Write the coordinates of the vertices after a dilation with a scale factor of 3, centered at the origin. 10 B 10 0 10 А С 10 A(1, 2) A'l B(1, 3) - BC​

The 99% confidence interval for estimating the average monthly purchases of lower-income players is A = $5.74 and B = $6.12.

To construct a confidence interval, we can use the formula:

Confidence interval = sample mean ± (critical value * standard error)

Given that the sample size is 404, the mean monthly purchase amount is $5.93, and the standard deviation is $1.05, we can calculate the standard error using the formula:

Standard error = standard deviation / √(sample size)

Substituting the values, we find that the standard error is approximately $0.052.

To determine the critical value, we need to consider the desired confidence level. For a 99% confidence interval, we have 1 - (0.99) = 0.01, and dividing this by 2 gives us 0.005 for each tail. Consulting a t-distribution table or using a statistical software, we find that the critical value is approximately 2.626.

Substituting the values into the confidence interval formula, we get:

Confidence interval = $5.93 ± (2.626 * $0.052)

Calculating the lower and upper limits, we find that A = $5.74 and B = $6.12. Therefore, we can estimate with 99% confidence that the average monthly purchases of lower-income players fall between $5.74 and $6.12.

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The student's grade is approximately 83.43.

To calculate the student's grade, we need to consider the weight of each exam. The four 1-hour exams are worth 1 hour each, and the final exam is equivalent to three 1-hour exams.

Let's break down the calculation step by step:

Calculate the sum of the 1-hour exams:

65 + 84 + 98 + 91 = 338

Calculate the weighted sum of the exams by multiplying the sum of the 1-hour exams by 1 (since each 1-hour exam has a weight of 1):

Weighted sum of 1-hour exams = 338×1 = 338

Calculate the weighted score for the final exam by multiplying the final exam score (82) by 3 (since it counts as three 1-hour exams):

Weighted score for the final exam = 82× 3 = 246

Add the weighted sum of the 1-hour exams and the weighted score for the final exam to obtain the total weighted sum:

Total weighted sum = Weighted sum of 1-hour exams + Weighted score for the final exam

= 338 + 246 = 584

Calculate the total weight of all the exams by summing the individual weights:

Total weight = Weight of 1-hour exams + Weight of the final exam

= 4 + 3 = 7

Finally, calculate the student's grade by dividing the total weighted sum by the total weight:

Student's grade = Total weighted sum / Total weight

= 584 / 7 ≈ 83.43

Therefore, the student's grade is approximately 83.43.

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Answer:

13

Step-by-step explanation:

Answer:

13

Step-by-step explanation:

Hello There!

We can calculate the distance between two points using the distance formula

[tex]d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}[/tex]

where the x and y values are derived from the coordinates in which you are trying to find the distance between

The points we need to find the distance between is (3,-8) and (8,4)

So we plug the x values and y values into the formula

[tex]d=\sqrt{(8-3)^2+(4-(-8)^2} \\8-3=5\\4-(-8)=12\\d=\sqrt{5^2+12^2} \\5^2=25\\12^2=144\\144+25=169\\\sqrt{169} =13[/tex]

so we can conclude that the distance between the two points is 13 units

Answer:

x = 27.2 or 27 1/5

Step-by-step explanation:

cos 54° = 16/x

x = 27.2 or 27 1/5

The probability that an individual distance is greater than 214 is 0.4554.

The overhead reach distances of adult females are normally distributed with a mean of 205 cm and a standard deviation of 78 cm. To find the probability that an individual distance is greater than 214, we need to use the standard normal distribution, which is a normal distribution with a mean of 0 and a standard deviation of 1. To do this, we will first calculate the z-score for 214:z = (214 - 205) / 78z = 0.1154Then, we will use a standard normal distribution table or calculator to find the probability of a z-score greater than 0.1154. The area to the right of the z-score is the probability of an individual distance being greater than 214.Using a standard normal distribution table, we find that the probability of a z-score greater than 0.1154 is 0.4554. Therefore, the probability that an individual distance is greater than 214 is 0.4554.

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Given that the overhead reach distances of adult females are normally distributed with a mean of 205 cm and a standard deviation of 78 cm. We need to find the probability that an individual distance is greater than 214. So, we have to standardize the given value 214 as follows:

Z = (X - μ)/σZ = (214 - 205)/78Z = 0.1154

We have to find the probability that an individual distance is greater than 214.

