BRAINLIEST!!! SOLVE ASAP!!!

The average temperature for a cat is 101.2° F, but it can vary by as much as 1.3° F. Write an inequality to represent the normal temperature range of a cat, where t represents body temperature.

A |t − 1.3| ≤ 101.2
B |t − 1.3| ≥ 101.3
C |t − 101.2| ≤ 1.3
D |t − 101.2| ≥ 1.3

The joint density function of X and Y is given by f(y1,y2)={1/8 y1 e^-(y1+y2)/2, y1<0, y2>0; 0, the probability that a component of type II will have a life length in excess of 400 hours is 0.

To find the probability that a component of type II will have a life length in excess of 400 hours, we need to integrate the joint density function over the region where Y exceeds 4 (since the length is given in hundreds of hours).

So, we have:

P(Y > 4) = ∫∫ f(x,y) dx dy, where the limits of integration are x=-∞ to x=∞ and y=4 to y=∞

Substituting the given joint density function, we have:

P(Y > 4) = ∫∫ (1/8) y1 e^-(y1+y2)/2 dx dy, where the limits of integration are x=-∞ to x=∞ and y=4 to y=∞

Since the joint density function is zero elsewhere, we can simplify the limits of integration to:

P(Y > 4) = ∫4∞ ∫0∞ (1/8) y1 e^-(y1+y2)/2 dx dy

Evaluating the inner integral with respect to x, we get:

P(Y > 4) = ∫4∞ [(1/8) y1 e^-(y1+y2)/2] dy

Using integration by parts, we can simplify this integral to:

P(Y > 4) = [1/8] (e^-2) ∫4∞ y1 e^(y1/2) dy

Solving this integral, we get:

P(Y > 4) = [1/8] (e^-2) (8e^2 - 16e^2) = 1/2e^2 ≈ 0.675

Therefore, the probability that a component of type II will have a life length in excess of 400 hours is approximately 0.675.
To find the probability that a component of type II will have a life length in excess of 400 hours, we will focus on variable Y, which represents the length of life for components of type II. Since we are looking for the probability of Y > 4 (because the length is in hundreds of hours), we need to calculate the following integral:

P(Y > 4) = ∫[∫(1/8 * y1 * e^-(y1+y2)/2) dy1] dy2

Since the joint density function f(y1, y2) is defined for y1 < 0 and y2 > 0, the integral limits for y2 will be from 4 to infinity, while the integral limits for y1 will be from -infinity to 0. Thus, we have:

P(Y > 4) = ∫[∫(1/8 * y1 * e^-(y1+y2)/2) dy1] dy2, where the limits for y2 are 4 to infinity and the limits for y1 are -infinity to 0.

However, as we integrate over y1, we realize that y1 is always negative, which causes the joint density function to be 0, rendering the entire integral 0.

Therefore, the probability that a component of type II will have a life length in excess of 400 hours is 0.

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The given statement "GROUPING SETS is extension to GROUP BY clause and is used to specify multiple groupings of data." is true because GROUPING SETS is an extension of the GROUP BY.

GROUPING SETS is an extension of the GROUP BY clause in SQL that allows for multiple groupings of data to be specified, but still provide a single result set. This feature was introduced in SQL Server 2008 and is now supported by many other database management systems.

When using GROUPING SETS, multiple grouping expressions can be listed within a single GROUP BY clause, and the result set will contain one row for each combination of the grouping expressions. This can be useful for generating summary reports or comparing different levels of aggregation.

For example, consider a table containing sales data for a retail store, with columns for date, product, and sales amount. A GROUPING SETS query could be used to generate a report showing total sales by day, by product, and overall, all in a single result set.

The GROUP BY clause could specify grouping sets for (date), (product), and (), which would provide three levels of aggregation in the report.

Overall, GROUPING SETS provides a powerful and flexible tool for grouping data in SQL queries, allowing for multiple levels of aggregation to be specified and combined into a single result set.

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The standard error of the sample mean is $25,000, the probability that the sample mean exceeds $968,000 is 0.0735, or 7.35%.c. Using the same formula as in part b, we find: , the probability that the sample mean exceeds $935,000 is 0.8413, or 84.13%., the probability that the sample mean is between $930,000 and $963,000 is 0.4633, or 46.33%.

a. The standard error of the sample mean is calculated as:

SE = σ/√n

where σ is the standard deviation of the population, n is the sample size.

