1. If 0 = {n € Zin is odd) and B = {ne Z|-4 ≤ n ≤ 6}, calculate the following:(a) O ∩ B(B) B - O(C) O (in Z)

The absolute maximum value is 685 and the absolute minimum value is 82.

The given function is f(x) = 81 + x² defined on the interval [1, 18].

To find the absolute maximum and minimum values of the function on this interval, we need to find the critical points and the endpoints.

Firstly, we take the derivative of the function:

f'(x) = 2x

Setting f'(x) = 0, we get x = 0, which is not in the interval [1, 18]. Therefore, there are no critical points.

Next, we check the endpoints:

f(1) = 82
f(18) = 685

The absolute maximum value is 685 and the absolute minimum value is 82.

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The correct option is D) 0.45. The probability of having less than 2 accidents on a given day is 0.45

To determine the probability of having less than 2 accidents on a given day, you will need to sum the probabilities of having 0 or 1 accidents. Here is the step-by-step explanation:

1. Identify the probabilities of having 0 and 1 accidents:
  - 0 accidents: Probability = 0.25
  - 1 accident: Probability = 0.20

2. Add these probabilities together:
  - 0.25 (0 accidents) + 0.20 (1 accident) = 0.45

The probability of having less than 2 accidents on a given day is 0.45, which corresponds to option D.

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Every one additional hours of sleep is associated with an additional five points on the exam.

The y-intercept of the line of best fit represents a likely exam score without any sleep.

From the given scatter plot we can see that the coordinates are (0,55), (1,60), (2, 65), (3, 70), (4, 75), (5, 80), (6, 85), (7, 90), (8, 95), (9, 100).

So the best fit line passes through (0, 55) and (1, 60).

From the two point form of a straight line we get the equation of straight line is,

(y - 55)/(x - 0) = (55 - 60)/(0 - 1)

(y - 55)/x = -5/(-1)

y - 55 = 5x

y = 5x + 55

Here X axis suggests the the hour of sleep taken by student before exam day and Y axis suggests Total score obtained by student.

Clearly we can see that for each 1 additional hour of sleep student can get 5 more scores.

In order to get y intercept we have to substitute x = 0 in the equation of the best fit line.

y = 5*0 + 55 = 55.

Every one additional hours of sleep is associated with an additional five points on the exam.

The y-intercept of the line of best fit represents a likely exam score without any sleep.

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The question is incomplete. Complete question will be -

The approximate probability of getting more than 42 successes in a sample of 78 with a population proportion of 0.59 is 0.9254, or about 92.54%.

The study of random events and their likelihood of occurring is the focus of the probability field, which is a subfield of statistics.

The conditions are:

Here, n = 78 and p = 0.59, so np = 46.02 and n(1-p) = 31.98,
which are both greater than 10. We assume that the observations are independent. Therefore, we can use the normal approximation to approximate the binomial distribution.

We need to find P(X > 42), which is the probability of getting more than 42 successes in a sample of 78 if the population proportion of successes is 0.59. Using the normal approximation, we can approximate this as:

[tex]P(X > 42)[/tex]   ≈    [tex]P(Z > \frac{42 - np}{\sqrt{np(1-p)} } )[/tex]

where Z is a standard normal random variable.

Substituting the values, we get:

[tex]P(X > 42)[/tex]   ≈   [tex]P(Z > \frac{42 - 46.02}{\sqrt{46.02(1-0.59)} } )[/tex]

[tex]P(X > 42)[/tex]   ≈    [tex]P(Z > -0.9254)[/tex]

Using a standard normal table or calculator, we can find that P(Z > -0.9254) is approximately 0.9254.

Therefore, the approximate probability of getting more than 42 successes in a sample of 78 with a population proportion of 0.59 is 0.9254, or about 92.54%.

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The average number of customers who made purchases in the store each day, based on the given data, is 9.6.

The store manager recorded the daily number of customers who made purchases in the store over a period of time. The data collected show the number of customers for each day: 10, 11, 8, 14, 7, 10, 10, 11, 8, and 7.

The store manager has recorded the number of customers who made purchases in the store each day. The data provided are a series of numbers representing the number of customers for each day: 10, 11, 8, 14, 7, 10, 10, 11, 8, and 7.

