a big water tank contained 200 litres of water. during the day, 88 l 580 ml of water was used and 200 ml leaked. how much water was left in the water tank?

As the value of R-squared (R2) gets higher, the line will be a better fit for the data (Option A).

R-squared (R2) is a statistical measure that represents the proportion of the variance in the dependent variable that can be explained by the independent variable(s) in a regression model. It ranges from 0 to 1, with higher values indicating a better fit of the model to the data.

Option B is incorrect: The value of R2 can range from negative infinity to positive infinity, although it is commonly reported between 0 and 1. Negative R2 values occur when the regression model performs worse than a horizontal line, and values above 1 are not possible.

Option C is incorrect: R2 values above 1.0 are not possible as R2 represents the proportion of variance explained, which cannot exceed 100%.

Option D is incorrect: A value of 1.0 for R2 indicates that the regression model explains all the variance in the dependent variable, meaning there is no deviation of the data from the line. It does not indicate maximum deviation.

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To find out the number of groups of 1/5 in 3, we need to divide 3 by 1/5.

We can also write this as a fraction: 3 / (1/5)

To divide fractions, we flip the divisor and then multiply. This gives us:3 / (1/5) = 3 x 5/1 = 15So there are 15 groups of 1/5 in 3.To show this on a number line, we can first mark 0 and 3 on the number line.

Then we can draw 15 equally spaced tick marks between 0 and 3. Each tick mark represents 1/5, so 15 tick marks represent 15 groups of 1/5.

We can also label the tick marks with fractions to show that each tick mark represents 1/5.

The number line should look something like this:0 ------- 1/5 ------- 2/5 ------- 3/5 ------- 4/5 ------- 1 ------- 6/5 ------- 7/5 ------- 8/5 ------- 9/5 ------- 2 ------- 11/5 ------- 12/5 ------- 13/5 ------- 14/5 ------- 3

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The inner product of u and v is (-15).

In this problem, we are given two vectors, u and v, and asked to compute their inner product. The first step in calculating the inner product is to write the vectors in component form. We are given that

u = (1, -3) and v = (6, 4).

The next step is to compute the product of the corresponding components and sum them up. This gives us:

u · v = (1)(6) + (-3)(4) = 6 - 12 = -6

Therefore, the inner product of u and v is (-6).

Inner product is an important concept in linear algebra and has many applications in fields such as physics, engineering, and computer science. It is a way to measure the similarity between two vectors and can be used to find angles between vectors, project one vector onto another, and solve systems of linear equations.

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Answer:

x=9.3

Step-by-step explanation:

use SohCahToa

in this case u use cos

cos(41°)=7/x

x=7/cos(41)

x=9.275090953

x=9.3

To determine whether the matrix is in echelon form, reduced echelon form, or neither, let's first write the given matrix clearly:

[1 0 5 4]
[0 1 -5 -3]
[0 0 0 0]
[0 0 0 0]

Now, let's analyze its form:

a) Echelon form requires:
1. All nonzero rows are above any rows of all zeros.
2. The leading coefficient (pivot) of a nonzero row is always to the right of the pivot of the row above it.

This matrix satisfies both conditions, so it is in echelon form.

b) Reduced echelon form requires:
1. The matrix is in echelon form.
2. The pivot in each nonzero row is 1.
3. Each pivot is the only nonzero entry in its column.

This matrix fulfills the first two conditions, but the third condition is not met due to the presence of '5' in the first row and the same column as the pivot '1' in the second row.

Therefore, the matrix is in echelon form (option b) but not in reduced echelon form.

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The general antiderivative of n(x) = x⁸ + 5x⁴ + x⁵ is N(x) = (1/9)x⁹ + (1/5)x⁵ + (1/6)x⁶ + C.

To find the antiderivative of n(x) = x⁸ + 5x⁴ + x⁵, we apply the power rule for integration, which states that ∫x^n dx = (xⁿ⁺¹)/(n+1) + C, where C is the constant of integration.

1. For the first term, x⁸, integrate using the power rule: ∫x⁸ dx = (1/9)x⁹ + C₁.
2. For the second term, 5x⁴, integrate: ∫5x⁴ dx = 5(1/5)x⁵ + C₂ = x⁵ + C₂.
3. For the third term, x⁵, integrate: ∫x⁵ dx = (1/6)x⁶ + C₃.