In this case, σ = $250,000 and n = 100.

SE = $250,000/√100 = $25,000

Therefore, the standard error of the sample mean is $25,000.

b. To calculate the probability that the sample mean exceeds $968,000, we need to standardize the sample mean using the formula:

z = (x - μ) / (σ / √n)

where x is the sample mean, μ is the population mean, σ is the population standard deviation, and n is the sample size.

In this case, x = $968,000, μ = $950,000, σ = $250,000, and n = 100.

z = ($968,000 - $950,000) / ($250,000 / √100) = 1.44

Using a standard normal distribution table or calculator, we find that the probability of a z-score being greater than 1.44 is approximately 0.0735.

Therefore, the probability that the sample mean exceeds $968,000 is 0.0735, or 7.35%.c. Using the same formula as in part b, we find:

z = ($935,000 - $950,000) / ($250,000 / √100) = -1

The probability of a z-score being less than -1 is approximately 0.1587. However, we are interested in the probability that the sample mean exceeds $935,000, which is equivalent to the probability that the z-score is greater than -1.

Using the symmetry of the normal distribution, we can subtract the probability of a z-score being less than -1 from 1 to find the probability of a z-score being greater than -1:

P(z > -1) = 1 - P(z < -1) = 1 - 0.1587 = 0.8413

Therefore, the probability that the sample mean exceeds $935,000 is 0.8413, or 84.13%.

d. To calculate the probability that the sample mean is between $930,000 and $963,000, we need to standardize both values:

z1 = ($930,000 - $950,000) / ($250,000 / √100) = -0.8

z2 = ($963,000 - $950,000) / ($250,000 / √100) = 0.52

Using a standard normal distribution table or calculator, we find the probability of a z-score being between -0.8 and 0.52 is approximately 0.4633.

Therefore, the probability that the sample mean is between $930,000 and $963,000 is 0.4633, or 46.33%.

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Using a t-test with a one-tailed distribution, the calculated t-value is (15.7-14)/(6/√42) = 2.69. Using a t-distribution table with 41 degrees of freedom (42-1), the p-value for this test is 0.0069. This means that there is strong evidence to suggest that the weight increase in the second trimester for expectant mothers is greater than 14 pounds.

To find the p-value for the hypothesis test, we will follow these steps:

1. State the null and alternative hypotheses:
  Null hypothesis (H₀): μ = 14 (The average weight increase in the second trimester is 14 pounds)
  Alternative hypothesis (H₁): μ > 14 (The average weight increase in the second trimester is greater than 14 pounds)
Based on the given information, the study conducted by CHDS found that the average weight increase in the second trimester for expectant mothers is 14 pounds. However, a random sample of 42 expectant mothers in 2015 showed that the mean weight increase in the second trimester is 15.7 pounds, with a standard deviation of 6.0 pounds. To determine if there is evidence that weight increase in the second trimester is greater than 14 pounds, a hypothesis test is conducted.
The null hypothesis (H0) is that the weight increase in the second trimester is equal to 14 pounds, while the alternative hypothesis (Ha) is that the weight increase in the second trimester is greater than 14 pounds.

2. Calculate the test statistic using the sample data:
  Test statistic = (Sample mean - Population mean) / (Sample standard deviation / √Sample size)
  Test statistic = (15.7 - 14) / (6.0 / √42)
  Test statistic ≈ 2.047

3. Determine the p-value using the test statistic and the standard normal distribution (Z-distribution):
  Since the alternative hypothesis is one-tailed (μ > 14), we will find the area to the right of the test statistic.
  Using a Z-table or calculator, we find the p-value for a Z-score of 2.047.
  p-value ≈ 0.0207

So, the p-value for the hypothesis test is approximately 0.0207, rounded to 4 decimal places.

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The unit vector in the same direction as v = 5i - 12j is (5i - 12j)/13 and the magnitude of this new unit vector is 1 is verified.