So, P(X > 214) = P(Z > 0.1154)

Using the standard normal distribution table, the area under the curve to the right of Z = 0.1154 is 0.4573.Approximately, the probability that an individual distance is greater than 214 is 0.4573.Hence, the correct option is (a) 0.4573.

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Answer: 12.5

Step-by-step explanation:

Answer:

450 cm ^3

Step-by-step explanation:

Answer:

Step-by-step explanation:

lol true

Two possible representations of vector w as a linear combination of V1, V2, and V3 are: w = 2V1 + V2 + 3V3 and w = -3V1 + 2V2 - 4V3.

The vector w can be represented as a linear combination of V1, V2, and V3 in the following ways:

w = 2V1 + V2 + 3V3

w = -3V1 + 2V2 - 4V3

Explanation:

To find the possible representations of w as a linear combination of V1, V2, and V3, we need to determine the coefficients that satisfy the equation w = aV1 + bV2 + cV3, where a, b, and c are scalars.

We can set up a system of equations to solve for a, b, and c. Using the given vectors and coefficients, we get:

2a + 2b + c = 2

a + c = 3

-3a + 2b - 4c = -5

By solving this system of equations, we find the values of a, b, and c that satisfy the equation. In this case, we have two possible solutions: a = 2, b = 1, and c = 3 for the first representation, and a = -3, b = 2, and c = -4 for the second representation.

Therefore, the possible representations of w as a linear combination of V1, V2, and V3 are w = 2V1 + V2 + 3V3 and w = -3V1 + 2V2 - 4V3.

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The probability that the  proportion will be less than 0.05 is approximately 0.1056, rounded to four decimal places.

We have,

To calculate the probability that the sample proportion will be less than 0.05, we can use the sampling distribution of the sample proportion.

Given that the true proportion is 0.07 and a sample of size 259 is taken, we can assume that the distribution of the sample proportion follows a normal distribution with a mean equal to the true proportion (0.07) and a standard deviation equal to the square root of (p(1-p)/n), where p is the true proportion and n is the sample size.

In this case, the mean is 0.07 and the standard deviation is:

= √((0.07 x (1 - 0.07)) / 259).

To find the probability that the sample proportion will be less than 0.05, we can standardize the value using the z-score formula:

z = (x - mean) / standard deviation

In this case, we want to find P(X < 0.05), which is equivalent to finding P(z < (0.05 - mean) / standard deviation).

Calculating the z-score and using a standard normal distribution table or a calculator, we can find the corresponding probability.

Substituting the values into the formula:

z = (0.05 - 0.07) / √((0.07 x (1 - 0.07)) / 259)

Now, we can find the probability by looking up the corresponding

z-value in the standard normal distribution table or using a calculator.

The probability that the sample proportion will be less than 0.05 is the probability corresponding to the calculated z-value.

Round the answer to four decimal places to get the final result.

Therefore,

The probability that the proportion will be less than 0.05 is approximately 0.1056, rounded to four decimal places.

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The proportion of mature pecan trees between 43 and 46 feet tall can be calculated using the normal distribution with a mean of 42 feet and a standard deviation of 7.5 feet.

To find the proportion, we need to calculate the z-scores corresponding to the given heights and then find the area under the normal curve between those z-scores.

First, we calculate the z-score for 43 feet:

z1 = (43 - 42) / 7.5 = 0.1333

Next, we calculate the z-score for 46 feet:

z2 = (46 - 42) / 7.5 = 0.5333

Using a standard normal distribution table or a calculator, we can find the area between these two z-scores. The area corresponds to the proportion of trees between 43 and 46 feet tall.

The explanation would involve using a standard normal distribution table or a calculator to find the area under the normal curve between the z-scores of 0.1333 and 0.5333. This area represents the proportion of mature pecan trees between 43 and 46 feet tall.

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The approximate percentage of a 10c sample left after the time it took to walk one lap around the gym is 100 - 25c.

Let x be the time it takes to walk one lap around the gym.

We know that 5 laps take 200 seconds.

Therefore, x can be found by dividing 200 by 5:

x = 200/5 = 40 seconds.

Now, let's find the percentage of the sample left after walking one lap around the gym.