Now, add the results of each integration and combine the constants: N(x) = (1/9)x⁹ + x⁵ + (1/6)x⁶ + (C₁ + C₂ + C₃). Since the constants are arbitrary, we can represent them as a single constant, C: N(x) = (1/9)x⁹ + (1/5)x⁵ + (1/6)x⁶ + C.

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The correct equation to represent this situation is D-95=16.50, where D represents the amount Ms. Park had to start with.

Let's assume that Ms. Park had D dollars to start with. After depositing $95, her balance would be D + $95.

But now, her balance is in debt by $16.50. So we can set up an equation:D + $95 = $16.50

The left side of the equation represents the amount Ms. Park has after depositing $95, while the right side of the equation represents the amount she owes.

We can then isolate D by subtracting $95 from both sides:D + $95 - $95 = $16.50 - $95D = $78.50

Therefore, Ms. Park had $78.50 to start with.

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Kim Barney's insurance plan has a $290.00 annual premium and a $500 deductible. The insurance company covers 80% of the remaining medical expenses after the deductible is met.

To calculate the amount Kim would pay out of pocket, we need to consider the deductible and the insurance company's coverage. The deductible is the initial amount Kim must pay before the insurance coverage kicks in. In this case, Kim's deductible is $500.00.

After paying the deductible, Kim's remaining expenses amount to $2,500.00 - $500.00 = $2,000.00. The insurance company covers 80% of this remaining expense, which is equal to 0.80 * $2,000.00 = $1,600.00.

Therefore, Kim would be responsible for paying the remaining 20% of the expense, which is equal to 0.20 * $2,000.00 = $400.00.

In summary, Kim would pay a total of $500.00 (deductible) + $400.00 (20% of the remaining expense) = $900.00 out of pocket for $2,500.00 in medical expenses.

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The correct degrees of freedom is 14. The two-sample t statistics for comparing the two population means has value -4.04. The p-value < 0.01. Margin of error is 2.14, 95% confidence interval is (12.86, 21.14). So, the correct answer is B).

The degrees of freedom for the t-test comparing the means of the two groups is df = 14, which is the sum of the sample sizes (8+8) minus two.

The two-sample t-statistic for comparing the two population means is

t = (X1 - X2) / (s_p * √(1/n1 + 1/n2))

where X1 is the sample mean of the high-capacity rats, X2 is the sample mean of the low-capacity rats, s_p is the pooled standard deviation, n1 is the sample size of the high-capacity rats, and n2 is the sample size of the low-capacity rats.

Using the given values, we get

t = (89 - 105) / (√(((8-1)*9² + (8-1)*13²) / (8+8-2)) * √(1/8 + 1/8))

t = -4.04

The p-value for testing the null hypothesis that the mean blood pressure of the high-capacity rats is equal to the mean blood pressure of the low-capacity rats is less than 0.01. So, the correct option is B).

The margin of error for a 95% confidence interval for the difference between the mean blood pressures of the two groups is approximately:

ME = t*(s_p * √(1/n1 + 1/n2))

Using the values from the previous calculations, we get

ME = 2.14

So the 95% confidence interval for the difference between the mean blood pressures of the two groups is approximately (105 - 89) ± 2.14, or (12.86, 21.14).

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The surface area of the cylindrical rod and the cost of a tin of paint indicates that minimum cost of painting the rod is £38.4

The surface area of the cylindrical rod can be found using the formula;

A = 2 × π × D²/4 × π × D × h

Where;

D = The diameter of the rod = 9 m

h = The height of the rod = 11 m

Therefore;

Surface area of the rod = 2 × π × 9²/4 × π × 9 × 11 ≈ 438.25

Surface area of the rod ≈ 438.25 m²

The area a tin of paint covers = 60 m²

The number of tins of paint required = 438.25/60 ≈ 7.3

Rounding up, we get;

The number of tins of paint required = 8 tins

Cost of the paint required = £4.80 × 8 = £38.4

The minimum cost of painting the rod is about £38.4

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Kevin and Randy have 34 quarters and 37 nickels in the jar.

