Bijan wants to go running during his family’s vacation to New York City. To do so, he will run a neighborhood block 20 times. Bijan runs a total of 8 miles. Use the formula for the perimeter of the neighborhood block and the reciprocal to find the width w of the city block

The 90% confidence interval for the difference between the proportions of all users of the two toothpaste who will never switch is (0.009, 0.101).

To construct a 90% confidence interval for the difference between the proportions of all users of the two toothpaste who will never switch, we can use the following formula:

CI = (p1 - p2) ± Zα/2 * √(p1(1-p1)/n1 + p2(1-p2)/n2)

where p1 and p2 are the sample proportions, n1 and n2 are the sample sizes, and Zα/2 is the critical value from the standard normal distribution for the desired confidence level.

Plugging in the values given in the problem, we get:

p1 = 90/450 = 0.20

p2 = 80/550 = 0.145

n1 = 450

n2 = 550

α = 0.10 (since we want a 90% confidence interval, which corresponds to a significance level of 0.10)

Zα/2 = 1.645 (from the standard normal distribution table)

Substituting these values into the formula, we get:

CI = (0.20 - 0.145) ± 1.645 * √((0.20 * 0.80 / 450) + (0.145 * 0.855 / 550))

Simplifying, we get:

CI = 0.055 ± 0.046

Therefore, the 90% confidence interval for the difference between the proportions of all users of the two toothpaste who will never switch is (0.009, 0.101).

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Answer: Answer below in pic

Step-by-step explanation:

:)

The correct interpretation of the given confidence interval is C. We are 98% confident that the proportion of obese clients losing weight under strategy 1 is between 13% less and 9% more than the proportion of obese clients losing weight under strategy 2.

The confidence interval is given as (-0.13, 0.09), which represents the range within which the true difference in proportions (p1 - p2) is likely to fall with a confidence level of 98%.

The negative value of -0.13 indicates that the proportion of obese clients losing weight under strategy 1 may be 13% less than the proportion under strategy 2.

The positive value of 0.09 indicates that the proportion of obese clients losing weight under strategy 1 may be 9% more than the proportion under strategy 2.

Since the confidence interval includes both positive and negative values, it suggests that the true difference in proportions could be either positive or negative.

The confidence level of 98% means that if we were to repeat the study and construct 100 different confidence intervals, about 98 of those intervals would capture the true difference in proportions (p1 - p2).

Therefore, we can conclude that we are 98% confident that the proportion of obese clients losing weight under strategy 1 is between 13% less and 9% more than the proportion of obese clients losing weight under strategy 2.

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The probability that between 124 and 125​, inclusive are on time is approximately 0.0655.

Given:

The probability of a flight arriving on time is 0.88

Number of flights selected randomly = 145

Let X be the number of flights arriving on time.

(a) P(exactly 128 flights are on time)

Using the normal approximation to the binomial distribution, we have:

Mean, µ = np = 145 × 0.88 = 127.6

Standard deviation, σ = sqrt(np(1-p)) = sqrt(145 × 0.88 × 0.12) = 3.238

P(X = 128) can be approximated using the standard normal distribution:

z = (128 - µ) / σ = (128 - 127.6) / 3.238 = 0.1234

P(X = 128) ≈ P(z = 0.1234) = 0.4511

Therefore, the probability that exactly 128 flights are on time is approximately 0.4511.

(b) P(at least 128 flights are on time)

P(X ≥ 128) can be approximated as:

z = (128 - µ) / σ = (128 - 127.6) / 3.238 = 0.1234

P(X ≥ 128) ≈ P(z ≥ 0.1234) = 0.4515

Therefore, the probability that at least 128 flights are on time is approximately 0.4515.

(c) P(fewer than 124 flights are on time)

P(X < 124) can be approximated as:

z = (124 - µ) / σ = (124 - 127.6) / 3.238 = -1.1154

P(X < 124) ≈ P(z < -1.1154) = 0.1326

Therefore, the probability that fewer than 124 flights are on time is approximately 0.1326.