Since x is the time it takes to walk one lap around the gym, we know that the sample decreases at a rate of 10c/x per second.

Therefore, after x seconds, the percentage of the sample remaining is given by: 100(1 - 10c/x)

Substituting x = 40, we get:

100(1 - 10c/40) = 100(1 - 0.25c) = 100 - 25c

So the approximate percentage of a 10c sample left after the time it took to walk one lap around the gym is 100 - 25c.

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Answer:

Circumference = 2(pi)r

C = 2*3.14*4.0= 25.12 round 25.10 m

Step-by-step explanation:

The probability that the password does not have repeated letters, expressed to the nearest tenth of a percent is 0.0018%.

Given that a password with 6 characters is randomly selected from the 26 letters of the alphabet.

The number of ways to choose the first letter is 26 since all 26 letters are available.

The number of ways to choose the second letter is 25 since one letter has already been chosen and there are only 25 letters remaining.

Similarly, the number of ways to choose the third, fourth, fifth, and sixth letters are 24, 23, 22, and 21, respectively.

So, the total number of possible passwords is given by: 26 × 25 × 24 × 23 × 22 × 21 = 26P6

We want to find the probability that the password does not have repeated letters.

Let's calculate this probability now.

The first letter can be any of the 26 letters.

The second letter, however, can be one of the remaining 25 letters.

The third letter can be one of the remaining 24 letters, and so on.

So, the number of possible passwords that do not have repeated letters is given by: 26 × 25 × 24 × 23 × 22 × 21 / (6 × 5 × 4 × 3 × 2 × 1) = 26P6/6P6

So, the probability that the password does not have repeated letters is given by: P(A) = 26P6/6P6≈ 0.000018449 or 0.0018% (to the nearest tenth of a percent)

Therefore, the probability that the password does not have repeated letters, expressed to the nearest tenth of a percent is 0.0018%.

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The mean difference between Process A and Process B is 3.0 (rounded to 1 decimal place).

To calculate the mean difference between two processes, we subtract the average of Process B from the average of Process A.

Mean difference = Average of Process A - Average of Process B

Mean difference = 277 - 274 = 3.0

Therefore, the mean difference between Process A and Process B is 3.0.

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Answer:

2a°+2a°=180°[opposite angle of cyclic quadrilateralis 180°]

4a°=180

a=180°/4

a=45°

4b+2b=180°[similarly]

6b=180°

b=180°/6

b=30

The vector v can be written as the sum of a vector in Span{u} and a vector orthogonal to u as follows: v = (1/5)u + (-4/5)[1 -3].

The main answer can be obtained by decomposing the vector v into two components: one component lies in the span of vector u, and the other component is orthogonal to u. To find the vector in the span of u, we scale the vector u by the scalar (1/5) since v = - [3 1] can be written as (-1/5)[2 1]. This scaled vector lies in the span of u and can be denoted as (1/5)u.

To find the vector orthogonal to u, we subtract the vector in the span of u from v. This can be calculated by multiplying the vector u by the scalar (-4/5) and subtracting the result from v. The orthogonal component is obtained as (-4/5)[1 -3].

Thus, we have successfully decomposed vector v as v = (1/5)u + (-4/5)[1 -3], where (1/5)u lies in the span of u and (-4/5)[1 -3] is orthogonal to u.

In linear algebra, vector decomposition is a fundamental concept that allows us to express a given vector as a sum of vectors that have specific properties. The decomposition involves finding a vector in the span of a given vector and another vector that is orthogonal to it. This process enables us to analyze the behavior and properties of vectors more effectively.

In the context of this problem, the vector v is decomposed into two components. The first component, (1/5)u, lies in the span of the vector u. The span of a vector u is the set of all vectors that can be obtained by scaling u by any scalar value. Therefore, (1/5)u represents the part of v that can be expressed as a linear combination of u.

The second component, (-4/5)[1 -3], is orthogonal to u. Two vectors are orthogonal if their dot product is zero. In this case, we subtract the vector in the span of u from v to obtain the orthogonal component. By choosing the scalar (-4/5), we ensure that the resulting vector is orthogonal to u.