To find the unit vector in the same direction as a given vector, we first need to find the magnitude of the vector. The magnitude of a vector is the square root of the sum of the squares of its components. For the given vector v = 5i - 12j, the magnitude is:

|v| = √(5² + (-12)²) = √(25 + 144) = √169 = 13

To find the unit vector in the same direction as v, we divide v by its magnitude:

u = v/|v| = (5i - 12j)/13

This gives us the unit vector in the same direction as v. To verify that the magnitude of this new unit vector is 1, we need to find its magnitude:

|u| = √[(5/13)² + (-12/13)²] = √(25/169 + 144/169) = √(169/169) = 1

Therefore, the magnitude of the new unit vector is indeed 1, which confirms that it is a unit vector in the same direction as v.

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P(100 <= X <= 115) = sum[(800 choose i) * 0.15^i * (1 - 0.15)^(800 - i)] for i = 100 to 115

= 0.0349 (using a calculator or software)

This is a binomial distribution problem where the probability of success (having gluten sensitivity) is 0.15 and the number of trials (sample size) is 800.

a) The probability of exactly 115 individuals having gluten sensitivity is:

P(X = 115) = (800 choose 115) * 0.15^115 * (1 - 0.15)^(800 - 115)

= 0.0066 (using a calculator or software)

b) The probability of at least 107 individuals having gluten sensitivity is:

P(X >= 107) = 1 - P(X < 107)

= 1 - P(X <= 106)

= 1 - sum[(800 choose i) * 0.15^i * (1 - 0.15)^(800 - i)] for i = 0 to 106

= 0.1428 (using a calculator or software)

c) The probability of at most 100 individuals having gluten sensitivity is:

P(X <= 100) = sum[(800 choose i) * 0.15^i * (1 - 0.15)^(800 - i)] for i = 0 to 100

= 0.0002 (using a calculator or software)

d) The probability of between 100 and 115 individuals having gluten sensitivity is:

P(100 <= X <= 115) = sum[(800 choose i) * 0.15^i * (1 - 0.15)^(800 - i)] for i = 100 to 115

= 0.0349 (using a calculator or software)

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c. =BINOM.DIST(3, 5, 0.2, TRUE)

1. This is a binomial probability problem since we have multiple-choice questions with a fixed probability of success (1 out of 5 or 0.2) for each question.

2. The Excel function for binomial probability is BINOM.DIST().

3. We want the probability of guessing less than four answers correctly, which means we need the cumulative probability of guessing 0, 1, 2, or 3 answers correctly.

4. Therefore, we use the formula "=BINOM.DIST(3, 5, 0.2, TRUE)", where 3 is the number of successes, 5 is the number of trials (questions), 0.2 is the probability of success (1/5), and TRUE indicates that we want the cumulative probability.

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The sum of two rational numbers 1/2 and 5/2 is 3.

What is rational expression?

A rational expression is the ratio of two polynomials.

To add or subtract two rational expressions with the same denominator, we simply add or subtract the numerators and write the result over the common denominator.

If the denominators are not the same, we need to manipulate them to make them the same. In other words, we need to find a common denominator.

There are a few steps to follow when adding or subtracting rational expressions with different denominators.

To add or subtract rational expressions with different denominators,

Here given two rational numbers are 1/2 and 5/2.

We want to add them.

So,

[tex] \frac{1}{2} + \frac{5}{2} [/tex]

Here LCM of denominators of two numbers is 2.

So,[tex] \frac{1 + 5}{2} [/tex]

We are adding 1 and 5,

[tex] \frac{6}{2} = 3[/tex]

Therefore, sum of two rational numbers 1/2 and 5/2 is 3.

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Correct question is "Add the rational expression 1/2 and 5/2".

The change of variables X = x, Y = ln(y), C = e^B is used to linearize the exponential regression model so that it can be solved using linear regression.

To find the least-squares curve y = f(x) = Ce^Ax using the change of variables X = x, Y = ln(y), C = e^B for the given data set, we can use the following Matlab code:

X = [-1 0 1 2 3];

Y = [6.62 3.94 2.17 1.35 0.89];

Ylog = log(Y);

A = [X' ones(size(X'))];

B = Ylog';

coeffs = A\B;

C = exp(coeffs(2));

A = coeffs(1);

x = linspace(-1,3);

y = C*exp(A*x);

plot(X,Y,'o',x,y)

The code first defines the x and y values of the data set. Then, it takes the natural logarithm of the y values and defines the matrix A as [X 1]. The matrix B is defined as the transpose of the natural logarithm of y. We use the backslash operator to solve the linear equation Ax = B for the coefficients A and B. We then calculate C as e^B and redefine A as the first element of the coefficients vector. Finally, we define a range of x values and calculate the corresponding y values using the equation y = Ce^Ax. The code then plots the original data points as circles and the least-squares curve as a line.

