You are about to take a major assessment, and each question is one two types. Multiple-choice questions are worth 2 points each, and short-response questions are worth 4 points each. The assessment is worth 100 total points. Let x represent the number of multiple-choice questions, and let y represent the number of short-response questions.

       For C1(Q) = 3 - 2Q.

        For C2(Q) =  4.

     2. The AVC function

           For C1(Q) = 5/Q + 3 + 20/Q - Q.

           For C2(Q) = 3/Q + 4 + 2/Q.

     3. The AFC function

              For C1(Q)= 5/Q - 20/(5 + 3Q + 20/Q - Q)

              For C2(Q) = 0.

     4.  To find the AC function

        For C1(Q) = (5 + 3Q + 202 - Q^2)/Q + 5/Q - 20/(5 + 3Q + 20/Q - Q).

         For C2(Q) = (3 + 4Q + 2)/Q + 3/Q + 4 + 2/Q.

    5.To find the ATC function

    For C1(Q)= 5/Q² + 3/Q + 20/Q² - Q/Q + 5/Q - 20/(5Q + 3Q² + 20 - Q²)

     For C2(Q)= 3/Q² + 4/Q + 2/Q² + 3/Q + 4/Q + 2/Q.

1. To find the MC function, we take the derivative of the cost functions with respect to Q.

For C1(Q) = 5 + 3Q + 202 - Q^2, MC1(Q) = 3 - 2Q.

For C2(Q) = 3 + 4Q + 2, MC2(Q) = 4.

2. To find the AVC function, we divide the cost functions by Q.

For C1(Q), AVC1(Q) = (5 + 3Q + 202 - Q^2)/Q = 5/Q + 3 + 20/Q - Q.

For C2(Q), AVC2(Q) = (3 + 4Q + 2)/Q = 3/Q + 4 + 2/Q.

3. To find the AFC function, we subtract the AVC function from the ATC function.

For C1(Q), AFC1(Q) = (5 + 3Q + 202 - Q^2)/Q - (5 + 3Q + 202 - Q^2)/(5 + 3Q + 20/Q - Q)

= 5/Q - 20/(5 + 3Q + 20/Q - Q).

For

C2(Q), AFC2(Q) = (3 + 4Q + 2)/Q - (3 + 4Q + 2)/(3/Q + 4 + 2/Q) = 0.

4. To find the AC function, we add the AVC function to the AFC function.

For

C1(Q), AC1(Q) = (5 + 3Q + 202 - Q^2)/Q + 5/Q - 20/(5 + 3Q + 20/Q - Q).

For

C2(Q), AC2(Q) = (3 + 4Q + 2)/Q + 3/Q + 4 + 2/Q.

5. To find the ATC function, we divide the AC function by Q.

For

C1(Q), ATC1(Q) = [(5 + 3Q + 202 - Q²)/Q + 5/Q - 20/(5 + 3Q + 20/Q - Q)]/Q

= 5/Q² + 3/Q + 20/Q² - Q/Q + 5/Q - 20/(5Q + 3Q² + 20 - Q²).

For

C2(Q), ATC2(Q) = [(3 + 4Q + 2)/Q + 3/Q + 4 + 2/Q]/Q

= 3/Q² + 4/Q + 2/Q² + 3/Q + 4/Q + 2/Q.

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Answer:

The expression to find how many hours earlier Sanya had first checked the thermometer is: \(\frac{21}{1.4}\).

Step-by-step explanation:

The rate at which the temperature was falling is, 1.4 degrees per hour.

Sanya checked the temperature at 6:00 AM and found out that the temperature had fallen 21 degrees.

Compute the number of hours it took for the temperature to fall to 21 degrees as follows:

\(\text{Number of hours}=\frac{\text{Final Temperature}}{\text{Rate of decrease}}\)

                           \(=\frac{21}{1.4}\\\\=15\)

It took 15 hours for the temperature to fall to 21 degrees.

Then 15 hours before 6:00 AM is, 3:00 PM.

Thus, the expression to find how many hours earlier Sanya had first checked the thermometer is: \(\frac{21}{1.4}\).

Answer:

-21 divided by -1.4

Step-by-step explanation:

I took the test and I checked my answers. It said that "A" or -21 divided by -1.4 was the correct answer

Given function:y = √(1 – x³)We need to find dy/dx.As per the chain rule of differentiation, if y = f(u) and u = g(x), then the derivative of y with respect to x is given by dy/dx = dy/du * du/dx.