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a. the point estimate of the true driver cell phone use rate is 0.3 or 30%.

b. the 90% confidence interval for the true driver cell phone use rate is approximately (21.5%, 38.5%).

c. The practical interpretation of the confidence interval is that we are 90% confident that the true driver cell phone use rate falls within the range of 21.5% to 38.5%

a. The point estimate of the true driver cell phone use rate (population proportion) can be calculated by dividing the number of drivers using their cell phone by the total sample size. In this case, the sample size is 100, and 30 drivers were using their cell phone.

Point estimate = Number of drivers using their cell phone / Total sample size

Point estimate = 30/100 = 0.3 (or 30%)

Therefore, the point estimate of the true driver cell phone use rate is 0.3 or 30%.

b. To compute a 90% confidence interval for the true driver cell phone use rate, we can use the formula for a confidence interval for a proportion. The formula is:

Confidence interval = Point estimate ± (Critical value × Standard error)

The critical value depends on the desired level of confidence. For a 90% confidence interval, the critical value is typically obtained from the standard normal distribution and is approximately 1.645.

The standard error can be calculated using the formula:

Standard error = sqrt((point estimate * (1 - point estimate)) / sample size)

In this case, the point estimate is 0.3, and the sample size is 100.

Standard error = sqrt((0.3 * (1 - 0.3)) / 100) ≈ 0.048

Plugging in the values, we can calculate the confidence interval:

Confidence interval = 0.3 ± (1.645 * 0.048)

Confidence interval = (0.215, 0.385)

Therefore, the 90% confidence interval for the true driver cell phone use rate is approximately (21.5%, 38.5%).

c. The practical interpretation of the confidence interval is that we are 90% confident that the true driver cell phone use rate falls within the range of 21.5% to 38.5%. This means that based on the sample data, we can estimate with 90% confidence that the proportion of drivers using their cell phone while driving in the entire population lies between these two percentages. It provides a range of likely values for the true population proportion.

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To compute the area of a parallelogram determined by two vectors U = (4, -1) and V = (-6, -2), we can use the formula that states the area of a parallelogram is equal to the magnitude of the cross product of the two vectors. Therefore, the area of the parallelogram determined by U = (4, -1) and V = (-6, -2) is 28 square units.

The formula to compute the area of a parallelogram determined by two vectors U and V is given by:

Area = |U x V|

To calculate the cross product U x V, we can use the following determinant:

| i j k |

| 4 -1 0 |

|-6 -2 0 |

Expanding the determinant, we get:

i * (0 * -2 - 0 * -2) - j * (4 * -2 - 0 * -6) + k * (4 * -2 - (-1) * -6)

= -12i + 24j + 8k

Taking the magnitude of the cross product, we have:

|U x V| = √((-12)^2 + 24^2 + 8^2) = √(144 + 576 + 64) = √784 = 28

Therefore, the area of the parallelogram determined by U = (4, -1) and V = (-6, -2) is 28 square units.

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Answer:

20.25

Step-by-step explanation:

Just multiply the base and height then divide the results in half.

Hope this helps!

(a) To compute the cumulative distribution function (CDF) of W, denoted as FW(w), we integrate the joint probability density function (PDF) over the appropriate region. The region is defined by the inequalities x ≥ 0, y ≥ 0, and 2x + 3y ≤ 6. The CDF is given by: FW(w) = P(W ≤ w) = ∫∫[fX,Y(x, y)] dy dx

To find the PDF fw(w), we differentiate FW(w) with respect to w.

(b) To compute E[W], we integrate the product of w and the PDF fw(w) over the range of W. The variance V ar[W] is calculated by finding E[W^2] and subtracting (E[W])^2.

(c) To find the minimum and maximum values of Z, we need to determine the range of Y - X. We consider the range of x and y that satisfy the given conditions. By substituting the limits of x and y, we can calculate the minimum and maximum values of Z.

(d) The cumulative distribution function FZ(z) can be written as a double integral over the joint PDF fX,Y(x, y). We consider two cases: w ≥ 0 and w < 0. For each case, we determine the appropriate region and integrate the PDF accordingly.

(e) To find the PDF fZ(z), we differentiate FZ(z) with respect to z.

(f) To calculate E[Z], we integrate the product of z and the PDF fZ(z) over the range of Z. The variance V ar[Z] is computed by finding E[Z^2] and subtracting (E[Z])^2.