(d) P(between 124 and 125​, inclusive are on time)

P(124 ≤ X ≤ 125) can be approximated as:

z1 = (124 - µ) / σ = (124 - 127.6) / 3.238 = -1.1154

z2 = (125 - µ) / σ = (125 - 127.6) / 3.238 = -0.7388

P(124 ≤ X ≤ 125) ≈ P(-1.1154 ≤ z ≤ -0.7388) = P(z ≤ -0.7388) - P(z < -1.1154)

P(124 ≤ X ≤ 125) ≈ 0.1981 - 0.1326 = 0.0655

Therefore, the probability that between 124 and 125​, inclusive are on time is approximately 0.0655.

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The area of the shape based on the information will be 90 inches ².

The area of a shape simply means the total space that is taken by the shape. It simply expresses the extent of the region on a particular plane as well as a curved surface.

The area of the shape based on the information will be:

= 1/2 b × h

= 1/2 × 15 × 12

= 15 × 6

= 90 inches ².

In conclusion, the area of the shape based on the information will be 90 inches ².

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Answer:

Plot the points on the graphing calculator, and then generate the linear regression equation:

y = 18.1x + 104.1

2022 is 11 years since 2011, so

y = 18.1(11) + 104.1 = 303.2

The projected profit for 2002 is about

$303 thousand.

By using the area of the square There are 1,400 plants per square meter.

To calculate this, we first need to determine the area of the square meter in terms of the row spacing and number of plants per meter. Since the row spacing is 25 cm, or 0.25 meters, we can fit 4 rows per meter (1 meter / 0.25 meters = 4 rows). Therefore, the area of the square meter would be 4 rows x 1 meter = 4 square meters.

Next, we need to determine the total number of plants in this 4 square meter area. Since there are 35 plants per linear meter, there are 35 plants per row. So, the total number of plants in the 4 square meter area would be 35 plants/row x 4 rows = 140 plants.

Finally, to determine the number of plants per square meter, we divide the total number of plants by the area of the square meter:

Therefore, there are 1,400 plants per square meter.

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The graph of the reflected rhombus P'Q'R'S' is shown below.

We know that the formula for the reflection of point (a,b) with respect to line y = k is point (a, 2k-b)

i.e., the coordinates of point A(x, y) changes to (x, 2k - y)

Here, the rhombus PQRS with vertices P(-8, 6), Q(-4, 8), R(0, 6), and S(-4, 4) reflected over the line y = 2.

This means that the value of k = 2

P(-8,6) ⇒ P′(-8,2⋅2-6)

            = P′(-8,-2)

Q(-4, 8)⇒ Q′(-4,2⋅2-8)

            = Q′(-4,-4)

R(0, 6) ⇒ R′(0, 2⋅2-6)

             = R′(0,-2)

S(-4,4) ⇒ S′(-4,2⋅2-4)

           = S′(-4,0)

Therefore, the coordintes of reflected rhombus P'Q'R'S' are:

P′(-8,-2), Q′(-4,-4), R′(0,-2), S′(-4,0)

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The complete question is:

Rhombus PQRS with vertices P(-8, 6), Q(-4, 8), R(0, 6), and S(-4, 4) REFLECTED over the line y = 2.

Graph the reflected rhombus.

We reject the null hypothesis and conclude that the data supports the claim that the mean height of high school students who play basketball on the school team is less than 73 inches.

Null Hypothesis: The mean height of high school students who play basketball on the school team is 73 inches or greater.

Alternative Hypothesis: The mean height of high school students who play basketball on the school team is less than 73 inches.

We are given a sample size of 40 players with a sample mean of 71 inches and a sample standard deviation of 1.5 inches.

To test our hypothesis, we will use a one-sample t-test with a significance level of 0.05.