Now, let u = 1 – x³. Then, y = √u ⇒ y = u^(1/2).

dy/dx = dy/du * du/dxdy/du = 1/(2√u) = 1/(2√(1 – x³))du/dx = -3x²Therefore, dy/dx = dy/du * du/dx = 1/(2√(1 – x³)) * (-3x²) = (-3x²)/(2√(1 – x³)).

Therefore,  (A) 3(1 - x²)².Option (A) is the correct answer.

The given function is y = √(1 – x³) and we need to find dy/dx. As per the chain rule of differentiation, if y = f(u) and u = g(x), then the derivative of y with respect to x is given by dy/dx = dy/du * du/dx.

Let u = 1 – x³. Then, y = √u ⇒ y = u^(1/2).

We need to find dy/dx.dy/dx = dy/du * du/dxdy/du = 1/(2√u) = 1/(2√(1 – x³))du/dx = -3x².

Therefore, dy/dx = dy/du * du/dx = 1/(2√(1 – x³)) * (-3x²) = (-3x²)/(2√(1 – x³)).

Thus, dy/dx = (-3x²)/(2√(1 – x³)).Therefore, the correct option is (A) 3(1 - x²)².  

The derivative of the given function y = √(1 – x³) with respect to x is (-3x²)/(2√(1 – x³)) and the correct option is (A) 3(1 - x²)².

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the larger the sample, the more confidence you have when you generalize your sample findings to the population facts.

The larger the sample size, the more confidence you can have when you generalize your sample findings to the population facts. This is because larger sample sizes provide more accurate representation of the population facts due to the increased number of data points collected. As a result, the more data points you have, the more reliable your findings are in terms of representing the population facts. The larger the sample size, the more confidence you can have in generalizing the findings to the population facts.

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The measure of the angle between points AC is equal to 66°

From the question, we observe that the larger angle between AC consists of the two angles between AB and BC, so we shall sum the two small angles and equate the result to the larger angle to solve for the value of x as follows:

10x + 6 = 5x + 4x + 12

10x - 5x - 4x = 12 - 6 {collect like terms}

10x - 9x = 6

x = 6

so we can evaluate for the angle between AC as follows:

10x + 6 = 10(6) + 6

10x + 6 = 60 + 6

10x + 6 = 66°

In conclusion, the measure of the angle between points AC is equal to 66°

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Answer:

The outcomes of this binomial distribution that would be considered unusual is {0, 1, 2, 8}.

Step-by-step explanation:

The outcomes provided are:

(A) 0, 1, 2, 6, 7, 8

(B) 0, 1, 2, 7, 8

(C) 0, 1, 7, 8

(D) 0, 1, 2, 8

Solution:

The random variable X can be defined as the number of employees who judge their co-workers by cleanliness.

The probability of X is:

P (X) = 0.65

The number of employees selected is:

n = 8

An unusual outcome, in probability theory, has a probability of occurrence less than or equal to 0.05.

Since outcomes 0 and 1 are contained in all the options, we will check for X = 2.

Compute the value of P (X = 2) as follows:

\(P(X=2)={8\choose 2}(0.65)^{2}(0.35)^{8-2}\)

                \(=28\times 0.4225\times 0.001838265625\\=0.02175\\\approx 0.022<0.05\)

So X = 2 is unusual.

Similarly check for X = 6, 7 and 8.

P (X = 6) = 0.2587 > 0.05

X = 6 not unusual

P (X = 7) = 0.1373 > 0.05

X = 7 not unusual

P (X = 8) = 0.0319

X = 8 is unusual.

Thus, the outcomes of this binomial distribution that would be considered unusual is {0, 1, 2, 8}.

Answer:

C. five samples.

Step-by-step explanation:

Answer:

x + 6

Step-by-step explanation:

(f ÷ g)(x)

= \(\frac{x^2+9x+18}{x+3}\) factorise the numerator

= \(\frac{(x+3)(x+6)}{x+3}\) ← cancel the factor (x + 3) on numerator/ denominator

= x + 6

Answer: like math equations that equal twenty?

Step-by-step explanation:

.5 times 40

4 times 5

2 times 10

3 times 6.2/3

Just to name a few

Answer:

4 times 5

the square root of 400

10 plus 10

2 to the 2nd power plus 4

30 minus 10

40 minus 20

50 minus 30

the absolent value of -20

5 to the 2nd power minus 5.

Step-by-step explanation:

Hope it Helps?

In each scenario, the choice between mean and median as a representative measure of central tendency depends on the nature of the data and the specific context..

1. Scenario: Income distribution of a population

- If the income distribution is skewed or contains extreme values (outliers), the median would be a better representation of the central tendency. This is because the median is not influenced by outliers and provides a more robust estimate of the "typical" income level. However, if the income distribution is approximately symmetric without outliers, the mean can also be an appropriate measure.