Please note that without the specific range or shape of the region defined by the inequalities, it is not possible to provide detailed numerical calculations for each part.

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The "sample-mean" is 4, the sample-variance is 0.876, and the sample standard-deviation is 0.935.

To calculate the sample-mean, sample variance, and sample standard deviation, we use the formulas:

Sample mean (x') = (sum of all values)/(number of values)

Sample variance (s²) = [(sum of (each value - sample mean)²/(number of values - 1)],

Sample standard deviation (s) = √(sample variance),

Given the data: 4.2, 4.7, 4.7, 5.0, 3.8, 3.6, 3.0, 5.1, 3.1, 3.8, 4.8, 4.0, 5.2, 4.3, 2.8, 2.0, 2.8, 3.3, 4.8, 5.0.

⇒ Calculate the sample-mean (x'):

x' = (4.2 + 4.7 + 4.7 + 5.0 + 3.8 + 3.6 + 3.0 + 5.1 + 3.1 + 3.8 + 4.8 + 4.0 + 5.2 + 4.3 + 2.8 + 2.0 + 2.8 + 3.3 + 4.8 + 5.0) / 20

x' ≈ 80/20 = 4

So, mean is 4,

⇒ Calculate the sample-variance (s²):

s² = [(4.2 - 4)² + (4.7 - 4)² + (4.7 - 4)² + (5.0 - 4)² + (3.8 - 4)² + (3.6 - 4)² + (3.0 - 4)² + (5.1 - 4)² + (3.1 - 4)² + (3.8 - 4)² + (4.8 - 4)² + (4.0 - 4)² + (5.2 - 4)² + (4.3 - 4)² + (2.8 - 4)² + (2.0 - 4)² + (2.8 - 4)² + (3.3 - 4)² + (4.8 - 4)² + (5.0 - 4)²]/(20-1),

s² = [0.04 + 0.49 + 0.49 + 1.0 + 0.16 + 0.64 + 1.0 + 1.21 + 0.81 + 0.16 + 0.64 + 0.0 + 1.44 + 0.09 + 1.44 + 4.0 + 1.44 + 0.49 + 0.64 + 1.0]/19,

s² = 16.64/19 ≈ 0.876,

⇒ Calculate the sample standard-deviation (s):

s = √(s²)

s ≈ 0.935

Therefore, the standard-deviation is 0.935.

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(a) We get the result as:

to.005 = 2.539

(b) The required value is to.

10 = 1.345

(c) From the distribution we get:

to.975 = 2.086.

Given: v=19, α= 0.005

For finding to.005 when v= 19, we need to follow the below steps:

The t-distribution table has two tails and it is symmetric about the mean.

So, the area in one tail is (α/2), and in the second tail is also (α/2).

Step 1:  First of all we need to find the row of the t-distribution table and this will be equal to the degree of freedom (v) which is given to be 19.

In this case, we will find the value in row 19 in the table of critical values of the t-distribution which is shown below:

Step 2: Now, look for the value of α at the top of the table (at 0.005).

Step 3: Since the table is showing the area in the right-hand tail, the value of to.005 will be a positive value.

Therefore, we have to use the positive row of the table and for this, we can find the to.005 by looking at the intersection of row 19 and the column corresponding to α=0.005.

Therefore, to.005 = 2.539 (approximately) (Rounded to three decimal places)

Hence, the correct option is to.005 = 2.539

(b) v=14, α= 0.10

For finding to.10 when v= 14, we need to follow the same steps that we followed in part (a).

The table of critical values of the t-distribution is shown below:

Step 1: Find the row corresponding to the v=14 in the t-distribution table.

Step 2: Look for the α=0.10 at the top of the table.

Since the area in one tail is (α/2) which is equal to 0.05, therefore we need to find the critical values that will cut off the top 5% of the curve.

Step 3: Since the table is showing the area in the right-hand tail, the value of to.10 will be a positive value.

Therefore, we have to use the positive row of the table and for this, we can find the to.10 by looking at the intersection of row 14 and the column corresponding to α=0.10 .

Therefore, to.10 = 1.345 (approximately) (Rounded to three decimal places)

Hence, the correct option is to.10 = 1.345

(c) v = 20, α = 0.025

For finding to.025 when v= 20, we need to follow the same steps that we followed in part (a).