Using a t-distribution table with 39 degrees of freedom (n-1), we find the critical t-value to be -1.686.

We calculate the test statistic as:

t = (sample mean - hypothesized mean) / (sample standard deviation / sqrt(sample size))

t = (71 - 73) / (1.5 / sqrt(40)) = -4.38

Using a t-distribution table with 39 degrees of freedom, we find the p-value to be almost zero (less than 0.0001).

Since the p-value is less than the significance level of 0.05, we reject the null hypothesis and conclude that the data supports the claim that the mean height of high school students who play basketball on the school team is less than 73 inches.

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The percent decrease in the mower's price to the nearest tenth of a percent is 25.3%.

We have,

We need to calculate the initial discount given by the 23% off sale:

Discount

= 0.23 x $425

= $97.75

After the first discount, the mower's price is:

New price

= $425 - $97.75

= $327.25

Then, the store manager discounted it by another $10, so the final price is:

Final price

= $327.25 - $10

= $317.25

The total decrease in price is:

= $425 - $317.25

= $107.75

The percent decrease in the mower's price is:

Percent decrease

= (107.75 / 425) x 100%

= 25.3%

Therefore,

The percent decrease in the mower's price to the nearest tenth of a percent is 25.3%.

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The particular solution for the given differential equation when y(1) = 1 is:
arctan(y) = x + π/4 - 1

The given equation is:
dy/dx = y² + 1

To solve this first-order, nonlinear, ordinary differential equation, we can use the separation of variables method. Here are the steps:

1. Rewrite the equation to separate variables:
dy/(y² + 1) = dx

2. Integrate both sides:
∫(1/(y² + 1)) dy = ∫(1) dx

On the left side, the integral is arctan(y), and on the right side, it's x + C:
arctan(y) = x + C

Now, we'll find the particular solution using the initial condition y(1) = 1:
arctan(1) = 1 + C

Since arctan(1) = π/4, we can solve for C:

π/4 = 1 + C
C = π/4 - 1

So, the particular solution for the given differential equation when y(1) = 1 is:

arctan(y) = x + π/4 - 1

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A one-tailed test has a P-value of 0.01 and a two-tailed test has a P-value of 0.053.

We must utilize the F-distribution table or calculator to address this question. The F test follows an F-distribution since it is a ratio of two variances.

a. To calculate the P-value for a one-tailed test, we must first calculate the likelihood that the F-statistic is greater than or equal to 7.46.

Using a table, we can find the area under the F-distribution curve to the right of 7.46 with 5 degrees of freedom in the numerator and 7 degrees of freedom in the denominator.

Assuming an alpha level of 0.05, we reject the null hypothesis if the P-value is less than 0.05.

For the given F-statistic of 7.46, the P-value for a one-tailed F-test with five degrees of freedom in the numerator and seven degrees of freedom in the denominator is approximately 0.01.

b. For a two-tailed test, we must calculate the likelihood of seeing an F-statistic that is 7.46 or more in either direction. This means we must calculate the area under the F-distribution curve in the distribution's two tails.

To find the p-value for a two-tailed F-test with 5 and 7 degrees of freedom and a test statistic of 7.46, we need to use an F-distribution table.

Using a table, we would look for the intersection of the row with 5 degrees of freedom and the column with 7 degrees of freedom, which gives us the critical F-value for a significance level of 0.05 (assuming equal variances in the two populations being compared). The critical F-value is 4.03.

Since the test statistic of 7.46 is greater than the critical F-value of 4.03, the p-value for the two-tailed test is less than 0.05. Specifically, the p-value is the probability of observing an F-statistic as extreme or more extreme than 7.46, which is the probability in the right tail of the F-distribution plus the probability in the left tail:

p-value = P(F > 7.46) + P(F < 1/7.46)

       = 0.052 + 0.001

       ≈ 0.053

Therefore, the p-value for the two-tailed test is approximately 0.053.

