What is the slope of the line that passes through the points (-3, 5) and (1,7)?
01
1/2
02

1) The confidence interval will get narrower if the variance decreases. The statement is True.
2) A point estimate includes a measure of variability. The statement is False.

1) The statement "The confidence interval will get narrower if the variance decreases" is True. When the variance (a measure of variability) of a sample decreases, the confidence interval around the point estimate will become narrower, as there is less variability in the data.
2) The statement "A point estimate includes a measure of variability" is False. A point estimate is a single value that represents an estimate of a population parameter (e.g., mean, median), and it does not include a measure of variability. Variability is typically assessed using confidence intervals or measures like variance and standard deviation, which are separate from point estimates.

The complete question is:-

1) The confidence interval will get narrower if the variance decreases. True O False O

2) A point estimate includes a measure of variability. True O False O

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The person will gain Rs. 5,500 after 3 years by lending the whole amount to a shopkeeper at the same rate of compound Interest for the same time.

Given, Principal amount (P) = Rs. 16,000

Rate of Interest (R) = 12.5%

Time (T) = 3 years(a)

The amount to be paid to the bank as simple interest can be calculated as follows:

Simple Interest (SI) = (P × R × T) / 100= (16000 × 12.5 × 3) / 100= Rs. 6,000Therefore, the amount to be paid to the bank as simple interest is Rs. 6,000.

(b) The amount to be received by the person from the shopkeeper as compound interest can be calculated as follows: Compound Interest (CI) = P × [1 + (R/100)]T – P= 16000 × [1 + (12.5/100)]3 – 16000= Rs. 21,500

Therefore, the amount to be received by the person from the shopkeeper as compound interest is Rs. 21,500.(c) Gain = CI - P= 21500 - 16000= Rs. 5,500

Therefore, the person will gain Rs. 5,500 after 3 years by lending the whole amount to a shopkeeper at the same rate of compound interest for the same time.

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B. No, because the Central Limit Theorem states that the sampling distribution of x is approximately normally distributed only if the sample size is large enough.

The Central Limit Theorem (CLT) is a fundamental concept in statistics that states that as the sample size increases, the sampling distribution of the sample means will approach a normal distribution. However, this is only true if certain conditions are met, one of which is having a large enough sample size.
The CLT states that the sampling distribution of x will be approximately normally distributed if the sample size is large enough (usually greater than 30). If the sample size is small, the sampling distribution may not be normally distributed. In such cases, other statistical techniques like the t-distribution should be used.
Furthermore, the CLT assumes that the population being sampled is not necessarily normally distributed, but it does require that the population has a finite variance. This means that even if the population is not normally distributed, the sampling distribution of x will still be approximately normal if the sample size is large enough.
In conclusion, the answer is B, as the Central Limit Theorem states that the sampling distribution of x is approximately normally distributed only if the sample size is large enough.

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Hello~

The answer is

-8884

Brainliest please!

Ary~

Step-by-step explanation:

\(600 - 9484 \\ = - 8884\)

\(\circ \: \: { \underline{ \boxed{ \sf{ \color{green}{Happy\:learning.}}}}}∘\)

Answer:

last option is the answer

Answer:

9<x<23

Step-by-step explanation:

I'm not sure if this is what it wants, but theres this theorem that says a side needs to be less than the other two combined.

x+7>16

x+16>7

16+7>x

=>

x>9

x>-9

x<23

=>

9<x<23

 The solution to the system of equations is (x, y) = (-5/3, -4/3).

What is system of linear equations?

A system of linear equations is a set of two or more equations with two or more variables that are to be solved simultaneously. Each equation in the system is linear, meaning it can be written in the form of ax + by + cz + ... = d, where a, b, c, and d are constants and x, y, z, and other variables are unknowns.

The goal of solving a system of linear equations is to find the values of the variables that satisfy all of the equations in the system. The solution of a system of linear equations is a set of values for the variables that make all of the equations true.

There are different methods to solve systems of linear equations, such as substitution method, elimination method, and matrix method. These methods involve manipulating the equations in the system to isolate one variable, substitute its value into another equation, and eventually find the values of all the variables.