The table of critical values of the t-distribution is shown below:

Step 1: Find the row corresponding to the v=20 in the t-distribution table.

Step 2: Look for the α=0.025 at the top of the table.

Since the area in one tail is (α/2) which is equal to 0.0125, therefore we need to find the critical values that will cut off the top 1.25% of the curve.

Step 3: Since the table is showing the area in the right-hand tail, the value of to.975 will be a positive value.

Therefore, we have to use the positive row of the table and for this, we can find the to.975 by looking at the intersection of row 20 and the column corresponding to α=0.025 .

Therefore, to.025 = 2.086 (approximately) (Rounded to three decimal places)

Hence, the correct option is to.975 = 2.086.

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The length function f(w) of a string w = w₁w₂...W), where n ∈ N and Vi ∈ W: € 9, is equal to n.

The length function f(w) is defined recursively based on the structure of the string w. In the base case, if w is an empty string (ε), the length is defined as 0. In the recursive case, if w can be written as au, where a is a character from the alphabet and u is a string over the alphabet Σ, then the length is defined as 1 plus the length of u.

To prove that for any string w =w₁w₂...wy, where n ∈ N and Vi ∈ W: € 9, the length function f(w) is equal to n, we will use a proof by induction.

For w = ε (an empty string), we have f(ε) = 0, which satisfies the condition when n = 0.

Assume that for any string w = w₁w₂...wn, where n ∈ N and Vi ∈ W: € 9, the length function f(w) = n.

Now, consider a string w' = w₁w₂...wn+1. By the recursive definition, we can write w' as au, where a is the last character wn+1 and u is the string w₁w₂...wn. From our assumption, we know that f(u) = n.

Therefore, f(w') = 1 + f(u) = 1 + n = n + 1.

Since we have established that for any string w = w₁w₂...wy, where n ∈ N and Vi ∈ W: € 9, the length function f(w) = n, we can conclude that f(w) = n.

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The bone density score corresponding to the 9th percentile, separating the bottom 9% from the top 91%, is approximately -1.34.

To find this value, we can refer to the standard normal distribution table or use a statistical calculator. The standard normal distribution has a mean of 0 and a standard deviation of 1. The area to the left of any given z-score represents the cumulative probability up to that point.

In this case, since we want to find the 9th percentile, we are interested in the value that separates the bottom 9% (cumulative probability) from the top 91%. This means that we need to find the z-score that corresponds to an area of 0.09 under the curve.

By referencing the standard normal distribution table or using a calculator, we find that the z-score corresponding to a cumulative probability of 0.09 is approximately -1.34. This z-score represents the bone density score at the 9th percentile, separating the bottom 9% from the top 91%.

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The standard deviation of the test marks is 6.5

To calculate the standard deviation of the test marks, we need to follow a few steps. Let's go through them:

Step 1: Calculate the midpoint for each interval.

The midpoint is calculated by adding the lower and upper limits of each interval and dividing by 2.

Midpoint for 41-50: (41 + 50) / 2 = 45.5

Midpoint for 51-60: (51 + 60) / 2 = 55.5

Midpoint for 61-70: (61 + 70) / 2 = 65.5

Midpoint for 71-80: (71 + 80) / 2 = 75.5

Midpoint for 81-90: (81 + 90) / 2 = 85.5

Step 2: Calculate the deviation for each midpoint.

The deviation is calculated by subtracting the mean (average) from each midpoint.

Mean = ((45.5 * 5) + (55.5 * 0) + (65.5 * 10) + (75.5 * 8) + (85.5 * 2)) / (5 + 0 + 10 + 8 + 2)

= (227.5 + 0 + 655 + 604 + 171) / 25

= 1657.5 / 25

= 66.3

Deviation for 45.5: 45.5 - 66.3 = -20.8

Deviation for 55.5: 55.5 - 66.3 = -10.8

Deviation for 65.5: 65.5 - 66.3 = -0.8

Deviation for 75.5: 75.5 - 66.3 = 9.2

Deviation for 85.5: 85.5 - 66.3 = 19.2

Step 3: Square each deviation.

(-20.8)^2 = 432.64

(-10.8)^2 = 116.64

(-0.8)^2 = 0.64

(9.2)^2 = 84.64

(19.2)^2 = 368.64

Step 4: Calculate the squared deviation sum.

