y = x² - 4x +3
y = x - 1
If (x, y) is a solution to the system of equations
above, what is one possible value of the product
of x and y?

Answers

Answer 1

Answer:

Possible values:

0 and 12

Explanation:

To solve the system of equations, let's set y = x^2 - 4x + 3 and y = x - 1 equal to each other to first solve for x.

x^2 - 4x + 3 = x - 1 (add one to both sides)

x^2 - 4x + 4 = x (subtract x from both sides)

x^2 - 5x + 4 = 0

Next, we simply factor the expression. -1 and -4 add up to -5 and multiply to 4, so the factored expression looks like this:

(x - 1)(x - 4) = 0

x = 1 and x = 4 are both solutions to the factored equation. Now, let's plug the values back in to both equations.

(1)^2 - 4(1) + 3 = 0

(4)^2 - 4(4) + 3 = 3

(1) - 1 = 0

(4) - 1 = 3

When 1 is plugged into either equation, 0 is the value of y. For 4, 3 is the value of y. The possible values of product of x and y are below:

0 * 1 = 0

4 * 3 = 12

Brainliest, please :)


Related Questions

r = 2.24 + 0.06x
p= 2.89 +0.10x
In the equations above, r and p represent the price per gallon, in dollars, of regular and premium grades of gasoline, respectively, x months after January 1 of last
year. What was the cost per gallon, in dollars, of premium gasoline for the month in which the per gallon price of premium exceeded
the per gallon price of regular by $0.93? Please explain thoroughly

Answers

Based on the equations of the price, the price of the premium gasoline, p = $3.59

What is the equation of the prices?

An equation is an expression that describes two quantities as being equal.

The equation of the different prices of gasoline is given below:

Price of regular grade of gasoline, r = 2.24 + 0.06xPrice of premium grade of gasoline, p = 2.89 + 0.10x

Where x is the number of months after January 1 of last year.

When the per gallon price of premium exceeded the per gallon price of regular by $0.93, the two equations will be related as shown below:

p = r + 0.93 ---- (3)

r + 0.93 = 0.93 + 2.24 + 0.06x

r + 0.93 = 3.17 + 0.06x

Therefore, p = 3.17 + 0.06x --- (4)

Also, p = 2.89 +0.10 x --- (2)

solving for x by equating (1) and (4)

3.17 + 0.06x = 2.89 + 0.10x

0.10x - 0.06x = 3.17 - 2.89

0.04x = 0.28

x = 7

Then, substituting for x in (4)

p = 3.17 + 0.06 * 7

p = $3.59

In conclusion, the price of the premium gasoline is determined from the given equations of the price.

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Consider an experiment of tossing two fair dice and noting the outcome on each die. The whole sample space consists of 36 elements. Now, with each of these 36 elements associate values of two random variables, X and Y, such that X≡ sum of the outcomes on the two dice, Y ≡ |difference of the outcomes on the two dice|. Construct joint probability mass function.

Answers

The joint probability mass function based on the given experiment is as seen in the attached file.

How to construct a joint probability mass function?

The whole sample space consists of 36 elements, i.e.,

Ω = {ω_ij = (i, j) : i, j = 1, ....6}

Now, with each of these 36 elements associate values of two random variables, X₁ and X₂, such that;

X₁ ≡ sum of the outcomes on the two dice.

X₂ ≡ |difference of the outcomes on the two dice|

That is;

X(ω_ij) = X₁(ω_ij) + X₂(ω_ij) = (i + j, |i - j|) i, j = 1, 2, ...., 6

Then, the bivariate rv X = (X₁, X₂) has the following joint probability mass

function (empty cells mean that the pmf is equal to zero at the relevant values of the rvs).

The joint probability mass function is seen in the attached file.

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A manufacturer of potato chips would like to know whether its bag filling machine works correctly at the 426 gram setting. Based on a 13 bag sample where the mean is 433 grams and the standard deviation is 29, is there sufficient evidence at the 0.05 level that the bags are overfilled? Assume the population distribution is approximately normal.