System of equations for solving the problem is achieved using

the number of quarters as "q" and

the number of nickels as "n."

From the given information, we can set up the following equations

q + n = 71                            equation 1

0.25q + 0.05n = 10.35      equation 2

Multiply equation 1 by 0.05

0.05q + 0.05n = 0.05(71)

0.05q + 0.05n = 3.55        equation 3

Now, subtract equation 3 from equation 2

0.25q + 0.05n - (0.05q + 0.05n ) = 10.35 - 3.55

0.25q - 0.05q = 6.80

0.20q = 6.80

q = 6.80 / 0.20

q = 34

Substitute the value of q back into equation 1

34 + n = 71

n = 71 - 34

n = 37

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The lower and upper sums of the function f(x) = 8sin(x) is  Lf(P) = 8.1943 and Uf(P) = 14.7393.

To compute the lower and upper sums of the function f(x) = 8sin(x) for the partition P = {0, π/6, π/4, π/3, π/2}, we need to evaluate the function at each subinterval.

The length of the subintervals are:

Δx1 = π/6 - 0 = π/6

Δx2 = π/4 - π/6 = π/12

Δx3 = π/3 - π/4 = π/12

Δx4 = π/2 - π/3 = π/6

The lower sum is given by:

Lf(P) = f(0)Δx1 + f(π/6)Δx2 + f(π/4)Δx3 + f(π/3)Δx4

= 8sin(0)(π/6) + 8sin(π/6)(π/12) + 8sin(π/4)(π/12) + 8sin(π/3)(π/6)

= 0 + 2.3094 + 1.8849 + 4

Therefore, Lf(P) = 8.1943.

The upper sum is given by:

Uf(P) = f(π/6)Δx1 + f(π/4)Δx2 + f(π/3)Δx3 + f(π/2)Δx4

= 8sin(π/6)(π/6) + 8sin(π/4)(π/12) + 8sin(π/3)(π/12) + 8sin(π/2)(π/6)

= 2.3094 + 1.8849 + 4 + 6.5450

Therefore, Uf(P) = 14.7393.

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The slope for the regression equation is given by:

b = sp / ssx

where sp is the sum of products of deviations, and ssx is the sum of squared deviations of x scores.

Substituting the given values, we get:

b = 55 / 21

b ≈ 2.62

Rounding to 2 decimal places, we get the slope as 2.62.

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The limit of the Absolute value of the  rate is equal to 1, the rate Test is inconclusive.      

The confluence or divergence of the series

∑( n =  1 to  perpetuity)( 11n( n 6) ² ⋅ 6n 9),

we will use the rate Test.  The rate Test states that for a series

∑ aₙ, if the limit of the absolute value of the  rate of  consecutive terms is  lower than 1, the series converges absolutely.

However, the series diverges, If the limit is lesser than 1. still, the rate Test is inconclusive, and we need to consider other tests, If the limit equals 1 or the limit doesn't  live.  Let's apply the rate Test to the given series

 lim( n → ∞)|( aₙ ₊₁/ aₙ)|  where aₙ =  11n( n 6) ² ⋅ 6n 9.

To simplify the  computation, let's  estimate the  rate of  consecutive terms

|( aₙ ₊₁/ aₙ)| = |( 11( n 1)(( n 1) 6) ² ⋅ 6( n 1) 9)/( 11n( n 6) ² ⋅ 6n 9)|  

Simplifying  farther

( aₙ ₊₁/ aₙ)| = |( 11n 11)( n 7) ² ⋅ 6n 15/( 11n)( n 6) ² ⋅ 6n 9|  

Next, we take the limit as n approaches  perpetuity  

lim( n → ∞)|( aₙ ₊₁/ aₙ)| =  lim( n → ∞)|( 11n 11)( n 7) ² ⋅ 6n 15/( 11n)( n 6) ² ⋅ 6n 9|  

To  estimate this limit, we can simplify the expression inside the absolute value  lim( n → ∞)|( 11n 11)( n 7) ² ⋅

6n 15/( 11n)( n 6) ² ⋅ 6n 9|  =  lim( n → ∞)|( 11n 11)( n 7) ²/( 11n)( n 6) ²|  

Now, let's divide both the numerator and the denominator by n ²  

lim( n → ∞)|( 11 11/ n)( 1 7/ n) ²/( 11)( 1 6/ n) ²|  

Taking the limit as n approaches  perpetuity

lim( n → ∞)|( 11 11/ n)( 1 7/ n) ²/( 11)( 1 6/ n) ²|  = ( 11)( 1)( 1)/( 11)( 1)  =  1  

Since the limit of the absolute value of the  rate is equal to 1, the rate Test is inconclusive. thus, grounded on the rate Test, we know nothing about the confluence or divergence of the series. fresh tests,  similar as the Root Test or other confluence tests, may be  demanded to determine the behavior of the series.

