To receive credit fWhich of the following can you use during a quiz?

Question 4 options:

a)

Notes

b)

Textbook

c)

Both A or B

d)

You cannot use the textbook or notes on a quiz.or a course, your overall average must be above?

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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Answer:

19, 54 and 27

Explanation:

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

19+54+27=100

The mean of the temperature based on the information given is -3.2.

The difference in the temperature based on the information will be:

98.4 - 99.0 = -0.6

98.7 - 99.3 = -0.6

97.5 - 97.9 = -0.4

97.3 - 96.9 = 0.4

97.1 - 97.5 = -0.4

Total = -1.6

The mean will be:

= -1.6/5

= -3.2

Also, the standard deviation is 2.8. This simony illustrates how spread the numbers are in the information given.

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The standard error is 1.5.

The range of 95% confidence interval is 160 mg/dL and 220 mg/dL

The standard error of the distribution is calculated as follows:

where N is the number of participants = 100

From the empirical rule of a normal or symmetrical distribution;

Therefore,  175 mg/dl and 205 mg/dl lie one standard deviation from the mean.

One standard deviation = 175 - 190 or 205 - 190 = ±15

Standard deviation = 15

a. Solving for the standard error using the equation above;

Standard error = 15/√100

Standard error = 1.5

b. The range of 95% confidence interval is two standard deviations away from the mean.

Range of values two standard deviations from the mean = 190 - (2*15) and 190 + (2 * 15)

Range of values two standard deviations from the mean = 160 mg/dL and 220 mg/dL

Therefore, the range of 95% confidence interval is 160 mg/dL and 220 mg/dL

In conclusion, the standard error is calculated from the standard deviation and the sample size.

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Answer:

I don't know

Explanation:

I don't know

sorry for it ;

I am really sorry

Answer:

The probability of ball 5=1/6

Explanation:

This is because in 1 to 6, there is only one 5 numbered ball out of the total number of balls which is 6.

Astronauts in space can travel around the earth in about 12 hours.

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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Answer:

their

Explanation:

Lets highlight the keywords in this sentence.

All the actresses brought (her, their) scripts.

The words all and actresses are telling you that there are more than 1 actress, so we will refer to them as 'their'.

Below is the meaning of each term and how each relates to the scientific method:

It is a set of methods and techniques structured and organized in a complex way to test a theory, find new concepts and knowledge through research, study, experimentation and analysis.

Therefore, the scientific methodology helps the development of society through the production of reliable and tested knowledge in order to assist in social and technological progress.

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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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