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The numerical length of JK is 6 based on the expression of segments of JL, JK and KL.

The complete segment JL is made up of constituent small segments JK and KL. So, using this relation to find the length of JK by relaying the expression.

JL = JK + KL

4x + 2 = 3x + 5x - 6

Performing addition on Right Hand Side of the equation

4x + 2 = 8x - 6

Rewriting the equation

8x - 4x = 6 + 2

Performing subtraction and addition on Left and Right Hand Side of the equation

4x = 8

x = 8/4

Performing division

x = 2

So, the length of JK = 3×2

Length of JK = 6

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The marketing firm should sample at least 752 people to estimate the approval rating for their next advertising campaign with a margin of error of 3% and 90% confidence.

To estimate the sample size for a marketing firm's advertising campaign approval rating with a margin of error of 3% and a 90% confidence level, you should use the following steps:

Step 1: Identify the critical value for a 90% confidence level. For a 90% confidence level, the critical value (z-score) is approximately 1.645.

Step 2: Determine the population proportion (p) and its complement (q). Since we don't have a specific proportion, we can use the conservative estimate of p=0.5 and q=0.5, which will result in the largest required sample size.

Step 3: Calculate the required sample size (n) using the formula:

n = (Z^2 * p * q) / E^2

where Z is the critical value, p and q are the population proportions, and E is the margin of error.

n = (1.645^2 * 0.5 * 0.5) / 0.03^2
n ≈ 751.18

Since we cannot have a fraction of a person in a sample, we'll round up to the nearest whole number.

Therefore to achieve a margin of error of 3% with a 90% confidence level, the marketing firm should sample approximately 752 people for their next advertising campaign's approval rating.

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The area of the square is increasing at a rate of 64 cm²/s when the area of the square is 16 cm² and each side is increasing at a rate of 8 cm/s.

The area of the rectangle is increasing at a rate of 67 cm²/s when the length is 13 cm and the width is 4 cm, and the length and width are increasing at rates of 3 cm/s and 5 cm/s, respectively.

The volume is increasing at the rate of 14400π mm³/s when the diameter is 60 mm.

We have,

1)

Each side of a square is increasing at a rate of 8 cm/s.

Let's use the formula for the area of a square:

A = s², where s is the length of the side of the square.

We are given that ds/dt = 8 cm/s, where s is the length of the side of the square, and we want to find dA/dt when A = 16 cm^2.

Using the chain rule, we can find dA/dt as follows:

dA/dt = d/dt (s^2) = 2s(ds/dt)

When A = 16 cm²,

s = √(A) = √(16) = 4 cm.

When A = 16 cm²,

dA/dt = 2s(ds/dt) = 2(4)(8) = 64 cm^2/s

So the area of the square is increasing at a rate of 64 cm²/s when the area of the square is 16 cm² and each side is increasing at a rate of 8 cm/s.

2)

The length of a rectangle is increasing at a rate of 3 cm/s and its width is increasing at a rate of 5 cm/s.

Let's use the formula for the area of a rectangle:

A = lw, where l is the length and w is the width.

We are given that dl/dt = 3 cm/s and dw/dt = 5 cm/s, and we want to find dA/dt when l = 13 cm and w = 4 cm.

Using the product rule, we can find dA/dt as follows:

dA/dt = d/dt (lw) = w(dl/dt) + l(dw/dt)

When l = 13 cm and w = 4 cm, we have:

dA/dt = w(dl/dt) + l(dw/dt) = 4(3) + 13(5) = 67 cm²/s

So the area of the rectangle is increasing at a rate of 67 cm^2/s when the length is 13 cm and the width is 4 cm, and the length and width are increasing at rates of 3 cm/s and 5 cm/s, respectively.

3)

The radius of a sphere is increasing at a rate of 4 mm/s.

Let's use the formulas for the radius and volume of a sphere:

r = d/2 and V = (4/3)πr^3, where d is the diameter.

