use vectors to decide whether the triangle with vertices p(2, −1, −1), q(3, 2, −3), and r(7, 0, −4) is right-angled.

The missing values in the triangle are:

∠B = 90°

AB =  20.98

CB = 16.99

First, remember that the sum of the interior angles of any triangle is always equal to 180°, then we can write:

51 + 39 + ∠B = 180

∠B = 180 - 51 - 39 = 90

So we have a right triangle.

Now, to find the values of AB and CB, we can use trigonometric relations, we know that teh hypotenuse is 27 units, then we can use:

cos(51°) = CB/27

27*cos(51°) = CB = 16.99

And:

cos(39°) = AB/27

27*cos(39°) = AB = 20.98

These are the missing values.

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For any n ≥ 1, the factorial function, denoted by n!, is the product of all the positive integers through n, it is proved that for n ≥ 4, n! ≥ 2n.

To prove this statement, we can use mathematical induction. For the base case, n = 4, we have 4! = 24 and 2^4 = 16. Since 24 > 16, the statement holds for n = 4.

Now suppose the statement holds for some integer k ≥ 4, that is, k! ≥ 2k. We need to show that the statement holds for k+1. We have:

(k+1)! = (k+1)k!

≥ (k+1)2k (by the induction hypothesis)

≥ 2·2k (since k+1 > 2 for k ≥ 4)

= 2k+1.

Therefore, the statement holds for k+1. By the principle of mathematical induction, the statement holds for all n ≥ 4. Therefore, we have proved that n! ≥ 2n for n ≥ 4.

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The number of students at the mariemont middle school in whole number is 360 students.

Exponents refers to the power to which a number, symbol or expression is to be raised.

Number of students at the mariemont middle school = 2³ × 3² × 5

= (2 × 2 × 2) × (3 × 3) × 5

= 8 × 9 × 5

= 360 students

In conclusion, there are 360 total number of students in mariemont middle school.

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The given series 4^n / 2^n 3^n is convergent.

To see why, we can use the ratio test, which states that if the limit of the ratio of consecutive terms is less than 1, then the series converges. Applying the ratio test to the given series, we get:

lim n→∞ |(4^n+1 / 2^n+1 3^n+1) / (4^n / 2^n 3^n)|

= lim n→∞ |4 / 3(1 + 1/2n+1)|

= 4/3

Since the limit is less than 1, the series converges. To find its sum, we can use the formula for the sum of a convergent geometric series:

S = a / (1 - r)

where a is the first term and r is the common ratio. In this case, a = 4/6 = 2/3 and r = 2/3, so we get:

S = (2/3) / (1 - 2/3) = 2

Therefore, the sum of the series is 2.

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Simplifying this expression, we get the margin of error as approximately 0.042.

(a) The sample proportion is p = 367/500 = 0.734. To find the 99% confidence interval for the true proportion p, we use the formula: p ± zα/2 * sqrt((p(1-p))/n).

where zα/2 is the critical value from the standard normal distribution corresponding to a 99% confidence level, which is approximately 2.576. Substituting the given values, we have:

0.734 ± 2.576 * sqrt((0.734(1-0.734))/500)

Simplifying this expression, we get the 99% confidence interval for p as (0.692, 0.776). (b) The margin of error for this 99% confidence interval is given by: zα/2 * sqrt((p(1-p))/n)

Substituting the given values, we have: 2.576 * sqrt((0.734(1-0.734))/500) Simplifying this expression, we get the margin of error as approximately 0.042.

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To find the Maclaurin expansion of f(x) = -e^x - sin(x), we can use the Maclaurin series of e^x and sin(x) and combine them with appropriate coefficients.

The Maclaurin series of e^x is:

e^x = 1 + x + x^2/2! + x^3/3! + ...

And the Maclaurin series of sin(x) is:

sin(x) = x - x^3/3! + x^5/5! - ...