Systems of linear equations are used in many areas of mathematics, science, engineering, and economics to model real-world situations and solve practical problems.

To solve the system of equations:

x + y = -3 (Equation 1)

y = 2x + 2 (Equation 2)

We can substitute Equation 2 into Equation 1 for y and solve for x:

x + (2x + 2) = -3

3x + 2 = -3

3x = -5

x = -5/3

Now that we know x, we can substitute it into either Equation 1 or Equation 2 to find y. Let's use Equation 2:

y = 2x + 2

y = 2(-5/3) + 2

y = -10/3 + 6/3

y = -4/3

Therefore, the solution to the system of equations is (x, y) = (-5/3, -4/3).

Comparing this solution to the answer choices, we see that option (b) is the correct answer:

(a) 5/3, 4/3

(b) -5/3, -4/3 <--- Correct answer

(c) -3/5, -3/4

(d) 3/4, 3/5

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The partial derivatives are:
f_x = 2x - 6λ
f_y = 2y - 9λ
f_λ = -(6x + 9y - 10)

To find the partial derivatives of f(x, y, λ) = x^2 + y^2 - λ(6x + 9y - 10), we'll differentiate with respect to each variable (x, y, and λ) while treating the others as constants:

1. Partial derivative with respect to x (f_x):
f_x = ∂f/∂x = 2x - λ(6)

2. Partial derivative with respect to y (f_y):
f_y = ∂f/∂y = 2y - λ(9)

3. Partial derivative with respect to λ (f_λ):
f_λ = ∂f/∂λ = -(6x + 9y - 10)

So, the partial derivatives are:
f_x = 2x - 6λ
f_y = 2y - 9λ
f_λ = -(6x + 9y - 10)

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Let θ be an angle in quadrant iii such that = cosθ − 45, the exact values of csc θ and tan θ are 5/3 and -3/4 respectively.

As an angle θ is in quadrant III, and cosθ = -4/5

To identify the exact value of cscθ and tanθ, we will first calculate the value of sinθ by using the Pythagorean identity

sin²θ + cos²θ = 1

⇒ sin²θ + (-4/5)² = 1

⇒ sin²θ = 1 - (-4/5)² = 1 - 16/25 = 9/25

⇒ sinθ = √(9/25) = 3/5

Now, we can calculate the value of cscθ as

cscθ = 1/sinθ = 1/(3/5) = 5/3

Next, we can estimate the value of tanθ by using the identity

tanθ = sinθ/cosθ= (3/5) / (-4/5) = -3/4

Therefore, the exact values of cscθ and tanθ are

cscθ = 5/3 and tanθ = -3/4.

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Answer:

9.38

Step-by-step explanation:

First let's start with what we know, we know the circumference and formula to get a circumference. (Formula: C= 2 π r) We also know how to find the circumference if we have the diameter. So what if we did it backwards? What if we divided? Let's try that, 29.48 ÷ π. Notice we don't put the radius because we don't know it yet. So when we do that on a calculator, we get 9.3885350 etc. Now we need to dived That by 2. That gives us 4.694267515 etc, but we can shorten that to 4.68. We're almost there! 4.68 is the RADIUS, not the diameter. so if you have the radius, you can easily get the diameter by mulitpling by 2. 4.68 x 2, is 9.38.

Answer:

b^(-6)

Step-by-step explanation:

The numerator of this fraction is b^(-11).  The -11 is the sum of -4 and -7.  

The denominator is b^(-5).

Recall that b^a / b^c = b^(a - c).  Thus, b^(-11) / b^(-5) = b^(-6)

Answer:

scale factor of lengths = 3

Step-by-step explanation:

Given 2 similar figures with scale factor of lengths = k

Then scale factor of areas = k²

Given

scale factor of areas = 9 , then

scale factor of lengths = \(\sqrt{9}\) = 3

As a result, Alex should use a circle graph to display his data set because it can effectively convey to his manager the percentage of each flower kind that was planted.

To visualise information and data, use a circle graph or pie chart. Typically, a circle graph is used to quickly and proportionately display the findings of an investigation.