2. Scenario: Exam scores in a class

- If the exam scores are normally distributed without significant outliers, the mean would be a suitable measure as it takes into account the value of each score. However, if there are extreme scores that deviate from the majority of the data, the median may be a better representation. This is especially true if the outliers are indicative of errors or exceptional circumstances.

3. Scenario: Housing prices in a city

- In this case, the median would be a more appropriate measure to represent the central tendency of housing prices. This is because the housing market often exhibits a skewed distribution with a few high-priced properties (outliers). The median, being the middle value when the data is sorted, is not influenced by these extreme values and provides a better understanding of the typical housing price in the city.

Ultimately, the choice between mean and median depends on the specific characteristics of the data and the objective of the analysis. It is important to consider the distribution, presence of outliers, and the context in which the data is being interpreted.

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Answer:

28.5

Step-by-step explanation:

If you can add brainliest

Please

Answer:

2x-3=-x+3

Step-by-step explanation:

Answer:

k = 3/4

Step-by-step explanation:

Parallel lines have the same slope so first, we need to find the slope of 8x + 6 = 2y. To do so, we can do the following:

8x + 6 = 2y

2y = 8x + 6

y = 4x + 3 (This is now in slope intercept form - y = mx + b where m = slope and b = y-int.)

Therefore, the slope is 4. Now, we can do:

3x = ky - 10

3x - 10 = ky

(3x / k) - (10 / k) = y

y = (3x / k) - (10 / k)  (where k ≠ 0)

Now that the second line is in slope-intercept form, we can set the two x terms equal to each other, therefore:

3x / k = 4x

3x = 4x * k

3x / 4x = k

k = 3/4

The required, Mmusi borrowed R3000 from Aggie as per the given conditions.

Simple interest is a quick and simple formula for figuring out how much interest will be charged on loan.

Amount = Principal [1 + rate×time]

Let's call the amount Mmusi borrowed from Aggie "P".

After two years, he has to pay back
= P + P * 18% * 2
= P + 0.18 * P * 2
= P + 0.36P
= 1.36P

So if he has to pay back R4080, we can set up an equation to solve for P:

1.36P = 4080

To find P, we can divide both sides by 1.36:

P = 4080 / 1.36 = 3000

So Mmusi borrowed R3000 from Aggie.

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Answer: 190800

Step-by-step explanation:

10 • 10 = 100

Than when you multiply anything 100 you just add two 0s to the end.

100 • 1908 = 190800

Answer:

193

Step-by-step explanation:

Merry Christmas

The student used 30.0 mL of 3.00 M HCl to make the 50.0 mL sample of 1.80 M HCl.

To find the volume of 3.00 M HCl needed to make a 50.0 mL sample of 1.80 M HCl, we can use the equation:

M₁V₁ = M₂V₂

Where M₁ is the initial concentration, V₁ is the initial volume, M₂ is the final concentration, and V₂ is the final volume.

We are given M₁ = 3.00 M, M₂ = 1.80 M, and V₂ = 50.0 mL. We can rearrange the equation to solve for V₁:

V₁ = (M₂V₂) / M₁

V₁ = (1.80 M * 50.0 mL) / 3.00 M

V₁ = 30.0 mL

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The null we are interested in testing here is that the coefficients on age and age^2 are jointly zero. So, we would need to run a restricted version of the regression above where these two variables are omitted and calculate.

Given:

what other regression do you need to run to test the null hypothesis that, holding other factors fixed, age has no effect on sleeping.

The null we are interested in testing here is that the coefficients on age and age^2 are jointly zero.

\(F = R^2_{ur}-R^2_r / q / 1-R^2_{ur} (n-k-1)\)

Therefore the null we are interested in testing here is that the coefficients on age and age^2 are jointly zero. So, we would need to run a restricted version of the regression above where these two variables are omitted and calculate.

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Answer:

3\sqrt(5)

or 6.71

Step-by-step explanation:

Using Permutation formula,

Total 720 ways are possible to stack the six boxes in order .

We have given that,

First stack contains number of boxes = 3

Second stack contains number of boxes = 2

Third stack contains number of boxes = 1

Total number of boxes = 6

Number of ways to stack 1 box = ⁶P₁ = 6!/5!

= 6 ways

Number of ways to stack 2 boxes = ⁵P₂= 5!/(5-2)!

= 5!/3! = 20 ways

Number of ways to stack 3 boxes = ⁴P₃= 4!/1!

= 4 ways

Total ways to stack 6 boxes = ⁶P₁ ×⁵P₂×⁴P₃

= 20×6× 4 = 720 ways

Hence, we can stack the 6 boxes in 720 ways .

