]3 years ago

4 0

I believe it would be (b) (-2,1)

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The manufacturer of an airport baggage scanning machine claims it can handle an average of 530 bags per hour. (a-1) At α = .05 i

fenix001 [56]

Answer:

Null hypothesis: $\mu \geq 250$

Alternative hypothesis: $\mu < 250$

$p_v =P(t_{15}$

If we compare the p value and the significance level given $\alpha=0.05$ we see that $p_v>\alpha$ so we can conclude that we FAIL to reject the null hypothesis, and the the actual mean is not significantly lower than 530 at 5% of significance.

Step-by-step explanation:

1) Data given and notation

$\bar X=510$ represent the mean for the sample

$s=50$ represent the standard deviation for the sample

$n=16$ sample size

$\mu_o =530$ represent the value that we want to test

$\alpha$ 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)

2) State the null and alternative hypotheses.

We need to conduct a hypothesis in order to determine if the claim thet the handle on average is 530 bags per hour:

Null hypothesis: $\mu \geq 530$

Alternative hypothesis: $\mu < 530$

We don't know the population deviation and the sample size is lees than 30, so for this case is better apply a t test to compare the actual mean to the reference value, and the statistic is given by:

$t=\frac{\bar X-\mu_o}{\frac{\sigma}{\sqrt{n}}}$ (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".

3) Calculate the statistic

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

$t=\frac{510-530}{\frac{50}{\sqrt{16}}}=-1.60$

4) Calculate the critical value

We need to begin calculating the degrees of freedom

$df=n-1=16-1=15$

The critical value for this case would be :

$P(t_{15}$

The value of a that satisfy this on the normal standard distribution is a=-1.75 and would be the critical value on this case zc=-1.75

5) Calculate the P-value

Since is a one-side upper test the p value would be:

$p_v =P(t_{15}$

6) Conclusion

8 0

3 years ago

A metal cylinder can with an open top and closed bottom is to have volume 4 cubic feet. Approximate the dimensions that require

Aleksandr-060686 [28]

Answer:

$r\approx 1.084\ feet$

$h\approx 1.084\ feet$

$\displaystyle A=11.07\ ft^2$

Step-by-step explanation:

<u>Optimizing With Derivatives </u>

The procedure to optimize a function (find its maximum or minimum) consists in :

Produce a function which depends on only one variable
Compute the first derivative and set it equal to 0
Find the values for the variable, called critical points
Compute the second derivative
Evaluate the second derivative in the critical points. If it results positive, the critical point is a minimum, if it's negative, the critical point is a maximum

We know a cylinder has a volume of 4 $ft^3$ . The volume of a cylinder is given by

$\displaystyle V=\pi r^2h$

Equating it to 4

$\displaystyle \pi r^2h=4$

Let's solve for h

$\displaystyle h=\frac{4}{\pi r^2}$

A cylinder with an open-top has only one circle as the shape of the lid and has a lateral area computed as a rectangle of height h and base equal to the length of a circle. Thus, the total area of the material to make the cylinder is

$\displaystyle A=\pi r^2+2\pi rh$

Replacing the formula of h

$\displaystyle A=\pi r^2+2\pi r \left (\frac{4}{\pi r^2}\right )$

Simplifying

$\displaystyle A=\pi r^2+\frac{8}{r}$

We have the function of the area in terms of one variable. Now we compute the first derivative and equal it to zero

$\displaystyle A'=2\pi r-\frac{8}{r^2}=0$

Rearranging

$\displaystyle 2\pi r=\frac{8}{r^2}$

Solving for r

$\displaystyle r^3=\frac{4}{\pi }$

$\displaystyle r=\sqrt[3]{\frac{4}{\pi }}\approx 1.084\ feet$

Computing h

$\displaystyle h=\frac{4}{\pi \ r^2}\approx 1.084\ feet$

We can see the height and the radius are of the same size. We check if the critical point is a maximum or a minimum by computing the second derivative

$\displaystyle A''=2\pi+\frac{16}{r^3}$