Note that the least-squares curve y = f(x) = Ce^Ax is equivalent to the exponential regression model y = Ce^Ax, where C is the y-intercept and A is the rate of change. The change of variables X = x, Y = ln(y), C = e^B is used to linearize the exponential regression model so that it can be solved using linear regression.

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The thickness measurements of the metal pieces should be centered around a mean value, with the majority of the measurements falling within a certain range and then tapering off towards the tails of the distribution.

Measuring the thickness of pieces of metal with the same dimensions using a micrometer/caliper due to variability in the dimensions caused by errors during the manufacturing process. The distribution of the measurements taken by the entire lab section is expected to resemble a normal distribution, also known as a Gaussian distribution or bell curve.

The reason for this expectation is that the manufacturing process may introduce small, random errors that affect the dimensions of the metal pieces. The central limit theorem states that, given a large enough sample size, the distribution of these random errors will approximate a normal distribution.

The thickness measurements of the metal pieces should be centered around a mean value, with the majority of the measurements falling within a certain range and then tapering off towards the tails of the distribution.

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The solution to the differential equation is given by the equation arctan(x) + arctan(y) - ln|x²+y²+1| = C.

The differential equation is given as:

x(1+y²)dx-y(1+x²)dy=0

To solve this differential equation, we can start by rearranging the terms and separating the variables. We can start by dividing both sides by x(1+y²), which gives:

dx/(1+y²) - y(1+x²)/(x(1+y²)) dy = 0

Next, we can integrate both sides of the equation. On the left-hand side, we can use the substitution u = y² + 1, which gives du = 2y dy. The equation then becomes:

∫dx/(1+y²) - ∫(1+x²)/x du = C

where C is the constant of integration.

To solve the second integral on the right-hand side, we can use the substitution v = x², which gives dv = 2x dx. The equation then becomes:

∫dx/(1+y²) - ∫(1+v)/v dv = C

To solve the first integral, we can use the substitution y = tanθ, which gives dy = sec²θ dθ. The equation then becomes:

∫dx/cos²θ - ∫(1+v)/v dv = C

We can simplify the first integral using the trigonometric identity sec²θ = 1 + tan²θ. The equation then becomes:

∫dx/(1+ tan²θ) - ∫(1+v)/v dv = C

The first integral can be evaluated using the substitution x = tanφ, which gives dx = sec²φ dφ. The equation then becomes:

∫sec²φ dφ/(1+tan²θ) - ∫(1+v)/v dv = C

Simplifying the first integral using the identity sec²φ = 1 + tan²φ, the equation becomes:

∫(1+tan²θ) dθ/(1+tan²θ) - ∫(1+v)/v dv = C

The first integral simplifies to ∫dθ, which is just θ + K, where K is another constant of integration. Substituting back the variables, we have:

arctan(x) + arctan(y) - ln|v| = C

where v = x² and C = K - ln|D|, where D is a constant of integration.

Finally, we can substitute back the variables u = y² + 1 and v = x² to obtain the solution to the differential equation:

arctan(x) + arctan(y) - ln|x²+y²+1| = C

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Complete Question:

Solve the differential equation:

x(1+y²)dx-y(1+x²)dy=0

In the study conducted by Wilkinson et al. (2021), the researchers investigated the secondary attack rate of COVID-19 among household contacts in the Winnipeg Health Region, Canada. The study included 28 individuals from 102 unique households. The findings of this research contribute to our understanding of COVID-19 transmission within households and can inform public health strategies to prevent further spread of the virus.

A study conducted by Wilkinson et al. in 2021 on the secondary attack rate of COVID-19 in household contacts in the Winnipeg Health Region in Canada. The authors of the study included 28 individuals from a total of 102 households in their analysis.

To provide a bit more context, the secondary attack rate refers to the proportion of individuals who develop COVID-19 after being exposed to a person with a confirmed case of the disease. In the case of household contacts, this would refer to individuals who live with someone who has tested positive for COVID-19.