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If the Durbin-Watson statistic is 2, we can conclude that there is no serial correlation. i.e. the option c) there is negative serial correlation

This statistic tests for the presence of autocorrelation between successive error terms of a regression model. Autocorrelation is when the value of a variable at a given point in time is correlated to the value of the same variable at an earlier or later point in time. If the Durbin-Watson statistic is 2, this means that there is no statistically significant serial correlation between the error terms.

In addition to a lack of serial correlation, a Durbin-Watson statistic of 2 also indicates that the error terms have constant variances and that there is no multicollinearity. Constant variance means that the magnitude of the error terms remains unchanged over time, and multicollinearity is when two or more independent variables are highly correlated. The lack of multicollinearity suggests that the independent variables are not driving the results in a predictable manner.

Overall, a Durbin-Watson statistic of 2 indicates that there is no serial correlation, the error terms have constant variances, and there is no multicollinearity. i.e. the option c) there is negative serial correlation

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\(\textit{Logarithm Change of Base Rule} \\\\ \log_a b\implies \cfrac{\log_c b}{\log_c a}\qquad \qquad c= \begin{array}{llll} \textit{common base for }\\ \textit{numerator and}\\ denominator \end{array} \\\\\\ \begin{array}{llll} \textit{Logarithm Cancellation Rules} \\\\ log_a a^x = x\qquad \qquad \stackrel{ \textit{we'll use this one} }{a^{log_a (x)}=x} \end{array} \\\\[-0.35em] \rule{34em}{0.25pt}\)

\(\log_5(62)\implies \cfrac{\log_8(62)}{\log_8(5)} ~~ \approx ~~ 2.56433 ~~ \approx ~~ \boxed{2.564} \\\\[-0.35em] ~\dotfill\\\\ \stackrel{\textit{now we can say that}}{2.56433=\log_8(y)}\implies 8^{2.56433}=8^{\log_8(y)}\implies 8^{2.56433}=y\implies 206.930\approx y \\\\\\ \textit{so we can write that in logarithmic form as}\qquad \boxed{\log_8(206.93)}\)

Answer:

30

Step-by-step explanation:

Answer:

six crore four lakh fifty seven thousand nine hundred eighty one

Answer: choices 1, 4, and 5.

Step-by-step explanation: uhhhh the answers basically explain why. 1 because it's a terminating decimal, 4 because it's a repeating number, and 5 because the square root of 11 is not a perfect square, and is not a repeating or terminating decimal.

There are 4,960 different ways that a new curriculum committee of 3 faculty members could be formed from a mathematics department of 32 faculty members.

Given that,

A mathematics department consists of 32 faculty members.

If a new curriculum committee of 3 faculty members needs to be formed, there are different ways this could happen.

The formula for combinations is

C(n,r) = n! / (r!(n-r)!),

where n represents the total number of items and r represents the number of items being chosen at a time.

In this case, there are 32 faculty members and we need to choose a committee of 3 members.

Therefore, the formula becomes: C(32,3) = 32! / (3!(32-3)!) = (32 * 31 * 30) / (3 * 2 * 1) = 4960.

There are 4,960 different ways that a new curriculum committee of 3 faculty members could be formed from a mathematics department of 32 faculty members.

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The precise quadratic equation, according to the provided assertion, is y = (1/9)x² + (2/9)x - 1/3.

Three elements decide a quadratic function of the formula f(x) = ax² + bx + c. A quadratic function can be determined by getting a, b, and c algebraically given three locations on the graph. In order to do this, a system of eqs with three unknowns must be solved.

To find the quadratic function that passes through the point (-8, 5) and has zeros at x = -3 and x = 1, we can start by using the factored form of a quadratic function:

y = a(x - r)(x - s)

where r and s are the zeros of the function. Substituting in the given zeros, we get:

y = a(x + 3)(x - 1)

To find the value of a, we can use the fact that the function passes through the point (-8, 5). Substituting these values into the equation, we get:

5 = a(-8 + 3)(-8 - 1)

Simplifying this expression, we get:

5 = a(-5)(-9)

5 = 45a

a = 1/9

Substituting this value of a into the equation we get:

y = (1/9)(x + 3)(x - 1)

Expanding the equation, we get:

y = (1/9)(x² + 2x - 3)

Multiplying both sides by 9 to eliminate the fraction, we get:

9y = x² + 2x - 3

So the quadratic function that passes through the point (-8, 5) and has zeros at x = -3 and x = 1 is:

9y = x² + 2x - 3

or

y = (1/9)x² + (2/9)x - 1/3.

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