A pie chart, often known as a circle chart, is a visual representation of the various values of a given variable or a means to summarise a set of nominal data. This kind of diagram consists of a circle with numerous segments.

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The valid conclusion is  A.  We can be 90% ... the mean amount ... is between $217 and $677.

The confidence interval used by the student researcher refers to the probability that the university's population parameter (average amount that students spent on sporting events last year) will fall between $217 and $677 for 90% of the time.

A confidence interval is the mean of your estimate plus and minus the variation in that estimate. This is the range of values you expect your estimate to fall between if you redo your test, within a certain level of confidence. Confidence, in statistics, is another way to describe probability.

Thus, the interval measurement gives the researcher some degree of certainty that the calculated sample mean falls within the population mean most of the time.

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The complete question is:-

marketing student is estimating the average amount of money that students at a large university spent on sporting events last year. He asks a random sample of 50 students at one of the university football games how much they spent on sporting events last year. Using this data he computes a 90% confidence interval, which turns out to be ($217, $677). Which one of the following conclusions is valid? We can be 90% confident that the mean amount of money spent at sporting events last year by all the students at this university is between $217 and $677. 90% of the sample said they spent between $217 and $677 at sporting events last year. No conclusion can be drawn.

Since both second partial derivatives are negative, the point (-10, -5) is a local maximum.

We need to find the points where the partial derivatives are equal to zero.

The partial derivatives of f(x, y) are:

∂f/∂x =\(3x^2 - 300\)

∂f/∂y = \(3y^2 - 75\)

Setting the partial derivatives equal to zero:

\(3x^2 - 300 = 0 ----(i)\)

\(3y^2 - 75 = 0 ----(ii)\)

Solving equations (i) and (ii) simultaneously:

From equation (i):

\(3x^2 = 300\)

\(x^2 = 100\)

x = ±10

From equation (ii):

\(3y^2 = 75\)

\(y^2 = 25\)

y = ±5

The critical points are (x, y) = (10, 5), (10, -5), (-10, 5), and (-10, -5).

We must utilize the second derivative test to ascertain the type of these important sites. To do this, we must determine f(x, y)'s second partial derivatives:

∂\(^2f/∂x^2 = 6x\)

∂\(^2f/\)∂x\(^2 = 6x\)

Now let's evaluate the second partial derivatives at each critical point:

For (x, y) = (10, 5):

∂\(^2f/\)∂\(x^2 = 6(10) = 60\) (positive)

∂\(^2f/\)∂\(y^2 = 6(5)\) = 30 (positive)

Since both second partial derivatives are positive, the point (10, 5) is a local minimum.

For (x, y) = (10, -5):

∂\(^2f/\)∂\(x^2 = 6(10) = 60\) (positive)

∂\(^2f/\)∂\(y^2 = 6(-5) = -30\) (negative)

Since the second partial derivative with respect to y is negative, the point (10, -5) is a saddle point.

For (x, y) = (-10, 5):

∂\(^2f/\)∂\(x^2 = 6(-10) = -60\) (negative)

∂\(^2f/\)∂\(y^2 = 6(5) = 30\)(positive)

Since the second partial derivative with respect to x is negative, the point (-10, 5) is a saddle point.

For (x, y) = (-10, -5):

∂\(^2f/\)∂\(x^2 = 6(-10) = -60\) (negative)

∂\(^2f/\)∂\(y^2 = 6(-5) = -30 (\)negative)

Since both second partial derivatives are negative, the point (-10, -5) is a local maximum.

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Answer:

7x+40=180

Step-by-step explanation:

(5x+40)+(2x)=180

7x+40=180

if you were to solve

subtract 40 from each side

7x=140

divide both sides by 7

x=20

check your answer

5(20)+40+2(20)=180

100+40+40=180

this is correct

hope this helps..and with future questions if you need to solve it

To find BCD

we know that C is 40 degrees and we know that  triangle is also 180

we can also see that B is a right triangle or 90 degree

so 90+40+A=180

130+A=180

subtract 130 from 180

A=50

Answer:

Step-by-step explanation:To solve the system of equations, we can set the two expressions for y equal to each other:

−x + 2 = 3x + 1

Solving for x, we get:

4x = 1

x = 1/4

Substituting this value of x into either of the original equations, we get:

y = −(1/4) + 2

y = 7/4

Therefore, the solution to the system of equations is the point (1/4, 7/4), which is the point of intersection of the two lines.