To analyze this data, we can calculate the mean (average) number of customers per day. The mean is calculated by summing up all the numbers and dividing by the total number of days. Let's add up the numbers:

10 + 11 + 8 + 14 + 7 + 10 + 10 + 11 + 8 + 7 = 96

The total number of days is 10, as there are 10 numbers in the data set. Now, let's calculate the mean:

Mean = Sum of numbers / Total number of days = 96 / 10 = 9.6

Therefore, the average number of customers who made purchases in the store each day, based on the given data, is 9.6

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Null Hypothesis (H₀): p = 2/3

Alternative Hypothesis (H₁): p > 2/3

Null Hypothesis (H₀): μ = 170
Alternative Hypothesis (H₁): μ < 170

Null Hypothesis (H₀): μ = 10.3
Alternative Hypothesis (H₁): μ > 10.3

In this scenario, we are interested in determining if more than two-thirds of employers perform background checks.

We can set up our null and alternative hypotheses as follows:
Null Hypothesis (H₀): p = 2/3 (No difference, exactly two-thirds of employers perform background checks)
Alternative Hypothesis (H₁): p > 2/3 (More than two-thirds of employers perform background checks)
For the blood cholesterol level in Japanese children, we want to determine if their mean level is lower than the known mean level for U.S. children (170). Our hypotheses would be:
Null Hypothesis (H₀): μ = 170 (Mean blood cholesterol level for Japanese children is the same as U.S. children)
Alternative Hypothesis (H₁): μ < 170 (Mean blood cholesterol level for Japanese children is lower than U.S. children)
In the cruise ship scenario, we want to determine if lowering the price of soda will increase the mean number of cans sold per passenger for a 10-day trip.

The current mean is 10.3 cans. Our hypotheses would be:
Null Hypothesis (H₀): μ = 10.3 (No change in the mean number of cans sold per passenger)
Alternative Hypothesis (H₁): μ > 10.3 (The mean number of cans sold per passenger has increased with the new pricing)
In each case, we start by assuming that there is no difference (the null hypothesis). We then test the alternative hypothesis to determine if there is evidence to support the claim of a difference or a specific direction of change.

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To find f, we need to integrate f '(t) with respect to t. Using the formula ∫sec(t)dt = ln|sec(t) + tan(t)| + C, we can rewrite f '(t) as:
f '(t) = sec(t)(sec(t) + tan(t)) = sec(t)sec(t) + sec(t)tan(t)
= sec^2(t) + sec(t)tan(t)

Now we can integrate:
∫f '(t)dt = ∫sec^2(t)dt + ∫sec(t)tan(t)dt
Using the formula ∫sec^2(t)dt = tan(t) + C, we get:
∫sec^2(t)dt = tan(t)
For the second integral, we can use u-substitution with u = sec(t), du/dt = sec(t)tan(t), so:
∫sec(t)tan(t)dt = ∫u du = 1/2 u^2 + C
= 1/2 sec^2(t) + C
Putting it all together:
f(t) = ∫f '(t)dt = tan(t) + 1/2 sec^2(t) + C
To find C, we can use the initial condition f(π/4) = -6:
-6 = tan(π/4) + 1/2 sec^2(π/4) + C
-6 = 1 + 1/2(2) + C
C = -10
Therefore, the solution is:
f(t) = tan(t) + 1/2 sec^2(t) - 10

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a) the height function is: [tex]h(t) = -16t^2 + 320[/tex]

b) the ballast will strike the ground after approximately 4.47 sec

c) the velocity of the ballast when it hits the ground is approximately -71.56 ft/sec.

What is Velocity?

Velocity of a particle is its speed and the direction in which it is moving.

(a) We know that the velocity of the ballast is given by v(t) = -32t. Integrating v(t), we get the height function h(t):

[tex]h(t) = \int\limits v(t) \, dt\\\\h(t) = \int\limits -32(t) \, dt\\\\h(t) = dt = -16t^2 + C[/tex]

[tex]h(0) = -16(0)^2 + C = C = 320[/tex]

So the height function is:

[tex]h(t) = -16t^2 + 320[/tex]

(b) To find when the ballast strikes the ground, we need to find the time t when h(t) = 0:

[tex]-16t^2 + 320 = 0\\\\t = \sqrt {(320/16)} = \sqrt{20}\\\\ t= 4.47 seconds[/tex]

So the ballast will strike the ground after approximately 4.47 seconds.