Using these series, we can write the Maclaurin expansion of f(x) as:

f(x) = -e^x - sin(x) = -1 - x - x^2/2! - x^3/3! - ... - (x - x^3/3! + x^5/5! - ...)

Simplifying this expression, we get:

f(x) = -1 - 2x - 5x^2/2! - 19x^3/3! - 87x^4/4! - ...

Therefore, the first five nonzero terms of the Maclaurin expansion of f(x) are:

f(x) = -1 - 2x - 5x^2/2! - 19x^3/3! - 87x^4/4! + O(x^5)

This means that for small values of x, f(x) can be approximated by the polynomial -1 - 2x - 5x^2/2! - 19x^3/3! - 87x^4/4!, which becomes more accurate as more terms are added. The term O(x^5) represents the error in this approximation and means that the actual value of f(x) is within a certain range of this polynomial for values of x close to zero.

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To find the function f, we need to integrate f'(x) with respect to x. Thus, we have found the function f with the given derivative f'(x) and initial condition f(9) = 67.

f'(x) = (1/3)x

Integrating both sides with respect to x, we get:

f(x) = (1/3) * (x^2/2) + C

where C is the constant of integration. To find the value of C, we use the given initial condition that f(9) = 67:

f(9) = (1/3) * (9^2/2) + C = 67

Simplifying the equation, we get:

C = 67 - (1/3) * (81/2) = 67 - 13.5 = 53.5

Therefore, the function f is:

f(x) = (1/3) * (x^2/2) + 53.5

Thus, we have found the function f with the given derivative f'(x) and initial condition f(9) = 67.

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The sampling technique used in this scenario is stratified random sampling. Stratified random sampling is a useful technique for obtaining a representative sample from a population with identifiable subgroups, and can improve the validity and generalizability of survey results.

Stratified random sampling involves dividing the population into homogeneous groups, or strata, based on a specific characteristic, and then taking a random sample from each stratum. In this case, the population of gym members was divided into men and women, which are two distinct and easily identifiable strata. A simple random sample was then taken from each stratum to obtain a representative sample of both genders.

The use of stratified random sampling can increase the precision and accuracy of the sample by ensuring that each stratum is represented proportionally in the sample. This technique is commonly used when the population of interest exhibits a significant characteristic that may impact the outcome of the survey. For example, if the survey was investigating the effectiveness of a new exercise program, it would be important to ensure that both men and women were represented equally in the sample, as their physiological differences may impact their response to the program.

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The statements, reasons, situations that proves the congruence of the triangles are presented as follows;

1. 5. ∠POM ≅ ∠NOM [tex]{}[/tex]        5. Definition of angle bisector

6. ΔPMO ≅ ΔNMO [tex]{}[/tex]           6. SAS Congruence theorem

2. C. III only

3. SAS congruence rule

Triangles are congruent if they have that same size and shape.

The completed two column method to prove the congruence of the triangles can be presented as follows;

Statements        [tex]{}[/tex]                          Reasons

1. [tex]\overrightarrow{MO}[/tex] bisects ∠PMN [tex]{}[/tex]                 1. Given

2. ∠PMO ≅ ∠NMO [tex]{}[/tex]                    2. Definition of angle bisector

3. [tex]\overline{MO}[/tex] ≅ [tex]\overline{MO}[/tex]            [tex]{}[/tex]                  3. Reflexive property

4. [tex]\overrightarrow{OM}[/tex] bisects ∠PON                 4. Given

5. ∠POM ≅ ∠NOM [tex]{}[/tex]                    5. Definition of angle bisector

6. ΔPMO ≅ ΔNMO [tex]{}[/tex]                    6. SAS congruence theorem

2. The leg HL Theorem states that the if the hypotenuse and a leg in one triangle are congruent to a leg and an hypotenuse side in another triangle, then the two triangles are congruent.