Wilkinson et al.'s study aimed to investigate the factors that might affect the secondary attack rate in household contacts, such as the age and sex of the individuals involved, as well as any potential exposure to other sources of COVID-19 outside of the household. By analyzing data from a total of 102 households, the authors were able to provide valuable insights into the transmission dynamics of COVID-19 within households and the factors that might influence the likelihood of transmission occurring.

Overall, the study by Wilkinson et al. provides important information for public health officials and policymakers working to contain the spread of COVID-19, particularly in terms of understanding how the disease is transmitted within households and what factors might contribute to higher rates of transmission.

In the study conducted by Wilkinson et al. (2021), the researchers investigated the secondary attack rate of COVID-19 among household contacts in the Winnipeg Health Region, Canada. The study included 28 individuals from 102 unique households. The findings of this research contribute to our understanding of COVID-19 transmission within households and can inform public health strategies to prevent further spread of the virus.

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Complete question: Wilkinson et al. (2021) studied the secondary attack rate of COVID-19 in houschold contacts in the Winnipeg Health Region, Canada. In their study, the authors included 28 individu- als from 102 unique households (102 primary cases and 279 household contacts). A total of 41 contacts from 25 households developed COVID-19 symptom in the 11 days since last un- protected exposure to the primary case. Calculate the secondary attack rate of COVID-19.

Answer:

x = 8

Step-by-step explanation:

The sum of a number x and 4 equals 12

x + 4 = 12

x = 8

So, the number is 8

Answer: B 52%

Step-by-step explanation:

Percentages are usually out of 100%

so just subtract the other students vote percentage from 100

100% - 29% - 19% = 52%

The factors of Rice type (brown or white), Cheese type (cheddar or mozzarella), and Bean type (black or pinto) seem to have a significant impact on whether children enjoy the taco bowls on Taco Tuesday.

55 Based on the Pareto Chart of the Effects, the significant main effects that should be retained are Rice, Cheese, and Beans.

To determine the significant main effects, we will use Lenth's Pseudo Standard Error (PSE) as a reference point. In this case, the Length's PSE is 0.28125. Main effects with an effect greater than the PSE are considered significant.

From the Pareto Chart of the Effects, the main effects and their values are:

1. Cheese (C): 0.58
2. Sour Cream (B): Value not provided
3. Beans (A): Value not provided
4. Rice (D): Value not provided
5. Chips (E): Value not provided
6. Salsa (F): Value not provided

We can see that only the Cheese effect (0.58) is provided and is greater than Lenth's PSE (0.28125), so it is significant. Unfortunately, we do not have enough information to determine the significance of the other factors.

However, based on the provided information, the significant main effect that should be retained is: Cheese

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The probability of selecting a female or a mathematics major or both is 0.7 or 70%.

To find the probability that the selected student will be female or a mathematics major or both, we need to add the probabilities of each event happening separately and then subtract the probability of both events happening at the same time.

First, the probability of selecting a female student is 25/50 = 0.5.

Second, the probability of selecting a mathematics major is 15/50 = 0.3.

Third, the probability of selecting a female mathematics major is 10/50 = 0.2.

To find the probability of selecting either a female or a mathematics major, we add the probabilities of each event happening separately:

0.5 + 0.3 = 0.8.

To find the probability of selecting both a female and a mathematics major, we multiply the probabilities of each event happening together:

0.5 x 0.2 = 0.1.

To find the probability of selecting either a female or a mathematics major or both, we subtract the probability of selecting both events at the same time from the sum of the probabilities of each event happening separately:

0.8 - 0.1 = 0.7.

Therefore, the probability of selecting a female or a mathematics major or both is 0.7 or 70%.

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We can interpret this interval as follows: we are 98% confident that the true population variance falls within this interval.

To construct a 98% confidence interval for the population variance, we can use the following formula:

CI = [(n-1)s^2 / χ^2(α/2, n-1), (n-1)s^2 / χ^2(1-α/2, n-1)]

where n is the sample size, s is the sample standard deviation, χ^2(α/2, n-1) is the chi-squared value with α/2 degrees of freedom, and χ^2(1-α/2, n-1) is the chi-squared value with 1-α/2 degrees of freedom.

In this case, n = 35, s = 2.8, α = 0.02 (since we want a 98% confidence interval), and degrees of freedom = n-1 = 34.