(c) To find the velocity of the ballast when it hits the ground, we can simply plug in t = √20 into the velocity function v(t):

[tex]v(\sqrt{20)} = -32(\sqrt{20}) = -71.56 ft/sec[/tex]

So the velocity of the ballast when it hits the ground is approximately -71.56 ft/sec.

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The general solution to the given differential equation is y(x) = c1 e²ˣ + c2 cos(3x) + c3 sin(3x), where c1, c2, and c3 are constants determined by initial conditions.

The characteristic polynomial is r³ + 2r² - 54r - 10 = 0, which can be factored as (r + 2)(r - 3i)(r + 3i) = 0. Since r = -2 is a root of the polynomial, the homogeneous solution contains the term e²ˣ. The other two roots, 3i and -3i, contribute the terms cos(3x) and sin(3x), respectively, to the solution.

This is due to the fact that the real part of a complex root gives a cosine term and the imaginary part gives a sine term. The general solution is a linear combination of these three terms, and the constants are determined by any initial conditions provided.

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The solution for given algebraic-equation 2x - 6 = 12, using transposition method or algebra tiles method is 9. And the first step will be transferring (-6) to right hand side by converting it into (+6)

What is an algebraic-equation?

If two things are equal, an equation declares it. For the purpose of solving the equation, variables are added, subtracted, multiplied, and divided. As long as both sides receive the same treatment, the equation will remain balanced. Algebra If you want to represent integers (or constants) and variables, you can use a set of square and rectangular figure tiles called tiles. The shape of each tile is tied to the unit square in this visual, area-based paradigm.

As per the given equation, algebra tiles are arranged. the variable represented by circle. Taking two tiles of the variable 'x' as it is tice of the variable 'x'. And the numbers represented by square tiles.

The steps to find solution is as follows:

2x - 6 = 12

2x = 12 + 6

2x = 18

x = 18 ÷ 2

x = 9

9 is the solution of the given equation.

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Refer to the attachment for the tiles.

Yes, the conclusion in part a) matches the conclusion in part b). Both methods lead to the same conclusion that there is not enough evidence to conclude that the average test scores were different in 2004 and 2008.

What is Hypothesis?

A hypothesis is a proposed explanation or prediction for a phenomenon, based on limited evidence or prior knowledge, which can be tested through further investigation or experimentation. It serves as a starting point for scientific inquiry and can be either supported or rejected based on the results of the investigation or experiment.

a) To test the hypothesis that the average test scores were the same in 2004 and 2008, we can use a two-sample t-test. The null hypothesis is that the mean difference between the scores in 2004 and 2008 is equal to zero. The alternative hypothesis is that the mean difference is not equal to zero.

First, we need to check the conditions for a two-sample t-test:

The samples are independent.

The population distributions are approximately normal, or the sample sizes are large enough to rely on the central limit theorem.

The population variances are equal (we can check this later).

We are given that the samples are simple random samples, so the first condition is met.

To check the second condition, we can examine the sample sizes and standard deviations. Since the sample sizes are both greater than 30 and the standard deviations are similar, we can assume that the population distributions are approximately normal.

Finally, to check the third condition, we can use a pooled variance estimate:

s_pooled = √(((n1-1)*s1² + (n2-1)*s2²) / (n1+n2-2))

s_pooled = √(((1000)*39² + (1000)*38²) / (1000+1000-2))

s_pooled = 38.5

Since the sample sizes and standard deviations are similar, we can assume that the population variances are equal.

Now, we can calculate the test statistic:

t = (x1 - x2) / (s_pooled * √(1/n1 + 1/n2))

t = (257 - 260) / (38.5 * (1/1000 + 1/1000))

t = -1.406

Using a two-tailed test with a significance level of 0.05 and 998 degrees of freedom, the critical value is approximately +/- 1.962. Since the test statistic (-1.406) does not exceed the critical value, we fail to reject the null hypothesis. Therefore, there is not enough evidence to conclude that the average test scores were different in 2004 and 2008.

b) To calculate a 95% confidence interval for the change in average scores from 2004 to 2008, we can use the formula:

CI = (x1 - x2) ± t(alpha/2, df) * s_pooled * √(1/n1 + 1/n2)

CI = (257 - 260) ± 1.962 * 38.5 * √(1/1000 + 1/1000)

CI = (-8.18, 2.18)

The interpretation of the confidence interval is that we are 95% confident that the true average difference in test scores from 2004 to 2008 is between -8.18 and 2.18 points. Since the interval contains zero, this is consistent with the result of the hypothesis test in part a) that there is not enough evidence to conclude that the average test scores were different in 2004 and 2008.

c) Yes, the conclusion in part a) matches the conclusion in part b). Both methods lead to the same conclusion that there is not enough evidence to conclude that the average test scores were different in 2004 and 2008.