So the correct description of the solution to the system of equations is:

a

Line y = −x + 2 intersects line y = 3x + 1.

Answer:

A=3

Step-by-step explanation:

Step 1: Divide both sides by -6/

-6a/-6= -18/-6

a=3

The probability that an individual prefers biking given that he or she is 35 years old or older is 9/53.

From the definition of the probability we can get,

Conditional Probability to occur event A when B is already occurred is,

P(A|B) = P(A and B)/P(B)

Now let A be the event that an individual prefers biking and B be the event that individual is 35 years old or older.

Number of 35 years old and older individual is = 159

Number of 35 years old and older preferred biking = 27

Total number of individual = 516

So, P(A and B) = 27/516 and P(B) = 159/516

So the required probability that an individual prefers biking given that he or she is 35 years old or older is

= P(A | B)

= P(A and B)/P(B)

= (27/516)/(159/516)

= (27/516) * (516/159)

= 27/159

= 9/53

Hence, the required probability that an individual prefers biking given that he or she is 35 years old or older is 9/53.

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The question is incomplete. The complete question will be -

"The contingency table below provides frequencies for the preferred type of exercise for people under the age of 35, and those 35 years of age or older. Find the probability that an individual prefers biking given that he or she is 35 years old or older."

Answer:

1.06

Step-by-step explanation:

2.332 divided by 2.2

= 1.06

Hope this helped!

Answer:

1.06

Step-by-step explanation:

Answer:

0.542

Step-by-step explanation:

divide the liter by 1000 and get your answer! have a great day

542 liters is approximately equal to 0.542.

Given that a quantity, 542 L we need to convert it into kL,

To convert liters (L) to kiloliters (kL), you need to divide the given value in liters by 1,000 since there are 1,000 liters in 1 kiloliter.

In this case, to convert 542 L to kL, you can use the following formula:

kL = L / 1000

Plugging in the given value:

kL = 542 L / 1000

Dividing 542 by 1000, you get:

kL ≈ 0.542 kL

Therefore, 542 liters is approximately equal to 0.542.

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Answers:

===============================================

Explanation:

Let's convert the mixed number \(15 \frac{1}{4}\) to an improper fraction

\(a \frac{b}{c} = \frac{a*c+b}{c}\\\\15 \frac{1}{4} = \frac{15*4+1}{4}\\\\15 \frac{1}{4} = \frac{61}{4}\\\\\)

Do the same for the other mixed number.

\(a \frac{b}{c} = \frac{a*c+b}{c}\\\\4 \frac{3}{4} = \frac{4*4+3}{4}\\\\4 \frac{3}{4} = \frac{19}{4}\\\\\)

We'll divide those two improper fractions to find the number of pieces

\(15 \frac{1}{4} \div 4 \frac{3}{4}\\\\\frac{61}{4} \div \frac{19}{4}\\\\\frac{61}{4} \times \frac{4}{19}\\\\\frac{61}{19}\\\\\)

Then let's convert that to a mixed number like so:

\(\frac{61}{19}=\frac{57+4}{19}\\\\\frac{61}{19}=\frac{57}{19}+\frac{4}{19}\\\\\frac{61}{19}=3+\frac{4}{19}\\\\\frac{61}{19}=3\frac{4}{19}\\\\\)

Or you could note that 61/19 leads to 3 remainder 4.

The carpenter is able to get 3 whole pieces out, then the extra fractional part is 4/19 of a foot.

-----------------------------

Here's another way to look at it.

Grab a footlong ruler and split it into 4 equal smaller lengths. I'll refer to these smaller pieces as "quarters" from now on.

1 ruler = 4 quarters

15 rulers = 15*4 = 60 quarters

15 rulers + 1 quarter = 60+1 = 61 quarters

So this is another way to see how the mixed number \(15 \frac{1}{4}\) is the same as the improper fraction \(\frac{61}{4}\)

The same idea would apply to \(4 \frac{3}{4} = \frac{19}{4}\\\\\)

The overall entire board is 61 quarters long, and each piece is 19 quarters long. So that means 61/19 = 3 & 4/19 represents the number of pieces we can cut.