We are given that dr/dt = 4 mm/s when d = 60 mm, and we want to find dV/dt.

Using the chain rule, we can find dV/dt as follows:

dV/dt = d/dt [(4/3)πr^3] = 4πr^2(dr/dt)

When d = 60 mm, we have r = d/2 = 30 mm.

dV/dt = 4πr²(dr/dt) = 4π(30)²(4) = 14400π mm³/s

Thus,

The area of the square is increasing at a rate of 64 cm²/s when the area of the square is 16 cm² and each side is increasing at a rate of 8 cm/s.

The area of the rectangle is increasing at a rate of 67 cm²/s when the length is 13 cm and the width is 4 cm, and the length and width are increasing at rates of 3 cm/s and 5 cm/s, respectively.

The volume is increasing at the rate of 14400π mm³/s when the diameter is 60 mm.

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Answer:

C. zero

Step-by-step explanation:

The equation for this graph is y = 2

Because the slope is 0 and the y-intercept is 2, that is why the line runs across y = 2.

Answer: I believe your answer is number one

Step-by-step explanation:

The 20th percentile for incubation times is 22 days. The incubation times that make up the middle 95% are between 21 and 25 days.

(a) To determine the 20th percentile for incubation times, follow these steps:

1. Find the z-score corresponding to the 20th percentile using a standard normal distribution table or calculator.

For the 20th percentile, you'll look for an area of 0.20. The z-score is approximately -0.84.
2. Use the following formula to convert the z-score to the incubation time (X): X = μ + (z × σ),

where μ is the mean, z is the z-score, and σ is the standard deviation.
3. Plug in the values: X = 23 + (-0.84 × 1) = 23 - 0.84 = 22.16
4. Round to the nearest whole number: X ≈ 22 days
The 20th percentile for incubation times is 22 days.

(b) To determine the incubation times that make up the middle 95%, follow these steps:

1. Find the z-scores corresponding to the lower and upper bounds of the middle 95%. You'll look for areas of 0.025 and 0.975 in the standard normal distribution table. The z-scores are approximately -1.96 and 1.96.
2. Use the formula X = μ + (z × σ) to convert the z-scores to incubation times.
3. For the lower bound, plug in the values: X = 23 + (-1.96 × 1) = 23 - 1.96 = 21.04
4. For the upper bound, plug in the values: X = 23 + (1.96 × 1) = 23 + 1.96 = 24.96
5. Round to the nearest whole number: Lower bound ≈ 21 days, Upper bound ≈ 25 days

The incubation times that make up the middle 95% are between 21 and 25 days.

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The formula for finding the slope and length of a segment indicates;

Slope of [tex]\overline{QR}[/tex] = -7, length of [tex]\overline{QR}[/tex] = 5·√2

Slope of [tex]\overline{RS}[/tex] = -1, length of [tex]\overline{RS}[/tex] = 5·√2

Slope of [tex]\overline{ST}[/tex] = -7, length of [tex]\overline{ST}[/tex] = 5·√2

Slope of [tex]\overline{TQ}[/tex] = -1, length of [tex]\overline{TQ}[/tex] = 5·√2

The length of a segment on a coordinate plane can be found using the distance formula for finding the distance, d, between two points (x₁, y₁), and (x₂, y₂), which can be expressed as follows;

d = √((x₂ - x₁)² + (y₂ - y₁)²))