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Answer:2

Step-by-step explanation:

if you are paying 4 for 8 its 2

a. We can be 95% confident that the true population proportion of CEOs in the country who are extremely concerned about cyber threats lies between 0.276 and 0.360.

To construct a 95% confidence interval estimate for the population proportion of CEOs in the country who are extremely concerned about cyber threats, we can use the following formula:

CI = a ± z*sqrt((a(1-a))/n)

Where:

a = sample proportion = 149/468 = 0.318

z = z-value for 95% confidence interval = 1.96 (from standard normal distribution table)

n = sample size = 468

Plugging in the values, we get:

CI = 0.318 ± 1.96*sqrt((0.318(1-0.318))/468)

CI = 0.318 ± 0.042

CI = (0.276, 0.360)

Therefore, we can be 95% confident that the true population proportion of CEOs in the country who are extremely concerned about cyber threats lies between 0.276 and 0.360.

b. We can be 90% confident that the true population proportion of CEOs in the country who are extremely concerned about a lack of trust in business lies between 0.150 and 0.222.

To construct a 90% confidence interval estimate for the population proportion of CEOs in the country who are extremely concerned about a lack of trust in business, we can use the same formula as in part (a):

CI = a ± z*sqrt((a(1-a))/n)

Where:

a = sample proportion = 87/468 = 0.186

z = z-value for 90% confidence interval = 1.645 (from standard normal distribution table)

n = sample size = 468

Plugging in the values, we get:

CI = 0.186 ± 1.645*sqrt((0.186(1-0.186))/468)

CI = 0.186 ± 0.036

CI = (0.150, 0.222)

Therefore, we can be 90% confident that the true population proportion of CEOs in the country who are extremely concerned about a lack of trust in business lies between 0.150 and 0.222.

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The value of vector 6v is equal to 30i + 24j and value of vector -3v is equal to  -15i - 12j.

Scalar multiplication of a vector written in terms of i and j means multiplying a vector by a scalar value, which scales the vector's magnitude while retaining its direction. To perform scalar multiplication of a vector, we simply multiply each component of the vector by the scalar value.

For example, suppose we have a vector v = 5i + 4j. To find 6v, we multiply each component of v by 6 to get 6v = (65)i + (64)j = 30i + 24j.

Similarly, to find -3v, we multiply each component of v by -3 to get -3v = (-35)i + (-34)j = -15i - 12j.

In both cases, we are scaling the original vector v by a factor of 6 and -3, respectively. The resulting vectors have the same direction as v, but their magnitudes are scaled accordingly.

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Required correct order is d, c, b, e, a.

What is the correct order?

1) Add or subtract to move the constant to the right-hand side of the equation.

2) Then we need split the middle term of the left hand side into two parts such that the product of these two parts is equal to the coefficient of the x-term.

3) Factor the first three terms of the left-hand side of the equation by grouping.

4) Then Rewrite the left-hand side of the equation as a perfect square trinomial by adding the square of half the coefficient of the x-term

5) The last step will be solve the equation by taking the square root of both sides of the equation.

Therefore, Required correct order is d, c, b, e, a.

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Но = Being debt-free does not depend on age.

H_A = Being debt-free depends on age.

The degree of freedom 3 Critical Value 7.815]

A. Yes, there is enough evidence to conclude that being debt-free depends on age.

To test this hypothesis, we will use the chi-square test. The chi-square test is used to determine whether there is a significant association between two categorical variables. In this case, the categorical variables are age and debt-free status.

The chi-square test statistic is calculated using the formula:

X² = sigma (observed - expected)²/expected

In this scenario, the calculated chi-square value is 245.97, and the degrees of freedom are (4-1) x (2-1) = 3. Using a chi-square distribution table, the critical chi-square value at the 5% level of significance with 3 degrees of freedom is 7.815.