The carpenter is able to get 3 whole pieces out and then extra fractional part is 4/19 of a foot.

A fraction is a number is expressed as a "quotient and a numerator is divided by denominator".

According to the question,

A carpenter is cutting a 15 1/4 foot piece of lumber into smaller pieces.

The length of each piece is 4 3/4 feet.

Convert mixed fraction into improper fraction  

15 1/4 = 61/4

4 3/4 = 19/4

In order to find the whole piece and the fractional part divide 61/4 and 19/4

61/4 ÷ 19/4

= 61/19

= \(\frac{57+4}{19}\)

= \(\frac{57}{4}\) + \(\frac{4}{19}\)

= 3 + \(\frac{4}{19}\)

= \(3\frac{4}{19}\).

Hence, the carpenter is cutting 3 whole piece lumber and 4/19 fractional part.

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Answer:

maybe is number d trust me I did it rn for my test

Answer: b

Step-by-step explanation:

The probability density function (pdf) of the largest of the five waiting times is given by: f(x) = 4/10^5 * x^4, where x is a real number between 0 and 10. The expected value of the largest of the five waiting times is 8.33 minutes.

The pdf of the largest of the five waiting times can be found by considering the order statistics of the waiting times. The order statistics are the values of the waiting times sorted from smallest to largest.

In this case, the order statistics are X1, X2, X3, X4, and X5. The largest of the five waiting times is X5.

The pdf of X5 can be found by considering the cumulative distribution function (cdf) of X5. The cdf of X5 is given by: F(x) = (x/10)^5

where x is a real number between 0 and 10. The pdf of X5 can be found by differentiating the cdf of X5. This gives: f(x) = 4/10^5 * x^4

The expected value of X5 can be found by integrating the pdf of X5 from 0 to 10. This gives: E[X5] = ∫_0^10 4/10^5 * x^4 dx = 8.33

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The probability that a randomly selected product is produced by Line 1 can be calculated by multiplying the probability of selecting Product A (60%) with the probability of Product A being produced by Line 1 (40%), and similarly for Product B and Line 1.

P(Product from Line 1) = P(Product A) * P(Product A from Line 1) + P(Product B) * P(Product B from Line 1)

= 0.60 * 0.40 + 0.40 * 0.30

= 0.24 + 0.12

= 0.36

The probability that a randomly selected product is produced by Line 1 is 0.36, or 36%.

If a product is randomly selected from Line 1, the probability that it is Product B can be calculated by dividing the probability of selecting Product B (40%) with the probability of selecting a product from Line 1 (36%).

P(Product B from Line 1) = P(Product B) / P(Product from Line 1)

= 0.40 / 0.36

= 1.11 (rounded to two decimal places)

If a product is randomly selected from Line 1, the probability that it is Product B is approximately 1.11 or 111.11% (rounded to two decimal places). This means that there is a higher chance of selecting Product B from Line 1 compared to the overall production.

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The singular points of the differential equation are x = -6 and x = 0, which is an irregular singular point.

To find the singular points of the differential equation, we need to determine the values of x for which the coefficients of y'', y' and y become zero or infinite.

In this case, the coefficient of y'' is (x+6), which is zero only at x = -6. The coefficient of y' is -x^2, which is zero at x = 0. However, it is also infinite at x = 0, so we need to check if this is a regular or irregular singular point.

To do this, we can substitute y = (x^r) into the differential equation, where r is a constant. We get:

(x+6)r(r-1)x^(r-2) - x^r + 8x^r = 0

Simplifying, we get:

r(r-1) + 2r - 8 = 0

r^2 + r - 8 = 0

(r+2)(r-4) = 0

Thus, the possible values of r are -2 and 4. Substituting these back into y = (x^r), we get two solutions: y = x^(-2) and y = x^4.

Therefore, the singular points of the differential equation are x = -6 and x = 0, which is an irregular singular point.