The slope of [tex]\overline{QR}[/tex] = (3 - (-4))/(5 - 6) = -7

The length of [tex]\overline{QR}[/tex] = √((3 - (-4))² + (5 - 6)²) = 5·√2

The slope of [tex]\overline{RS}[/tex] = (8 - 3)/(0 - 5) = -1

The length of [tex]\overline{RS}[/tex] = √((8 - 3)² + (0 - 5)²) = 5·√2

The slope of [tex]\overline{ST}[/tex] = (8 - 1)/(0 - 1) = -7

The length of [tex]\overline{ST}[/tex] = √((8 - 1)² + (0 - 1)²) = 5·√2

The slope of [tex]\overline{TQ}[/tex] = (-4 - 1)/(6 - 1) = -1

The length of   [tex]\overline{TQ}[/tex]   = √((-4 - 1)² + (6 - 1)²) = 5·√2

The quadrilateral QRST can best be described as a rhombus

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To find the cumulative probability distribution function (CDF) F(y), we need to integrate the given PDF f(y) from 0 to y:

F(y) = integral of f(y) dy from 0 to y

= integral of c*y*(1-y)^2 dy from 0 to y   (substituting c=30 from part a)

= 30*integral of y*(1-y)^2 dy from 0 to y

To integrate this, we can use integration by substitution. Let u = 1 - y, then du/dy = -1 and y = 1 - u. Substituting, we get:

F(y) = 30*integral of (1-u)*u^2 * (-du) from 0 to 1-y

= 30*integral of u^2 - u^3 du from 0 to 1-y

= 30*[u^3/3 - u^4/4] evaluated at 0 and 1-y

= 10*(1 - (1-y)^3 - 3(1-y)^4/4),   0 <= y <= 1

Therefore, the cumulative probability distribution function (CDF) of Y is:

F(y) = {

        0,                             y < 0

        10*(1 - (1-y)^3 - 3(1-y)^4/4), 0 <= y <= 1

        1,                             y > 1

      }

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An ellipse is a geometric shape that looks like a flattened circle, with two focal points. The standard form of the equation of an ellipse with center at the origin is (x^2/a^2) + (y^2/b^2) = 1, where a and b are the lengths of the major and minor axes, respectively.

To find the standard form of the equation of an ellipse with given characteristics and center at the origin, we first need to identify the values of a and b. The major axis is the longer axis of the ellipse, while the minor axis is the shorter axis. If we know the length of the major and minor axes, we can easily find a and b.

Once we have identified a and b, we can plug them into the standard form equation and simplify it to find the equation of the ellipse. For example, if the length of the major axis is 8 and the length of the minor axis is 6, then a = 4 and b = 3. We can plug these values into the equation (x^2/4^2) + (y^2/3^2) = 1 and simplify it to get the standard form of the equation of the ellipse.

In conclusion, finding the standard form of the equation of an ellipse with given characteristics and center at the origin involves identifying the values of a and b, and then plugging them into the standard form equation and simplifying it.

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The answer is: a. It would converge to the shape of a normal distribution.

When you increase your sample size, more data points are included in the sample, resulting in a more accurate representation of the population. As a result, the distribution of the sample means will approach a normal distribution, known as the Central Limit Theorem. This means that the shape of the sampling distribution will become more symmetrical and bell-shaped as the sample size increases.

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It is important to interpret the results with caution and consider the potential limitations and sources of error in the data.

The Type of Dangerous Restraint Technique Training has a statistically significant impact on the number of lawsuits, injuries, and citizen complaints. The F-ratios for these three negative consequences are greater than the critical value, and their p-values are less than the alpha level of 0.05, indicating that the null hypothesis that there is no difference between the means of the groups can be rejected. This means that the Type of Dangerous Restraint Technique Training is associated with significant differences in the number of lawsuits, injuries, and citizen complaints.

The Type of Dangerous Restraint Technique Training does not have a statistically significant impact on the number of deaths. The F-ratio for the number of deaths is less than the critical value, and its p-value is greater than the alpha level of 0.05, indicating that the null hypothesis cannot be rejected. This means that the Type of Dangerous Restraint Technique Training is not associated with a significant difference in the number of deaths.

The Type of Dangerous Restraint Technique Training has its most statistically significant impact on the number of citizen complaints. The F-ratio for citizen complaints is the highest among all the negative consequences, and its p-value is the lowest, indicating that the Type of Dangerous Restraint Technique Training is associated with the most significant difference in the number of citizen complaints.