Since the calculated chi-square value (245.97) is greater than the critical chi-square value (7.815), we reject the null hypothesis and conclude that there is enough evidence to support the alternative hypothesis.

Therefore, the conclusion is A. Yes, there is enough evidence to conclude that being debt-free depends on age.

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The integral: S1 0 (-x³ - 2x² - x + 3)dx is -1/12

An integral is a mathematical operation that calculates the area under a curve or the value of a function at a specific point. It is denoted by the symbol ∫ and is used in calculus to find the total amount of change over an interval.

To evaluate the integral:

[tex]$ \int_0^1 (-x^3 - 2x^2 - x + 3)dx $[/tex]

We can integrate each term of the polynomial separately using the power rule of integration, which states that:

[tex]$ \int x^n dx = \frac{x^{n+1}}{n+1} + C $[/tex]

where C is the constant of integration.

So, we have:

[tex]$ \int_0^1 (-x^3 - 2x^2 - x + 3)dx = \left[-\frac{x^4}{4} - \frac{2x^3}{3} - \frac{x^2}{2} + 3x\right]_0^1 $[/tex]

Now we can substitute the upper limit of integration (1) into the expression, and then subtract the result of substituting the lower limit of integration (0):

[tex]$ \left[-\frac{1^4}{4} - \frac{2(1^3)}{3} - \frac{1^2}{2} + 3(1)\right] - \left[-\frac{0^4}{4} - \frac{2(0^3)}{3} - \frac{0^2}{2} + 3(0)\right] $[/tex]

Simplifying:

[tex]$ = \left[-\frac{1}{4} - \frac{2}{3} - \frac{1}{2} + 3\right] - \left[0\right] $[/tex]

[tex]$ = -\frac{1}{12} $[/tex]

Therefore,

[tex]$ \int_0^1 (-x^3 - 2x^2 - x + 3)dx = -\frac{1}{12} $[/tex]

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From a nutrition perspective, concerning the how healthy the cookies are, I will say D.Yes, homemade cookies will not contain preservatives.

Based on the given information, most of the home made cookies that is been made by the large companies are been preserved with preseratives because it is been produced in a large quantities which can be delivered to different regions and this implies that it may take somne days because it will be sold completely.

However too much of preservatives is not too good for our health, hence from  nutrition perspective home made can be considered to be more healthy.

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The binomial probability is 0.1933, 0.4015 and 0.1103.

The binomial probability formula, we have:

[tex]P(X = 1) = (9 choose 1) \times (0.20)^1 \times (0.80)^8[/tex]

[tex]= 9 \times 0.20 \times 0.1074[/tex]

= 0.1933 (rounded to 4 decimal places)

The binomial probability formula, we have:

[tex]P(X = 3) = (8 choose 3) \times (0.60)^3 \times (0.40)^5[/tex]

[tex]= 56 \times 0.216 \times 0.3277[/tex]

= 0.4015 (rounded to 4 decimal places)

The binomial probability formula, we have:

[tex]P(X = 4) = (12 choose 4) \times (0.60)^4 \times (0.40)^8[/tex]

[tex]= 495 \times 0.1296 \times 0.1678[/tex]

= 0.1103 (rounded to 4 decimal places)

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The partial derivative of w with respect to u (Wu) is 4u + 3v + 4uv.

To find the partial derivative of w with respect to u (Wu), we need to use the chain rule.

First, we'll find the partial derivatives of w with respect to x, y, and z:

∂w/∂x = y + z
∂w/∂y = x + z
∂w/∂z = x + y

Next, we'll find the partial derivatives of x, y, and z with respect to u:

∂x/∂u = 1
∂y/∂u = 2
∂z/∂u = v

Using the chain rule, we can now find Wu:

Wu = (∂w/∂x)(∂x/∂u) + (∂w/∂y)(∂y/∂u) + (∂w/∂z)(∂z/∂u)

Wu = (y + z)(1) + (x + z)(2) + (x + y)(v)

Substituting the expressions for x, y, and z:

Wu = [(2u - v) + (uv)] + [(u + 2v) + (uv)] + [(u + 2v) + (2u - v)](v)

Simplifying:

Wu = 4u + 3v + 4uv

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The interquartile range, which represents the range between the 25th and 75th percentiles of a dataset, was calculated for both male and female employees at First River Bank. The interquartile range of male employee ages was found to be different from that of female employees.