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The most appropriate statistical method to determine factors that affect COVID-19 rates would be multivariate regression analysis.

Multivariate regression analysis is a statistical method used to determine the relationship between a dependent variable and several independent variables. In the case of the CDC trying to determine factors that affect COVID-19 rates, the dependent variable would be the COVID-19 rates, and the independent variables would be various factors that could affect the rates, such as age, gender, race, socioeconomic status, vaccination rates, and so on.

The multivariate regression analysis would allow the CDC to examine the relationship between each of these independent variables and the COVID-19 rates while controlling for the effects of the other independent variables. The regression analysis would also provide a way to quantify the strength and direction of each variable's effect on the COVID-19 rates.

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Suppose the initial balance is $1800, and the interest rate in the first year is 8 percent. In the second year, the interest rate would rise to 23 percent. We need to determine the difference in the monthly interest payable in years 1 and 2 in this case.  the difference in the monthly interest payable during years 1 and 2 is $22.5.

Here is how to compute the monthly interest for both years:

Year 1:In the first year, the interest rate is 8 percent.

Therefore, the monthly interest payable can be calculated as follows:

Monthly interest = (Annual interest rate x Balance)/12

Monthly interest = (8/100 x 1800)/12

Monthly interest = $12

Year 2:

In the second year, the interest rate is 23 percent.

Therefore, the monthly interest payable can be calculated as follows:

Monthly interest = (Annual interest rate x Balance)/12Monthly interest

= (23/100 x 1800)/12

Monthly interest = $34.5

Thus, the difference in the monthly interest payable between years 1 and 2 is:

$34.5 - $12

= $22.5.

Therefore, the difference in the monthly interest payable during years 1 and 2 is $22.5.

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The second boxplot is correct representation of the number of customers at each Slurpee Sam's Spaghetti Shop.

Given the data set is,

24, 25, 29, 30, 31, 31, 32, 34, 34

Hence, Minimum value = 24

Maximum value = 34

And, First quartile (Q1) = 1/4(n+1)th term

Q1 = 1/4 x 10 = 10/4 = 2.5th term = (25+29)/2 = 27

Q2 = 1/2(n+1)th term = 1/2(10) = 5th = 31

Q3 = 3/4(n+1)th term = 3/4(10) = 30/4 = 7.5th term = (7th +8th) term = (32 + 34) / 2 = 66/2 = 33

Hence, correct boxplot should have :

Minimum value = 24

Q1 = 27

Q2 = 31

Q3 = 33

Maximum value = 34

Thus, The second boxplot is correct.

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The conditional statement "T → (8<5)" is true because the antecedent "T" is false, and by the truth table of a conditional statement, a conditional with a false antecedent is always true, regardless of the truth value of the consequent.

what is antecedent?

In logic, an antecedent is the first part of a conditional statement (if-then statement) that precedes the word "if." It is the statement that implies or asserts the truth of the consequent. For example, in the conditional statement "If it is raining, then I will stay inside," the antecedent is "it is raining."

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Among the given expressions and equations, two are equations represented by triangles (A), while the remaining three are expressions represented by boxes().

The first equation, "2x + 9 = 2," is represented by a triangle (A) because it contains an equal sign, indicating that both sides are equal. The second expression, "32 + 3 x 9) = 59," is represented by a box () as it does not have an equal sign, making it an arithmetic expression rather than an equation.

The third equation, "3k + 7 = 34," is an equation and represented by a triangle (A) because it has an equal sign, signifying an equality between two expressions. The fourth expression, "5 (b + 28) = 150," is an expression and represented by a box () because it lacks an equal sign. It involves arithmetic operations but does not establish an equality.  