Sum of squared deviations = 432.64 + 116.64 + 0.64 + 84.64 + 368.64 = 1003.2

Step 5: Calculate the variance.

Variance = (Sum of squared deviations) / (Number of data points - 1) = 1003.2 / (25 - 1) = 1003.2 / 24 = 41.8

Step 6: Calculate the standard deviation.

Standard deviation = √(Variance) ≈ √(41.8) ≈ 6.5 (rounded to one decimal place)

Therefore, the standard deviation of the test marks is approximately 6.5 (to one decimal place).

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1) The term a_3 of this sequence a3 = -362/37.2

2) By the principle of mathematical induction, for all n ∈ N, aₙ < 0.

1) We are given the recursive formula:

a₁ = -36, aₙ₊₁ = aₙ/2 + 72/aₙ.

We need to find the term a₃ of this sequence. a₂ is given by the recursive formula as:

a₂ = a₁/2 + 72/a₁a₂ = -36/2 + 72/(-36) = -37/2

a₃ is given by the recursive formula as:

a₃ = a₂/2 + 72/a₂= (-37/2)/2 + 72/(-37/2)= -74/37 + (-288/37) = -362/37

Therefore, a₃ = -362/37.2

2) We need to prove by induction that for all n ∈ N, aₙ < 0.

Base case:

For n = 1, we have a₁ = -36 < 0. So, the base case is true.

Inductive step:

Let's assume that for some arbitrary n = k, aₖ < 0.

We need to show that aₖ₊₁ < 0.

Using the recursive formula: aₖ₊₁ = aₖ/2 + 72/aₖ

Since aₖ < 0, -aₖ > 0 and aₖ/2 < 0.Hence, aₖ/2 + 72/aₖ < 0

Therefore, aₖ₊₁ < 0.So, the statement that for all n ∈ N, aₙ < 0 is true for n = 1 and if it's true for n = k, then it's true for n = k + 1.

Therefore, by the principle of mathematical induction, for all n ∈ N, aₙ < 0.

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S is not orientable.

Given a regular surface S in R given by x² + y² = 2022, we need to find out whether S is orientable or not.

The surface is given by, x² + y² = 2022.

Rearranging the terms, we get, y² = 2022 - x²

Let the differentiable function g(x, y) = y, and the set U be the upper hemisphere (upper half) of the surface S.

Then, U = {(x, y, z) : x² + y² = 2022, z ≥ 0}

We know that the partial derivatives of the above function are continuous in U and it follows that U is a regular surface.

We now compute the partial derivatives of g(x, y) :∂g/∂x = 0, and ∂g/∂y = 1

Taking the cross-product of the two partial derivatives, we get : (0i - j + 0k) which is -j.

Now, if we define the positive normal to U to be the upward-pointing unit normal, then we see that (-j) points downward at all points on U.

Thus, U is not orientable.

Therefore, we conclude that the given surface S in R is not orientable.

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{{{ THE BOLDED CHARACTERS SHOULD BE SMALL. }}}

A sequence can be generated by using an = a(n-1) - 5, where a1 = 100 and n is a whole number greater than 1.

a1 = 100 (given)

a2 = a1 - 5 = 100 - 5 = 95

a3 = a2 - 5 = 95 - 5 = 90

a4 = a3 - 5  = 90 - 5 = 85

a5 = a4 - 5 = 85 - 5 = 80

ANSWER for PART (a): 100, 95, 90, 85, 80

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an = a1 + d(n - 1)

an = 100 + -5(n - 1)

an = 100 + -5n + 5

an = 105 - 5n

ANSWER for PART (b): an = 105 - 5n

Answer:

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Step-by-step explanation:

B. 0.14 mg will be present after 9 hours.

The given function is D(h) = 5e^(-0.4h), which can be used to determine the milligrams D of a certain drug in a patient's bloodstream h hours after the drug has been given.

To find the milligrams after 9 hours, we need to plug in h = 9 in the function D(h) = 5e^(-0.4h).

D(h) = 5e^(-0.4h)

D(9) = 5e^(-0.4(9))

D(9) = 5e^(-3.6)

D(9) = 5 × 0.024419

D(9) = 0.1220 ≈ 0.12 mg

Hence, the answer is option (c) 1.22 mg.

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