To compare the interquartile ranges of male and female employees at First River Bank, we first calculated the 25th and 75th percentiles for each group, which represent the lower quartile (Q1) and upper quartile (Q3) respectively. The interquartile range (IQR) is then calculated as the difference between Q3 and Q1. By analyzing the ages of male and female employees separately, we found that the IQR of male employees was different from that of female employees.

Therefore, the interquartile ranges of the ages of male and female employees at First River Bank are not the same.

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27. The probability of winning a lottery with odds of 1 in 1,000 is 0.001 as a decimal and 0.1% as a percentage.

28. The probability of winning a lottery with odds of 1 in 1,000,000 is 0.000001 as a decimal and 0.0001% as a percentage.

Convert the fraction 1/1,000 to a decimal by dividing 1 by 1,000.

1 ÷ 1,000 = 0.001

Convert the decimal to a percentage by multiplying it by 100.

0.001 x 100 = 0.1%

So, the probability of winning a lottery with odds of 1 in 1,000 is 0.001 as a decimal and 0.1% as a percentage.

Convert the fraction 1/1,000,000 to a decimal by dividing 1 by 1,000,000.

1 ÷ 1,000,000 = 0.000001

Convert the decimal to a percentage by multiplying it by 100.

0.000001 x 100 = 0.0001%

So, the probability of winning a lottery with odds of 1 in 1,000,000 is 0.000001 as a decimal and 0.0001% as a percentage.

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1) The confidence interval will get narrower if the variance decreases. The statement is True.
2) A point estimate includes a measure of variability. The statement is False.

1) The statement "The confidence interval will get narrower if the variance decreases" is True. When the variance (a measure of variability) of a sample decreases, the confidence interval around the point estimate will become narrower, as there is less variability in the data.
2) The statement "A point estimate includes a measure of variability" is False. A point estimate is a single value that represents an estimate of a population parameter (e.g., mean, median), and it does not include a measure of variability. Variability is typically assessed using confidence intervals or measures like variance and standard deviation, which are separate from point estimates.

The complete question is:-

1) The confidence interval will get narrower if the variance decreases. True O False O

2) A point estimate includes a measure of variability. True O False O

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The absolute maximum value is log2 6 and the absolute minimum value is 0.

To find the absolute maximum and absolute minimum values of f(x) = log2 (2x2+2), -1≤x≤1, we need to find the critical points and endpoints of the function.

First, we find the derivative of f(x) using the chain rule:

f'(x) = (1/ln2)(1/(2x2+2))(4x)

Setting f'(x) = 0 to find the critical points, we get:

1/(2x2+2) = 0
2x2+2 = ∞
x = ±1

Checking the endpoints, we get:

f(-1) = log2 (2(-1)2+2) = log2 4 = 2
f(1) = log2 (2(1)2+2) = log2 6

Now, we can compare the function values at the critical points and endpoints to find the absolute maximum and absolute minimum values:

f(-1) = 2
f(1) = log2 6
f(±1) = 0

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The image of each vertex as an ordered pair include the following:

A → A' (-5, -1).

B → B' (0, -6).

C → C' (-3, -9).

In Mathematics, the translation of a graph to the left is a type of transformation that simply means subtracting a digit from the value on the x-coordinate of the pre-image while the translation of a graph to the right is a type of transformation that simply means adding a digit to the value on the x-coordinate of the pre-image.

By translating the pre-image of triangle ABC horizontally right by 2 units and vertically down 5 units, the coordinates of triangle ABC include the following:

(x, y)                               →                  (x + 2, y - 5)

A (-7, 4)                        →                  (-7 + 2, 4 - 5) = A' (-5, -1).

B (-2, -1)                        →                  (-2 + 2, -1 - 5) = B' (0, -6).

C (-5, -4)                        →                  (-5 + 2, -4 - 5) = C' (-3, -9).

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Missing information:

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

The number of ways to distribute 10 identical red balls and 10 identical blue balls into 4 distinct urns, with each urn having at least 1 ball, is 36300.

Let A be the event that the first urn has at least 1 red ball and at least 2 blue balls.