Using a chi-squared table or calculator, we can find χ^2(α/2, n-1) to be 19.196 and χ^2(1-α/2, n-1) to be 53.984.

Plugging in the values, we get:

CI = [(n-1)s^2 / χ^2(α/2, n-1), (n-1)s^2 / χ^2(1-α/2, n-1)]

= [(34)(2.8^2) / 19.196, (34)(2.8^2) / 53.984]

= [3.662, 8.676]

Therefore, the 98% confidence interval for the population variance is (3.662, 8.676). We can interpret this interval as follows: we are 98% confident that the true population variance falls within this interval.

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The result of the equation 4.53 x 10⁵ + 2.2 x 10⁶ is 2.653 x 10⁶.

To solve this given equation,

One first needs to take the common exponent out in both numbers

i.e. we need to take common from 4.53 x 10⁵ and 2.2 x 10⁶ which comes out to be 10⁵

Therefore, using the distributive property of multiplication that states ax + bx = x (a+b)

we have, 4.53 x 10⁵ + 2.2 x 10⁶ = 10⁵ (4.53 + 2.2 x 10)

= 10⁵ (4.53 + 22)

= 10⁵ (26.53)

=26.53 x 10⁵

We convert this into proper decimal notation, and we get

=2.653 x 10⁶

Therefore, we get 2.653 x 10⁶ as the result of the given equation.

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(a) The model for C(t), the concentration of the drug after t hours is      [tex]C(t) = 300 * (0.65^t)[/tex]. (b) The concentration of the drug after 6 hours is      35 mg/mL.

We need to find a model for C(t), the concentration of the drug after t hours, and determine the concentration after 6 hours, using the information provided.

(a) Since the concentration is 65% of the level of the previous hour, we can represent this as a decay model. The general form of an exponential decay model is [tex]C(t) = C_0 * (r^t)[/tex], where [tex]C_0[/tex] is the initial concentration and r is the decay rate.

In this case, the initial concentration [tex]C_0[/tex] is 300 mg/mL, and the decay rate r is 65% or 0.65 (as a decimal). So, our model for C(t) is:

[tex]C(t) = 300 * (0.65^t)[/tex]

(b) To determine the concentration of the drug after 6 hours, we need to plug t = 6 into the model:

[tex]C(6) = 300 * (0.65^6)[/tex]

C(6) ≈ 34.68 mg/mL

Rounding to the nearest whole number, the concentration of the drug after 6 hours is approximately 35 mg/mL.

In summary, the model for the concentration of the drug after t hours is [tex]C(t) = 300 * (0.65^t)[/tex], and the concentration after 6 hours is approximately 35 mg/mL.

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The length of the missing side in the right triangle is 6.5

We can see that it is a right triangle, thus, we can use the Pythagorean's theorem.

It says that the sum of the squares of the legs is equal to the square of the hypotenuse.

So if x is the missing side, we can write:

x² + 3.6² = 7.4²

Solving that for x we will get.

x = √(7.4² - 3.6²) = 6.5

That is the length of the missing side.

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The given rational number 7.84848484848...  can be represented in the form of the fraction [tex]\frac{868}{9801}.[/tex]

To express 7.84848484848... as a rational number, we can represent it as an infinite repeating decimal:

7.84848484848... = 7.84 + 0.004848484848...

Let x = 0.004848484848...

Then 100x = 0.4848484848...

Subtracting the two equations gives:

99x = 7.84848484848... - 7.84 = 0.00848484848...

Simplifying:

x = 0.00848484848... / 99

Now, we need to express 0.00848484848... as a fraction. Let y = 0.00848484848...

Then 100y = 0.848484848...

Subtracting the two equations gives:

99y = 0.84

Simplifying:

y = 0.84 / 99

Substituting back into the first equation:

x = (0.84 / 99) / 99

Simplifying:

x = 0.84 / (99^2)

Now we can add the two fractions:

7.84848484848... = 7.84 + x = 7.84 + 0.84 / (99^2)

Therefore, the rational representation of 7.84848484848... is:

p = 784 + 84 = 868

q = 99²

So, 7.84848484848... = 868/9801.

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For n=3, value of n are the line y = 3x + 1 and y = nx - 4 perpendicular.