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The Taylor polynomials T2 and T3 centered at x = 5 for the function f(x) = x^4 - 2x are T2(x) = 545 + 190(x - 5) + 150(x - 5)^2 and T3(x) = 545 + 190(x - 5) + 150(x - 5)^2 + 120(x - 5)^3.

a) For the function f(x) = sin(x), the Taylor polynomials T2 and T3 centered at a = 0 can be calculated as follows:

The Taylor polynomial of degree 2 for f(x) = sin(x) centered at x = 0 is:

T2(x) = f(0) + f'(0)x + (f''(0)/2!)x^2

= sin(0) + cos(0)x + (-sin(0)/2!)x^2

= x

The Taylor polynomial of degree 3 for f(x) = sin(x) centered at x = 0 is:

T3(x) = f(0) + f'(0)x + (f''(0)/2!)x^2 + (f'''(0)/3!)x^3

= sin(0) + cos(0)x + (-sin(0)/2!)x^2 + (-cos(0)/3!)x^3

= x - (1/6)x^3

Therefore, the Taylor polynomials T2 and T3 centered at x = 0 for the function f(x) = sin(x) are T2(x) = x and T3(x) = x - (1/6)x^3.

b) For the function f(x) = x^4 - 2x, the Taylor polynomials T2 and T3 centered at a = 5 can be calculated as follows:

The Taylor polynomial of degree 2 for f(x) = x^4 - 2x centered at x = 5 is:

T2(x) = f(5) + f'(5)(x - 5) + (f''(5)/2!)(x - 5)^2

= (5^4 - 2(5)) + (4(5^3) - 2)(x - 5) + (12(5^2))(x - 5)^2

= 545 + 190(x - 5) + 150(x - 5)^2

The Taylor polynomial of degree 3 for f(x) = x^4 - 2x centered at x = 5 is:

T3(x) = f(5) + f'(5)(x - 5) + (f''(5)/2!)(x - 5)^2 + (f'''(5)/3!)(x - 5)^3

= (5^4 - 2(5)) + (4(5^3) - 2)(x - 5) + (12(5^2))(x - 5)^2 + (24(5))(x - 5)^3

= 545 + 190(x - 5) + 150(x - 5)^2 + 120(x - 5)^3

Therefore, the Taylor polynomials T2 and T3 centered at x = 5 for the function f(x) = x^4 - 2x are T2(x) = 545 + 190(x - 5) + 150(x - 5)^2 and T3(x) = 545 + 190(x - 5) + 150(x - 5)^2 + 120(x - 5)^3.

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The given expressions, when A=1, B=0, and C=1, X evaluates to 1.001; when A=B=C=1, X evaluates to 1.111; and when A=B=C=0, X evaluates to 0.000. These evaluations are based on the given values of A, B, and C, and the notation ¯¯¯¯¯¯BC represents the complement of BC.

To evaluate the expression X = A.¯¯¯¯¯¯BC, we substitute the given values of A, B, and C into the expression.

(a) For A = 1, B = 0, and C = 1:

X = 1.¯¯¯¯¯¯01

To find the complement of BC, we replace B = 0 and C = 1 with their complements:

X = 1.¯¯¯¯¯¯01 = 1.¯¯¯¯¯¯00 = 1.001

(b) For A = B = C = 1:

X = 1.¯¯¯¯¯¯11

Similarly, we find the complement of BC by replacing B = 1 and C = 1 with their complements:

X = 1.¯¯¯¯¯¯11 = 1.¯¯¯¯¯¯00 = 1.111

(c) For A = B = C = 0:

X = 0.¯¯¯¯¯¯00

Again, we find the complement of BC by replacing B = 0 and C = 0 with their complements:

X = 0.¯¯¯¯¯¯00 = 0.¯¯¯¯¯¯11 = 0.000

In conclusion, when A = 1, B = 0, and C = 1, X evaluates to 1.001. When A = B = C = 1, X evaluates to 1.111. And when A = B = C = 0, X evaluates to 0.000. The evaluation of X is based on substituting the given values into the expression A.¯¯¯¯¯¯BC and finding the complement of BC in each case.

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