Let A1, A2, A3, and A4 be the events that urns 1, 2, 3, and 4, respectively, have no ball we can count the number of outcomes that satisfy each individual condition as follows:

The number of outcomes in A1 is the number of ways to distribute 20 balls into 3 urns, which is C(20 + 3 - 1, 3 - 1) = C(22, 2) = 231.

Similarly, the number of outcomes in A2, A3, and A4 is also 231.

Similarly, the number of outcomes in A13, A14, A23, A24, and A34 is also 63.

The number of outcomes in A123 is the number of ways to distribute 20 balls into 1 urn, which is 1. However, we need to multiply this by 3, since we can choose any of the remaining 3 urns to have 2 balls, and multiply by 3 again, since we can choose any of the remaining 2 urns to have 1 ball. Therefore, the number of outcomes in A123 is 9.

Similarly, the number of outcomes in A124, A134, and A234 is also 9.

Using the inclusion-exclusion principle, the number of outcomes that satisfy at least one condition is:

(number of outcomes) = |A1 ∪ A2 ∪ A3 ∪ A4| = |A1| + |A2| + |A3| + |A4| - |A12| - |A13| - |A14| - |A23| - |A24| - |A34| + |A123| + |A124| + |A134| + |A234|

= 231 + 231 + 231 + 231 - 63 - 63 - 63 - 63 - 63 - 63 + 9 + 9 + 9 + 9

= 36300

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The critical region for a size-alpha test of H0: sigma1^2 = sigma2^2 vs H1: sigma1^2 /= sigma2^2 in the two random samples of normal distribution is defined as follows:

Reject H0 if the test statistic F = S1^2/S2^2 (where S1^2 is the sample variance of the first sample and S2^2 is the sample variance of the second sample) is greater than the upper alpha/2 percentile of the F-distribution with (n1-1) and (n2-1) degrees of freedom or less than the lower alpha/2 percentile of the F-distribution with (n1-1) and (n2-1) degrees of freedom.

In other words, the critical region for a size-alpha test is given by the rejection region: F > F(1-alpha/2; n1-1, n2-1) or F < F(alpha/2; n1-1, n2-1) where F(alpha/2; n1-1, n2-1) and F(1-alpha/2; n1-1, n2-1) are the lower and upper alpha/2 percentiles, respectively, of the F-distribution with (n1-1) and (n2-1) degrees of freedom.

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To solve these we can use fractional multiplication, the answer to 3/8 of 48 is 18 and the answer to 4/5 of 15 is 12.

Fractional multiplication is a process of multiplying two or more fractions (or mixed numbers) together. This is done by multiplying the numerators together and multiplying the denominators together.

The first problem we need to solve is 3/8 of 48. To solve this we can use fractional multiplication.

When multiplying fractions, we multiply the numerators together and the denominators together.

So for the first problem,

3 x 48 = 144 and 144/8 = 18.

Therefore, the answer to 3/8 of 48 is 18.

The second problem we need to solve is 4/5 of 15. Again, we can use fractional multiplication.

4 x 15 = 60 and 60/5 = 12.

Therefore, the answer to 4/5 of 15 is 12.

We can see that the answers to both problems are the same as the fractions used in the problems. This is because when we use fractional multiplication, the numerator and denominator of the answer are the products of the numerator and denominator of the original fractions.

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According to unitary method, the cost of one cup of cake is N6.

Let's assume that the cost of one cup of cake is "x" Naira. According to the problem, a bowl of ice cream and a cup of cake cost N11. This can be expressed as:

1 bowl of ice cream + 1 cup of cake = N11

We can use this equation to express the cost of one bowl of ice cream in terms of the cost of one cup of cake. To do this, we need to isolate the cost of one bowl of ice cream on one side of the equation. This gives us:

1 bowl of ice cream = N11 - 1 cup of cake

Now, we can use this expression to find the cost of two bowls of ice cream and three cups of cake. According to the problem, this costs N28. This can be expressed as:

2 bowls of ice cream + 3 cups of cake = N28

We can substitute the expression we derived earlier for the cost of one bowl of ice cream into this equation. This gives us:

2(N11 - x) + 3x = N28

Simplifying this equation gives us:

22 - 2x + 3x = N28

x = N6

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Answer:

x=12

Step-by-step explanation:

perimeter=addition and 48/4=12