Given that,

the line y = 3x + 1 and y = nx - 4

now, we have to find the value of n, for which the lines are perpendicular.

so, for line- 1:

y = 3x + 1

slope is: m1 = 3

for line -2:

y = nx - 4

slope is : m2 = n

we know that,

two lines are perpendicular to each other is, their slopes are equal.

i.e. m1 = m2

so, we get,

n = 3

Hence, For n=3, value of n are the line y = 3x + 1 and y = nx - 4 perpendicular.

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National Electric incurred a total cost of $7,950 in producing the first 500 coffeemakers/day.

To find the total cost incurred by National in producing the first 500 coffeemakers/day, we need to calculate the sum of the fixed cost and the variable cost for producing 500 units.

The fixed cost is given as $700.

To find the variable cost, we first need to calculate the marginal cost function, which is the derivative of the cost function:

[tex]C'(x) = 0.00003x^2 - 0.03x + 24[/tex]

The variable cost of producing x units is given by integrating the marginal cost function from 0 to x:

[tex]C(x) = \int [0, x] C'(t) dt[/tex]

[tex]C(x) = \int [0, x] (0.00003t^2 - 0.03t + 24) dt[/tex]

[tex]C(x) = 0.00001t^3 - 0.015t^2 + 24 [0, x][/tex]

[tex]C(x) = 0.00001x^3 - 0.015x^2 + 24x[/tex]

So, the variable cost of producing 500 units is:

[tex]C(500) = 0.00001(500)^3 - 0.015(500)^2 + 24(500) = $7,250[/tex]

Therefore, the total cost incurred by National in producing the first 500 coffeemakers/day is:

Total cost = Fixed cost + Variable cost

Total cost = $700 + $7,250

Total cost = $7,950.

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The directional derivative of the function w = y2 + xz at point P(1, 2, 2) in the direction of vector v = 2i - j + 2k is 10/3.

To find the directional derivative of the function w = y2 + xz at point P(1, 2, 2) in the direction of vector v = 2i - j + 2k, we first need to find the gradient of the function at point P.
The gradient of a scalar-valued function is a vector that points in the direction of the maximum rate of increase of the function, and its magnitude is equal to the rate of increase in that direction.
So, the gradient of the function w = y2 + xz at point P is:
grad(w) = ∇w = [∂w/∂x, ∂w/∂y, ∂w/∂z]
Taking partial derivatives, we get:
∂w/∂x = z
∂w/∂y = 2y
∂w/∂z = x
Therefore, the gradient at point P(1, 2, 2) is:
∇w(P) = [2, 4, 1]
Next, we need to find the directional derivative in the direction of vector v. The directional derivative is the dot product of the gradient and the unit vector in the direction of v.
First, we need to find the unit vector in the direction of v:
|v| = √(2² + (-1)² + 2²) = √9 = 3
So, the unit vector in the direction of v is:
u = v/|v| = (2/3)i - (1/3)j + (2/3)k
Now, we can find the directional derivative:
Dv(w) = ∇w(P) · u = [2, 4, 1] · [(2/3)i - (1/3)j + (2/3)k]
= (2/3)(2) - (1/3)(-4) + (2/3)(1)

= 4/3 + 4/3 + 2/3
= 10/3

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The x and y component of the vector is -5 and 10 respectively (-5, 10).

The components of the vector is calculated as follows;

The initial position of the vector = (−3,4)

The final position of the vector =  (−2,6)

The sum of the x component of the vector is calculated as;

Vx = -3 + (-2) = -5

The sum of the y component of the vector is calculated as;

Vy = 4 + 6 = 10

= (-5, 10)

Thus, the x and y component of the vector is -5 and 10 respectively.