The specified dimensions of the triangle that indicates that the hypotenuse of the two triangles are congruent is the option III

The correct option is; C. III Only

3. The three angles in triangle ΔFDG are congruent to the three angles in triangle ΔFDE.

The reflexive property of congruence indicates; The side FD is congruent to itself (reflexive property of congruence)

The triangle ΔFDG is congruent to the triangle ΔFDE by the ASA congruence rule

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Thus, Taylor polynomial approximation for cos(x) gives values of x close to 0, and the value of x=12.

The fourth-degree Taylor polynomial for cos(x) about x=0 can be used to approximate the value of cos(12).

A Taylor polynomial is a polynomial that approximates a function by using the function's derivatives at a specific point. For cos(x), the Taylor polynomial about x=0 (also known as the Maclaurin series) is given by:
P(x) = Σ [(-1)^n * x^(2n)] / (2n)! , where the sum is from n = 0 to infinity.

Since we are interested in the fourth-degree Taylor polynomial, we will consider only the first three terms (n=0, 1, and 2):
P(x) ≈ 1 - x^2/2! + x^4/4!.

Now, we need to approximate the value of cos(12) using this polynomial:
P(12) ≈ 1 - (12^2)/2! + (12^4)/4! ≈ 1 - 72 + 20736/24 ≈ 1 - 72 + 864 ≈ 793.

However, it is important to note that the Taylor polynomial approximation for cos(x) works best for values of x close to 0, and the value of x=12 is relatively far from 0.

This means that the approximation might not be very accurate for cos(12). In practice, it's better to use a calculator or computer software to obtain a more precise value for cos(12).

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Answer:

0.18

Step-by-step explanation:

Therefore, the simplified expression is m³+3am² -3am² -2a³

We can first simplify the given expression below.

2a²m - 3am² + m³ am² - a²m-2a³​

Lets combine the like terms.

2a²m - 3am² + m³ am² - a²m-2a³​

m³ - 3a²m +3am² + 2a³

m³+3am² -3am² -2a³

Therefore, the simplified expression is m³+3am² -3am² -2a³

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The probability that a randomly selected 49- to 54-year-old mother who had a live birth in 2012 has had her fourth live birth is low, as it is rare for women in this age group to have more than three children.

However, the exact probability would depend on various factors, such as the woman's individual fertility, access to contraception, and cultural and social norms regarding family size.

The probability of a woman having a fourth live birth decreases as she gets older and approaches menopause. Women in their late forties and fifties have a higher risk of pregnancy complications and may have more difficulty conceiving than younger women.

Additionally, many women may choose to limit their family size or prioritize their careers and personal goals over having more children. However, there are exceptions, and some women may choose to have a fourth child or have an unplanned pregnancy. Ultimately, the probability of a woman having a fourth live birth is influenced by a variety of factors, and it is difficult to provide a precise estimate without additional information.

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So we've shown that every pair of vertices in S is connected by an edge in the subgraph induced by S. Therefore, the subgraph induced by S is a complete graph.

To start with, let's clarify what we mean by some of these terms. A graph is just a collection of vertices (or nodes) and edges connecting them. In this case, we're dealing with the complete graph kn, which means that there are n vertices and every possible edge connecting them is included in the graph. So there are a total of n(n-1)/2 edges in the graph.

Now, we're interested in subgraphs of kn. A subgraph is just a subset of the vertices and edges from the original graph. In this case, we're interested in subgraphs induced by nonempty subsets of the vertex set. So if we take some subset of the n vertices in kn, we can look at the edges connecting them and see if they form a complete graph.

So let's say we take some subset of the vertices and call it S. We want to show that the subgraph induced by S is a complete graph. In other words, every pair of vertices in S is connected by an edge.

To see why this is true, let's consider the complement of S, which we'll call S'. This is just the set of vertices in kn that are not in S. Since S is nonempty, S' is also nonempty.

Now, consider any pair of vertices in S. Call them v and w. Since v and w are both in S, they are not in S'. This means that there is an edge connecting v and w in kn, since kn is a complete graph. But since we're only looking at the vertices in S, this edge is also in the subgraph induced by S.