Finally, the fifth equation, "9a + 7 =," is an equation and represented by a triangle (A). Although it appears incomplete, it still contains an equal sign, indicating that the expression on the left side is equal to an unknown value on the right side.  

In summary, two equations are represented by triangles (A) because they contain equal signs and establish equalities between expressions, while the remaining three are expressions represented by boxes () as they lack equal signs and do not create equalities.

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The derivative of the function f(x) = 0 to x sec(7t) dt is sec^2(7x) * tan(7x).

The derivative of the function f(x) = 0 to x sec(7t) dt is sec(7x).

To see why, we use part one of the fundamental theorem of calculus, which states that if F(x) is an antiderivative of f(x), then the definite integral from a to b of f(x) dx is F(b) - F(a).

Here, we have f(x) = sec(7t), and we know that an antiderivative of sec(7t) is ln|sec(7t) + tan(7t)| + C, where C is an arbitrary constant of integration.

So, using the fundamental theorem of calculus, we have:

f(x) = 0 to x sec(7t) dt = ln|sec(7x) + tan(7x)| + C

Now, we can take the derivative of both sides with respect to x, using the chain rule on the right-hand side:

f'(x) = d/dx [ln|sec(7x) + tan(7x)| + C] = sec(7x) * d/dx [sec(7x) + tan(7x)] = sec(7x) * sec(7x) * tan(7x) = sec^2(7x) * tan(7x)

Therefore, the derivative of the function f(x) = 0 to x sec(7t) dt is sec^2(7x) * tan(7x).

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The statement is false. Every quantitative data set has a median, which is the middle value when the data is arranged in ascending or descending order. If there is an even number of data points, the median is the average of the middle two values.

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(a) The expected value of Y is :

E(Y) = (n+1)/3

(b) The value of E(Y^2) = (2n^2+5n+1)/6

(c) The variance of Y = (2n^2+5n+1)/6 - [(n+1)/3]^2

(d) P(X+Y=2) = 1/n

a) To find the expected value of Y, we use the law of total probability:

E(Y) = ∑ P(X=k)E(Y|X=k) for k=1 to n

Since Y is uniformly distributed on {1,2,...,X}, we have E(Y|X=k) = (k+1)/2.

Therefore,

E(Y) = ∑ P(X=k)(k+1)/2 for k=1 to n

To find P(X=k), note that X can take on any value from 1 to n with equal probability, so P(X=k) = 1/n for k=1 to n. Thus,

E(Y) = ∑ (k+1)/2n for k=1 to n

E(Y) = [1/2n ∑ k] + [1/2n ∑ 1] for k=1 to n

E(Y) = [1/2n (n(n+1)/2)] + [1/2n n]

E(Y) = (n+1)/3

b) To find E(Y^2), we use the law of total probability again:

E(Y^2) = ∑ P(X=k)E(Y^2|X=k) for k=1 to n

Since Y is uniformly distributed on {1,2,...,X}, we have E(Y^2|X=k) = (k^2+3k+2)/6. Therefore,

E(Y^2) = ∑ P(X=k)(k^2+3k+2)/6 for k=1 to n

Using the same values of P(X=k) as before, we get:

E(Y^2) = ∑ (k^2+3k+2)/6n for k=1 to n

E(Y^2) = [1/6n ∑ k^2] + [1/2n ∑ k] + [1/6n ∑ 1] for k=1 to n

E(Y^2) = [1/6n (n(n+1)(2n+1)/6)] + [1/2n (n(n+1)/2)] + [1/6n n]

E(Y^2) = (2n^2+5n+1)/6

c) The variance of Y is given by Var(Y) = E(Y^2) - [E(Y)]^2. Therefore,

Var(Y) = (2n^2+5n+1)/6 - [(n+1)/3]^2

d) To find P(X+Y=2), we note that X+Y=2 if and only if X=1 and Y=1. Since X is uniformly distributed on {1,2,...,n}, we have P(X=1) = 1/n. Since Y is uniformly distributed on {1,2,...,X}, we have P(Y=1|X=1) = 1. Therefore,

P(X+Y=2) = P(X=1)P(Y=1|X=1) = 1/n

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i) We can show that f is not continuous at x=0 using the method of Example 5.1.6. Consider the sequence {(−1)^n/n} which converges to 0 as n approaches infinity.