There are a few potential concerns about the data in the table. Firstly, the sample of police academies may not be representative of all police academies in the country, which may limit the generalizability of the findings. Secondly, the data may be subject to reporting bias or measurement error, which may affect the accuracy and reliability of the results. Thirdly, the ANOVA assumes that the data meet certain assumptions, such as normality and homogeneity of variances, which may not be met in this case. For example, the number of deaths is highly skewed towards the high end, and the variances of the groups may not be equal. These violations of assumptions may affect the validity and robustness of the ANOVA results. Therefore, it is important to interpret the results with caution and consider the potential limitations and sources of error in the data.

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Answer:

Area= Length × Width

Area= 45 × 15 (you get 45 as your Length because it says in the equation that the Length of the end table is 45 inches and you get 15 as the width because in the equation it says the width is 15 inches)

Answer = 45 × 15=675

At one time, the British advanced corporation tax system taxed British companies' foreign earnings at a higher rate than their domestic earnings.

This was put in place to discourage multinational corporations from artificially shifting profits earned in the UK to low-tax jurisdictions. The policy was aimed at preventing companies from avoiding tax by moving profits out of the UK and into tax havens. By imposing a higher tax rate on foreign earnings, the UK government hoped to make it less attractive for companies to engage in profit-shifting practices.

The policy was controversial and faced criticism from some business groups, who argued that it placed an unfair burden on companies operating overseas. However, the government defended the policy as necessary to ensure that companies paid their fair share of tax in the countries where they operated. Eventually, the policy was replaced by a territorial tax system, which only taxes companies on their profits earned in the UK. This change was made to simplify the tax system and make it more attractive for companies to invest in the UK.

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The solution is, : Factor out the greatest common factor, then solving the equation 4(2x – 1) + 8 = 4x + 24, we get, x=5.

Here, we have,

given that,

4(2x – 1) + 8 = 4x + 24.

Factor out a 4 from each side

4{ 2x-1 +2} = 4(x+6)

Cancel the 4 on each side

2x-1+2 = x+6

Combine like terms

2x+1 = x+6

Subtract x from each side

2x+1-x = x+6-x

x+1 = 6

Subtract 1 from each side

x+1-1 = 6-1

x = 5

Factor out the greatest common factor, then solving the equation 4(2x – 1) + 8 = 4x + 24, we get, x=5.

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complete question:

Solve the equation 4(2x – 1) + 8 = 4x + 24. Factor out the greatest common factor, then

solve.

a) The team can be chosen in 126 ways if all four members have the same role on the team.

b) The team can be chosen in 1260 ways if all four members have different roles on the team.

a) When all four members have the same role, we simply need to select 4 employees out of 9 eligible ones. This can be done in 9C4 ways, which is equal to 126.

b) When all four members have different roles, we need to select one employee for each of the four roles. The first employee can be chosen in 9 ways, the second in 8 ways (as one employee has already been chosen), the third in 7 ways, and the fourth in 6 ways.

Therefore, the total number of ways to select the team is 9 x 8 x 7 x 6, which is equal to 1260.

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The angles which must be right angles include the following:

∠PLJ

∠KLJ

∠PEJ

∠JEB

In Mathematics and Geometry, a perpendicular bisector is a line that bisects or divides a line segment exactly into two (2) equal halves and forms an angle that has a magnitude of 90 degrees at the point of intersection.

In Mathematics and Geometry, a right angle can be defined as a type of angle that is formed in a triangle by the intersection of two (2) straight lines at 90 degrees. This ultimately implies that, a right angled triangle has a measure of 90 degrees.

Based on the diagram shown in the image attached below, we can reasonably infer and logically deduce that angles formed at L and E must be right angles.

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Missing information:

The question is incomplete and the complete question is shown in the attached picture.