So we've shown that every pair of vertices in S is connected by an edge in the subgraph induced by S. Therefore, the subgraph induced by S is a complete graph.

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The coefficient on the intercept would change if the dependent variable, birth weight, was in ounces instead of pounds. It would be equal to 0.00, rounded to two decimal places.

The intercept coefficient represents the value of the dependent variable when all independent variables are equal to zero. In this case, it would represent the birth weight when all predictors are equal to zero. Since birth weight is measured in ounces, the intercept coefficient would represent the weight of a newborn when all predictors are equal to zero, which is not a meaningful or practical value. Therefore, the intercept coefficient would be equal to 0.00.

This result is expected since changing the unit of measurement of the dependent variable does not change the relationship between the dependent variable and the independent variables, only the scale of the coefficients. The regression equation would still provide useful information about the relationship between birth weight and the predictors, but the coefficients would need to be interpreted differently.

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The area of one leaf of the four-leaved rose is π/2 square units.

PART 1:

Using symmetry, we can find the area of one leaf of the four-leaved rose by integrating from 0 to π/4 and multiplying the result by 4. So we have:

Area of one leaf = 4 × ∫[0 to pi/4] 1/2 r^2 dt

= 4 × ∫[0 to pi/4] 1/2 (2cos2t)^2 dt

= 4 × ∫[0 to pi/4] 1/2 (4cos^2(2t)) dt

= 4 × ∫[0 to pi/4] 2cos^2(2t) dt

= 4 × ∫[0 to pi/4] (cos(4t) + 1) / 2 dt

= 4 × [1/8 sin(4t) + 1/2 t] evaluated from 0 to pi/4

= 4 × (1/8 sin(pi) + 1/2 (pi/4) - 1/8 sin(0) - 1/2 (0))

= 4 × (1/2 (pi/4))

= π/2

PART 2:

The area of one leaf of the four-leaved rose is π/2 square units.

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Answer:

The last one, bottom right corner

We conclude that there is not enough evidence to suggest that there is a significant difference among the mean meal prices for italian, seafood, and steakhouse restaurants.

to test whether there is a significant difference among the mean meal prices for italian, seafood, and steakhouse restaurants, we can use an analysis of variance (anova) test. the null hypothesis is that there is no significant difference among the means, while the alternative hypothesis is that there is a significant difference among the means.here are the steps to conduct the anova test:1. calculate the sample means for each category of restaurants:- italian: (12 + 16 + 24 + 13 + 18 + 19 + 15 + 17) / 8 = 17.25- seafood: (23 + 21 + 20 + 15 + 22 + 17 + 19 + 27) / 8 = 20.75

- steakhouse: (18 + 31 + 24 + 18) / 4 = 22.752. calculate the overall mean:(12 + 16 + 24 + 13 + 18 + 19 + 15 + 17 + 23 + 17 + 26 + 25 + 18 + 23 + 21 + 20 + 15 + 22 + 17 + 19 + 27 + 24 + 18 + 31) / 24 = 20.3753. calculate the sum of squares between groups (ssb):

ssb = 8 x (17.25 - 20.375)² + 8 x (20.75 - 20.375)² + 4 x (22.75 - 20.375)²    = 38.54. calculate the sum of squares within groups (ssw):ssw = (12 - 17.25)² + (16 - 17.25)² + ... + (18 - 22.75)² + (31 - 22.75)²    = 598.5

5. calculate the degrees of freedom for between groups (dfb):dfb = k - 1 = 3 - 1 = 26. calculate the degrees of freedom for within groups (dfw):dfw = n - k = 24 - 3 = 21

7. calculate the mean square between groups (msb):msb = ssb / dfb = 38.5 / 2 = 19.258. calculate the mean square within groups (msw):msw = ssw / dfw = 598.5 / 21 = 28.5

9. calculate the f-statistic:f = msb / msw = 19.25 / 28.5 = 0.6810. look up the critical f-value from an f-distribution table with dfb = 2 and dfw = 21 and a significance level of 0.05. the critical f-value is 3.10.