However, the image sequence {f((−1)^n/n)} oscillates between -1 and 1 and does not converge to f(0) which is 1. Hence, f is not continuous at x=0.

(ii) Since f(x) = 1 for x > 0, f^-1(1) is the set of all positive real numbers. Let c be any positive real number. Then, for any δ > 0, there exists a point x in the interval (c-δ, c+δ) such that f(x) = -1.

Hence, f is not continuous at any positive real number c. Therefore, f is not continuous on the entire real line R.

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The area enclosed by the given parametric curve and the y-axis is -4.5 square units.

To find the area enclosed by the given parametric curve and the y-axis, we can use the formula for calculating the area bounded by a parametric curve:

A = ∫ |x(t) dy/dt| dt

In this case, the parametric equations are:

x = t^2 - 3t

y = t

To calculate the derivative dy/dt, we differentiate y = t with respect to t:

dy/dt = 1

Now we can substitute the values into the area formula:

A = ∫ |(t^2 - 3t)(1)| dt

A = ∫ |t^2 - 3t| dt

To calculate the integral, we need to split it into two parts based on the absolute value:

A = ∫ (t^2 - 3t) dt (for t ≥ 0)

A = ∫ -(t^2 - 3t) dt (for t < 0)

Evaluating the integrals:

For t ≥ 0:

A = (1/3)t^3 - (3/2)t^2 + C1

For t < 0:

A = -(1/3)t^3 + (3/2)t^2 + C2

To find the specific bounds of integration, we need to determine the range of t that corresponds to the area enclosed by the curve and the y-axis. This can be done by finding the points where the curve intersects the y-axis.

Setting x = 0, we have:

0 = t^2 - 3t

t(t - 3) = 0

t = 0 or t = 3

Therefore, the bounds of integration will be from t = 0 to t = 3.

Substituting these bounds into the area formula, we get:

A = [(1/3)(3)^3 - (3/2)(3)^2] - [(1/3)(0)^3 - (3/2)(0)^2]

A = [(1/3)(27) - (3/2)(9)] - 0

A = 9 - 13.5

A = -4.5

The area enclosed by the given parametric curve and the y-axis is -4.5 square units. Note that the negative sign indicates that the curve is below the x-axis for part of the interval.

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Ryan needs to contribute $1000.07 per month.

To determine the monthly contribution needed, we will use the formula for monthly payment [tex]FV = P * [(1 + r)^n - 1] / r,[/tex]

Plugging values:

[tex]208,000 = P * [(1 + 0.078/12)^{11*12} - 1] / (0.078/12).\\208,000 = P * [1.0065^{132} - 1] / 0.0065.[/tex]

Rearranging to solve for P

[tex]P = 208,000 * 0.0065 / [1.0065^{132} - 1].[/tex]

P = 208,000 * 0.0065 / 1.35190003004

P = 1000.07394775

P = $1000.07

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The degrees of freedom for this two-mean non pooled hypothesis test is 15.

To find the degrees of freedom for a two-mean nonpooled hypothesis test, we use the following formula:

df = (s1^2/n1 + s2^2/n2)^2 / ( (s1^2/n1)^2 / (n1 - 1) + (s2^2/n2)^2 / (n2 - 1) )

Substituting the given values, we get:

df = (3^2/17 + 7^2/24)^2 / ( (3^2/17)^2 / (17 - 1) + (7^2/24)^2 / (24 - 1) )

= 14.97

Rounding to the nearest integer, we get:

df = 15

Therefore, the degrees of freedom for this two-mean non pooled hypothesis test is 15.

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To find the cartesian equation of a line tangent to r = 1-sinx, take the derivative of the polar equation to get the slope of the tangent line in terms of theta, convert the equation to cartesian coordinates using [tex]r = \sqrt{(x^2 + y^2) }[/tex] and  [tex]\theta = a$tan2(y,x),[/tex] substitute the values of r and theta at the point of tangency, and simplify the resulting equation.