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The magnitude of v(t) is √313 and the length of the indicated portion of the curve is π√313

To find the unit tangent vector of the curve, we need to first find the velocity vector v(t) and then divide it by its magnitude.

r(t) = (6sin 2t)i + (6 cos 2t)j + 5t K

v(t) = dr/dt = (12 cos 2t)i - (12 sin 2t)j + 5K

The magnitude of v(t) is:

|v(t)| =√(12 cos 2t)² + (-12 sin 2t)² + 5²)

|v(t)| = √(144 + 144 + 25)

|v(t)| = √313

The unit tangent vector T(t) is:

T(t) = v(t)/|v(t)|

= [(12 cos 2t)/√313]i - [(12 sin 2t)/√313]j + (5/√313)K

To find the length of the curve from t = 0 to t = pi, we use the formula:

[tex]L\:=\:\int _a^b\:\:|r'\left(t\right)|\:dt[/tex]

where a = 0 and b = pi.

|r'(t)| = |v(t)| = √313

Therefore, the length of the indicated portion of the curve is π√313

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find the curve's unit tangent vector. Also, find the length of the indicated portion of the curve.

r(t) = (6sin 2t)i +(6 cos 2t)j + 5t K   0 ≤ t ≤ pi

When calculating a 99% confidence interval with the same sample size (n) compared to a 95% confidence interval, the margin of error will be larger.

Confidence intervals are used to estimate the true population parameter based on a sample. The confidence level represents the probability that the true population parameter falls within the calculated interval. A 95% confidence interval means that there is a 95% probability that the true parameter lies within the interval, leaving a 5% chance of error. Similarly, a 99% confidence interval means that there is a 99% probability that the true parameter falls within the interval, leaving only a 1% chance of error.

To calculate a confidence interval, the margin of error is added and subtracted from the sample statistic (e.g., mean or proportion). The margin of error is influenced by the confidence level and the sample size. A higher confidence level requires a larger margin of error to account for the increased level of certainty.

As the confidence level increases from 95% to 99%, the margin of error also increases. This is because a higher confidence level requires a larger interval to be confident that the true parameter falls within it. Therefore, when calculating a 99% confidence interval with the same sample size (n) compared to a 95% confidence interval, the margin of error will be larger to accommodate the increased level of confidence.

Therefore, the margin of error will be larger when calculating a 99% confidence interval instead of a 95% confidence interval with the same sample size (n).

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Answer:60$

Step-by-step explanation:400*%15=60$

Answer:

60%

Step-by-step explanation:

400*%15=60$

Considering the given random sample of data: 12, 27, 29, 15, 23, 5, 8, 2, 110, 19

a) The median of the sample data is 17.

b) If the outlier is removed, 17 is the median of the remaining sample data.

a) To find the median of the sample data, we need to first arrange the numbers in order from smallest to largest: 2, 5, 8, 12, 15, 19, 23, 27, 29, 110. Then, we can see that there are 10 numbers in the sample, so the median will be the average of the 5th and 6th numbers in the list. So, the median is (15 + 19)/2 = 17.
b) If the outlier (110) is removed, then the remaining sample data is 2, 5, 8, 12, 15, 19, 23, 27, 29. Again, we arrange the numbers in order from smallest to largest: 2, 5, 8, 12, 15, 19, 23, 27, 29. Now, we can see that there are 9 numbers in the sample, so the median will be the average of the 5th and 6th numbers in the list. So, the median of the remaining sample data is (15 + 19)/2 = 17.

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The probability that at least one of the 3 randomly chosen respondents voted for pink is approximately 76.17%.

To find the probability that at least one of the three respondents chose pink, we need to calculate the probability that none of them chose pink and then subtract that from 1.

The probability that the first respondent did not choose pink is 0.62 (since 38% chose pink, the probability that the first respondent did not choose pink is 1-0.38=0.62). The probability that the second respondent did not choose pink is also 0.62, and the probability that the third respondent did not choose pink is also 0.62.

To find the probability that none of the three respondents chose pink, we multiply these probabilities together:

0.62 x 0.62 x 0.62 = 0.238328

So the probability that none of the three respondents chose pink is 0.238328.

To find the probability that at least one of them chose pink, we subtract this from 1:

1 - 0.238328 = 0.761672

So the probability that at least one of the three respondents chose pink is approximately 0.761672 or 76.17%.
To find the probability that at least one of the 3 randomly chosen respondents voted for pink, we can use the complementary probability method. First, let's find the probability that none of the 3 respondents chose pink.

The probability that a respondent did not choose pink is 1 - 0.38 = 0.62.

The probability that all 3 respondents did not choose pink is (0.62)^3 = 0.238328.

Now, we can find the complementary probability, which is the probability that at least one respondent chose pink: 1 - 0.238328 = 0.761672.

So, the probability that at least one of the 3 randomly chosen respondents voted for pink is approximately 76.17%.

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