11. compare the f-statistic to the critical f-value. since the f-statistic (0.68) is smaller than the critical f-value (3.10), we fail to reject the null hypothesis. the test   statistics    value from our analysis is f = 0.68.

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The net profit on the sale of these fruits is Rs 6,050.

To calculate the net profit on the sale of these fruits, we need to determine the total revenue generated from the sales and deduct the total expenses.

First, let's calculate the revenue from each type of fruit:

Revenue from apples: 100 kg × Rs 140/kg = Rs 14,000

Revenue from bananas: 30 dozen × Rs 75/dozen = Rs 2,250

Revenue from grapes: 50 kg × Rs 150/kg = Rs 7,500

Next, let's calculate the total revenue:

Total revenue = Revenue from apples + Revenue from bananas + Revenue from grapes

Total revenue = Rs 14,000 + Rs 2,250 + Rs 7,500

Total revenue = Rs 23,750

Now, let's calculate the total expenses:

Total expenses = Cost of apples + Cost of bananas + Cost of grapes + Transportation cost

Total expenses = Rs 9,000 + Rs 1,800 + Rs 6,000 + Rs 900

Total expenses = Rs 17,700

Finally, let's calculate the net profit:

Net profit = Total revenue - Total expenses

Net profit = Rs 23,750 - Rs 17,700

Net profit = Rs 6,050

Therefore, the net profit on the sale of these fruits is Rs 6,050.

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The surface area of the pyramid is 55ft²

The area occupied by a three-dimensional object by its outer surface is called the surface area.

The surface area of a pyramid is calculated by adding all the area of the faces

area of the 4 triangles

A = 1/2 bh

A = 1/2 × 3 × 5

A = 15/2 = 7.5 ft²

area for 4 triangle = 7.5 × 4

= 30ft²

Area of the square base = l²

= 5 × 5

= 25 ft²

Therefore area of the pyramid

= 25 + 30

= 55ft²

therefore the area of the pyramid is 55ft²

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The Taylor polynomial T3(x) for the function f centered at the number a is 1, -1.

To find the slope of the tangent line to the curve at a given point, we need to find the derivative of the curve and evaluate it at that point. So, let's find the derivative of the curve x(t) = cos^3(4t), y(t) = sin^3(4t):

x'(t) = 3cos^2(4t) * (-sin(4t)) * 4 = -12cos^2(4t)sin(4t)

y'(t) = 3sin^2(4t) * cos(4t) * 4 = 12sin^2(4t)cos(4t)

Now, let's evaluate these derivatives at t = pi/6:

x'(pi/6) = -12cos^2(2pi/3)sin(2pi/3) = -6sqrt(3)

y'(pi/6) = 12sin^2(2pi/3)cos(2pi/3) = 6sqrt(3)

So, the slope of the tangent line at t = pi/6 is:

y'(pi/6) / x'(pi/6) = (6sqrt(3)) / (-6sqrt(3)) = -1

Therefore, the answer is option 1, -1.

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In a regression analysis, the regression equation given is y = 12 - 6x. The correct option for the coefficient of correlation is b. -0.7.