To find the Cartesian equation of a tangent line to the curve given by polar equation  [tex]r=f(\theta)$ at a point $(r_0, \theta_0)$,[/tex] we can use the following steps:

Find the polar gradient of the curve, which is given by [tex]$dy/dx = (dy/d\theta)/(dx/d\theta) = (f'(\theta)\sin \theta + f(\theta)\cos \theta)/(f'(\theta)\cos \theta - f(\theta)\sin \theta)[/tex]

Evaluate the polar gradient at the point [tex]$(r_0, \theta_0)$[/tex] to obtain the slope of the tangent line.

Convert the polar coordinates of the point [tex]$(r_0, \theta_0)$[/tex]  to Cartesian coordinates  [tex]$(x_0, y_0)$[/tex] using the formulas [tex]x = r \cos \theta$ and $y = r \sin \theta[/tex]

Use the point-slope form of the equation of a line, which is given by [tex]y - y_0 = m(x - x_0)$,[/tex]

where m is the slope found in step 2.

Simplify the equation from step 4 to obtain the Cartesian equation of the tangent line.

Now, applying these steps to the given polar equation [tex]r = 1 - \sin \theta$,[/tex] we get:

[tex]$dy/dx = [(1 - \cos \theta) \cos \theta + (1 - \sin \theta) \sin \theta]/[(1 - \cos \theta) \sin \theta - (1 - \sin \theta) \cos \theta][/tex]

Evaluating the polar gradient at the point $[tex](r_0, \theta_0) = (1, \pi/2)$,[/tex] we get [tex]$dy/dx = -1[/tex].

Converting polar coordinates to Cartesian coordinates, we get [tex]$(x_0, y_0) = (0, 1)[/tex]

Using the point-slope form of the equation of a line, we get [tex]y - 1 = -1(x - 0)$.[/tex]

Simplifying the equation, we get [tex]$y = -x + 1$[/tex], which is the Cartesian equation of the tangent line.

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To find the Cartesian equation of a line tangent to the polar curve r=1-sin(theta), we need to first find the derivative of the equation with respect to theta using the polar derivative formula: dr/d(theta) = (dr/dx)(cos(theta)) + (dr/dy)(sin(theta)).

           Using this formula, we get: dr/d(theta) = -cos(theta)
Next, we need to find the value of theta at the point of tangency. For a curve in polar coordinates, the tangent line at a point with polar coordinates (r,theta) corresponds to the line through (r,theta) with slope -dr/d(theta). Therefore, the tangent line to r=1-sin(theta) at theta=t will have slope -cos(t).
Now, we can use the point-slope equation of a line to find the Cartesian equation of the tangent line: y-y1 = m(x-x1), where (x1,y1) is the point of tangency. The Cartesian equation of the line tangent to the polar curve r=1-sin(theta) at theta=t is therefore: y - (1-sin(t)) = -cos(t)(x - 0), or y = -cos(t)x + 1 + sin(t).

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Let's assume that the marked price of the article is "M" dollars. The marked price of the article is approximately $4941.18.

According to the problem statement, the dealer gives a discount of 10%, so the selling price (S) of the article is:

S = M - 0.10M = 0.90M

Now, the GST of 12% is applied on the marked price, so the amount of GST paid is:

GST = 0.12M

Therefore, the total amount paid by the consumer (C) is:

C = S + GST

C = 0.90M + 0.12M

C = 1.02M

We are given that the consumer pays $5040, so we can set up the equation:

1.02M = 5040

Solving for M, we get:

M = 5040 / 1.02

M ≈ 4941.18

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The probability of rolling an even number is the sum of the probabilities of rolling 2 and 4 and 6, which is:

1/12 + 1/12 + 3/12 = 5/12

Let p be the probability of rolling each of the numbers 1, 2, 3, and 4. Since 5 and 6 are three times as likely to come up as each of the other numbers, the probabilities of rolling 5 and 6 are 3p each. The sum of all probabilities must be equal to 1, so we have:

p + p + p + p + 3p + 3p = 1

Simplifying this equation, we get:

12p = 1

p = 1/12

Therefore, the probability distribution is:

Outcome 1 2 3 4 5 6

Probability 1/12 1/12 1/12 1/12 3/12 3/12

The probability of rolling an even number is the sum of the probabilities of rolling 2 and 4 and 6, which is:

1/12 + 1/12 + 3/12 = 5/12

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