Which of the following is the best way to improve human capital?

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

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

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

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

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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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.

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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Answer:

D

Explanation:

They are a tool for others to judge you and your mistake

Answer:

C. They are a learning tool for judging yourself

Explanation:

Examinations and assessments aren't meant to point out your failures or show what you don't know because not everyone makes mistakes on tests and exams, and if they do, those mistakes could have very well been careless. They also aren't a tool for others to judge you and your mistakes because no one is perfect, and people should judge you based on your overall personality and actions, not your test scores.

C is correct because tests and exams are a learning tool... If you do miss anything, you can refer back and study them to prepare for the next test or review what you missed. They allow you to judge yourself by letting you look at where you are now and compare it to where you should be in a healthy way.

Hope this helps! Brainliest, please :)

Answer:

19, 54 and 27

Explanation:

54 is 2 times of 27 and 27-8=19

19+54+27=100

The UCS is known to be the loud pop

The CR is known to be  a light flashes

The NS is the reaction  of bona

The UCR  flash of light.

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

The test statistic can be calculated by using this formula:

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

Where:

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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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.

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.

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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