The terms SSE (sum of squared errors) and SST (total sum of squares) are provided, with values 510 and 1000, respectively. To determine the coefficient of correlation (r), we need to first calculate the coefficient of determination (R²), which is given by the formula:
R² = (SST - SSE) / SST
Substituting the given values, we get:

R² = (1000 - 510) / 1000 = 490 / 1000 = 0.49
Now, we need to find the correlation coefficient (r), which is the square root of the coefficient of determination (R²). However, we need to determine the sign (positive or negative) based on the regression equation. Since the slope of the equation (in this case, -6) is negative, the correlation coefficient will also be negative. Therefore, we have:
r = -√0.49 = -0.7

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A differential equation whose general solution is y=c1e^6t c2e^-2t is  37y'' − 18y' + y = 0

To find a differential equation whose general solution is y=c1e^6t+c2e^−2t, we can differentiate both sides of the equation:

y = c1e^6t+c2e^−2t

y' = 6c1e^6t−2c2e^−2t

y'' = 36c1e^6t+4c2e^−2t

Substituting these expressions for y, y', and y'' into the standard form of a linear homogeneous differential equation:

ay'' + by' + cy = 0

we get:

36c1e^6t+4c2e^−2t + 6(6c1e^6t−2c2e^−2t) + c1e^6t+c2e^−2t = 0

Simplifying this equation, we get:

(37c1)e^6t+(c2) e^−2t=0

Since this equation must hold for all t, the coefficients of each exponential term must be zero. Therefore, we have the system of equations:

37c1 = 0

c2 = 0

Solving for c1 and c2, we get c1 = 0 and c2 = 0.

Since this implies that the differential equation has trivial solution, we need to modify the differential equation slightly. One way to do this is to add a constant to the exponent of one of the terms in the general solution, say e^−2t:

y = c1e^6t+c2e^(−2t+1)

Taking the first and second derivatives of y with respect to t, we have:

y' = 6c1e^6t−2c2e^(−2t+1)

y'' = 36c1e^6t+4c2e^(−2t+1)

Substituting these expressions into the standard form of a linear homogeneous differential equation, we get:

36c1e^6t+4c2e^(−2t+1) + 6(6c1e^6t−2c2e^(−2t+1)) + c1e^6t+c2e^(−2t+1) = 0

Simplifying this equation, we get:

(37c1)e^6t+(9c2)e^(−2t+1)=0

Since this equation must hold for all t, the coefficients of each exponential term must be zero. Therefore, we have the system of equations:

37c1 = 0

9c2 = 0

Solving for c1 and c2, we get c1 = 0 and c2 = 0.

Therefore, the modified differential equation is:

37y'' − 18y' + y = 0

Note that this differential equation has y=c1e^6t+c2e^(−2t+1) as its general solution.

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The largest diameter particle (in microns) that will pass through the filter is given as follows:

0.756 microns.

The z-score of a measure X of a normally distributed variable that has mean represented by [tex]\mu[/tex] and standard deviation represented by [tex]\sigma[/tex] is obtained by the equation presented as follows:

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

The mean and the standard deviation for this problem are given as follows:

[tex]\mu = 0.5, \sigma = 0.2[/tex]

The largest diameter is the 90th percentile, which is X when Z = 1.28, as 1.28 has a p-value of 0.9, hence:

1.28 = (X - 0.5)/0.2

X - 0.5 = 0.2 x 1.28

X = 0.756 microns.

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Bryant and his sister have a combined total of $30.50 to pay for a new tank for their pet. Their parents have offered to cover the remaining cost. The total cost of the tank is $139.99. Therefore, the parents will pay the difference between $139.99 and $30.50.

First Calculate the combined amount Bryant and his sister have contributed. Bryant has $30.50, and his sister also has $30.50. So, we need to add these amounts together: $30.50 (Bryant) + $30.50 (Sister) = $61.00. Bryant and his sister have a total of $61.00 to contribute towards the bigger tank for their pet lizard, Iggy. Now we have to determine the remaining cost for the bigger tank after their contribution. The total cost of the tank is $139.99. We will now subtract the $61.00 that Bryant and his sister have from the total cost: $139.99 (Total cost) - $61.00 (Bryant and Sister's contribution) = $78.99. and finally to Identify the amount the parents will pay. The remaining cost of the bigger tank after Bryant and his sister's contribution is $78.99. Since their parents offered to pay the remaining cost, they will pay:$78.99. In conclusion, Bryant and his sister will contribute $61.00 towards the new, bigger tank for Iggy, and their parents will cover the remaining cost, which is $78.99.