Answers

The conclusion of the research is that we reject the null hypothesis and we conclude that the bag filling machine does not work correctly at the 426 gram setting.

How to solve hypothesis testing?

We are given;

Population Mean; μ = 426

Sample mean; x' = 433

Sample size; n = 13

Standard deviation; s = 29

significance level; α = 0.05

Let us define the hypotheses;

Null Hypothesis; H₀: μ = 426 g

Alternative Hypothesis; H_a: μ < 426 g

Formula for the z-score here is;

z = (x' - μ)/(s/√n)

z = (433 - 426)/(29/√13)

z = 0.87

From online p-value from z-score calculator, we have;

p-value =  0.1922

This p-value is greater than the significance value and as such we reject the null hypothesis and we conclude that the bag filling machine does not work correctly at the 426 gram setting.

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in a clinical trial, 28 out of 888 patients taking a prescription drug daily complained of flulike symptoms. suppose that it is known that 2.8% of the patients taking competing drugs complain of flulike symptoms. is there sufficient evidence to conclude that more than 2.8% of this drug's user experience flulike symptoms as a side effect at the aplha 0.05 level of significance

Answers

The p-value (0.0001) is less than α = 0.05. Based on this, we should reject the null hypothesis.

What is a null hypothesis?

A null hypothesis (H₀) can be defined the opposite of an alternate hypothesis (H₁) and it asserts that two (2) possibilities are the same.

How to calculate value of the test statistic?

The test statistic can be calculated by using this formula:

[tex]t=\frac{x\;-\;u}{\frac{\delta}{\sqrt{n} } }[/tex]

Where:

x is the sample mean.u is the mean.is the standard deviation.n is the number of hours.

For this clinical trial (study), we should use a t-test and the null and alternative hypotheses would be given by:

H₀: p = 0.028

H₁: p > 0.028

For the sample proportion, we have:

Sample proportion, P = 28/888

Sample proportion, P = 0.032.

Next, we would calculate the t-test as follows:

[tex]t=\frac{0.032\;-\;0.028}{\frac{0.02 \times 0.028}{888 } }\\\\t=\frac{0.004}{\sqrt{\frac{0.00056}{888 } }}[/tex]

t = 0.004/0.00079

t-test, t = 5.06.

For the p-value, we have:

P-value = P(t > 5.06)

P-value = 1 - P(t < 5.06)

P-value = 1 - 0.9999

P-value = 0.0001.

Therefore, the p-value (0.0001) is less than α = 0.05. Based on this, we should reject the null hypothesis.

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bona has a stated reaction to a sound of a loud < >. immediately before the loud pop, a light flashes. after several times. Bona has started reaction to the flash of lighy. What is UCS? What is CR? What is NS? What is UCR?

Answers

The UCS is known to be the loud pop

What is CR?

The CR is known to be  a light flashes

What is NS?

The NS is the reaction  of bona

What is UCR?

The UCR  flash of light.

What is an unconditioned response?

In classical conditioning, an unconditioned response is known to be that behavior that is said to be unlearned response that takes place as a sort of natural  reaction to the unconditioned stimulus.

Hence, The UCS is known to be the loud pop.

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Before a site begins recruitment, a protocol monitor will conduct
a visit to review the data the research site has collected from
study participants.

Answers

Initiation visits occur before a study site begins participant recruitment for protocol participation and after the necessary IRB permissions are obtained. During a site initiation visit, study monitors review trial records to ensure that they are complete and in order.

What is Protocol Monitor?

The experimental Protocol Monitor displays the underlying requests and reactions happening over the DevTools protocol. For example, when considering some JavaScript via DevTools, observe the following notes on the protocol: Runtime.

How do I turn off Developer Tools on my phone?

You can disable the Developer options and hide the menu by opening Locations, and then tapping Developer options. Tap the switch at the top of the net to turn off Developer options.

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