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9.172 cm² is the area of the unshaded reason.

It is given that,

From the general formula of the area of the arc of the circle,

Area of the arc = (θ/360) x πr²

where A is the area of the arc, θ is the central angle of the arc (in degrees), and r is the radius of the circle.

The area of the shaded part is given = 56.87 cm²

Angle of the shaded arc = 360-50 = 310

So,

310/360* πr² = 56.87

πr²/360 = 56.87/310

For the 50° part,

Area of the unshaded part = 50/360* πr²

From the above value of the  πr²/360,

Area of the unshaded part = 50*56.87/310

Area of the unshaded part = 9.172 cm²

Therefore, the area of the unshaded reason is 9.172 cm².

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The answer is (a) 0.8494, to find the probability that the battery fails before the one year warranty expires,

we need to calculate the integral of the given pdf from 0 to 1, as X represents the length of the battery life in years.

So, P(X<1) = ∫(0 to 1) f(x) dx = ∫(0 to 1) (16/49) e^(-8x^2/49) dx ≈ 0.8494

Therefore, the answer is (a) 0.8494.

The given pdf describes the distribution of the length of the battery life, and we are interested in finding the probability that the battery fails before the one year warranty expires.

This can be found by integrating the pdf from 0 to 1, as the warranty lasts for one year.

Using the formula for the probability density function, we calculate the integral of the given pdf from 0 to 1, and get the answer as 0.8494.

This means that the probability of the battery failing before the one year warranty expires is about 84.94%.

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The 95% confidence interval centered on the mean that captures 95% of the sample means is (80.96, 85.04).

We can use the formula for the confidence interval for the mean of a normally distributed population:

CI = X ± z(α/2) * (σ/√n)

Where:

X = sample mean

z(α/2) = the z-score associated with the desired confidence level and calculated using the standard normal distribution table. For a 95% confidence level, α/2 = 0.025, and the corresponding z-score is approximately 1.96.

σ = population standard deviation

n = sample size

Substituting the given values, we get:

CI = 83 ± 1.96 * (5.2/√25)

CI = 83 ± 2.04

Therefore, the 95% confidence interval centered on the mean that captures 95% of the sample means is (80.96, 85.04).

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Answer:

17 units.

Step-by-step explanation:

To find the radius of the circle, we can use the distance formula between the center of the circle and a point on the circle.

Let's denote the center of the circle as (h, k) and the point on the circle as (x, y).

The distance formula is given by:

d = sqrt((x - h)^2 + (y - k)^2)

In this case, the center of the circle is (-7, -1) and a point on the circle is (8, 7).

Plugging these values into the distance formula:

d = sqrt((8 - (-7))^2 + (7 - (-1))^2)

= sqrt((8 + 7)^2 + (7 + 1)^2)

= sqrt(15^2 + 8^2)

= sqrt(225 + 64)

= sqrt(289)

= 17

Therefore, the radius of the circle is 17 units.

Now, to determine if the point (-15, y) lies on this circle, we can substitute the x-coordinate (-15) into the equation of the circle and solve for y.

Using the equation of a circle:

(x - h)^2 + (y - k)^2 = r^2

where (h, k) is the center of the circle and r is the radius, we have:

(-15 - (-7))^2 + (y - (-1))^2 = 17^2

(-15 + 7)^2 + (y + 1)^2 = 289

(-8)^2 + (y + 1)^2 = 289

64 + (y + 1)^2 = 289

(y + 1)^2 = 289 - 64

(y + 1)^2 = 225

y + 1 = ±√225

y + 1 = ±15

Solving for y, we have two possible values:

y + 1 = 15

y = 15 - 1

y = 14

y + 1 = -15

y = -15 - 1

y = -16

Therefore, the point (-15, 14) and (-15, -16) both lie on the circle with a radius of 17 units.