Which statement about the following equation is true? 3x2 â€"" 8x 5 = 5x2.

Answer: No. Collin was wrong.

Step-by-step explanation:

From the question, we are informed that the answer that Collin had on his paper was −13 − (15 + 58) and the correct answer, located in the back of the book, was −13 −15 + 58.

Collin is wrong because he didn't take into cognisance the effect of minus on the bracket. Here, −13 − (15 + 58) will be equal to -13 -15 -18. This is because minus multiplied by plus equals to minus. He shouldn't have put the numbers in the bracket.

Answer: $24000

Step-by-step explanation:

Depreciation for 2018 = ($74,000 - $18000) / 7

= $56000/7

= $8000

Depreciation for 2019 = ($74,000 - $18000) / 7

= $56000/7

= $8000

Depreciation up to 2019 = $8000 + $8000 = $16000

Book value at end of 2019 = $74000 - $16000 = $58000

Revised residual value = $10,000

Number of remaining years = 4-2 = 2 years

Depreciation expensed for 2020 will be:

= ($58000 - $10,000) / 2

= $48000/2

= $24000

Answer:

the correct answer is B

Step-by-step explanation:

$2.29 is the cost of a notebook.

Here, we have,

given that,

Jack bought 3 pencils and 4 notebooks and paid a total of $10.93. Sara bought 5 pencils and 2 notebooks and paid a total of $7.53.

let, the cost of a notebook = n

the cost of a pencil = p

we would solve this with simultaneous equations,

so if we write it as:

4n + 3p = 10.93

2n + 5p = 7.53

=> we get,

multiplying the equations by 5 and 3, we get,

20n + 15p = 54.65

6n + 15p = 22.59

(subtract)

14n = 32.06

divide both sides by 14,

14n ÷ 14 = 32.06÷ 14

n = 2.29

notebooks = 2.29

So the final answer is notebooks are $2.29.

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Answer:

Step-by-step explanation:

2x squared, denoted (2x)2, is equal to 4x2. In general, we can raise a product to a power using the following product rule for exponents: (ab)n = an

There is no question, just some graphs, I can not explain if I don't know the question

ok

y = 2x

This is a positive line that increases from left to right, because the number 2 of the equation is positive.

So from the graphs that you send me, the right answer is the second one, it is an increasing line.

A graph increases from left to rigth if it goes up

A graph decreases from left to right if it goes down

Variance and Standard deviation can tell us the variation of data within a group of samples.

What is a variance?

Variance is a term used in probability theory and statistics to measure how far a group of numbers are spread out from their mean value.

In the context of hypothesis testing, the variance of two groups of data is compared to see if they are statistically significant.

A smaller variance indicates that the data points are closer together, whereas a larger variance indicates that the data points are farther apart.

What is Standard deviation?

The standard deviation is the square root of the variance, and it is used to express the average distance that data points deviate from the mean in a group of samples.

It is used in statistics to measure how tightly a set of data is clustered around its mean.

What is Mean?

The average value of a group of numbers is referred to as the mean.

It is calculated by adding all of the numbers together and then dividing by the total number of numbers.

What is T-value?

The t-value is used to test the null hypothesis, which states that there is no significant difference between two groups of data.

The t-value is calculated by subtracting the mean of one group from the mean of the other group and then dividing by the standard deviation of the two groups combined.

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Answer:

It has to be the middle one. The 27...

Step-by-step explanation:

The solution to the system of equations using Cramer's rule is x1 = 5/4, x2 = -1/8, and x3 = 0.

To solve the system of equations using Cramer's rule, we need to first find the determinant of the coefficient matrix and the determinants of the matrices obtained by replacing each column of the coefficient matrix with the constant terms on the right-hand side of the equations.

The coefficient matrix is

| 1 1 2 |

| 0 2 -3 |

| 1 2 1 |

The determinant of the coefficient matrix is:

det(A) = 1(21 - 2-3) - 1(01 - 21) + 2(02 - 21) = 8

Now, we replace the first column of the coefficient matrix with the constant terms on the right-hand side of the equations:

| 0 1 2 |

| 1 2 -3 |

| 1 1 1 |

The determinant of this matrix is

det(A1) = 1(21 - 2-3) - 1(11 - 2-3) + 2(11 - 22) = 10

Next, we replace the second column of the coefficient matrix with the constant terms on the right-hand side of the equations:

| 1 0 2 |

| 0 1 -3 |

| 1 1 1 |

The determinant of this matrix is

det(A2) = 1(11 - 21) - 0(11 - 2-3) + 2(01 - 11) = -1

Finally, we replace the third column of the coefficient matrix with the constant terms on the right-hand side of the equations:

| 1 1 0 |

| 0 2 1 |

| 1 2 1 |

The determinant of this matrix is

det(A3) = 1(21 - 12) - 1(01 - 11) + 0(02 - 21) = 0

Using Cramer's rule, we can now find the solution for the system of equations

x1 = det(A1)/det(A) = 10/8 = 5/4

x2 = det(A2)/det(A) = -1/8

x3 = det(A3)/det(A) = 0

Therefore, the solution for the system of equations is x1 = 5/4, x2 = -1/8, x3 = 0.

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========================================================

Explanation:

We have a triangle with sides a,b,c where a = 3 and b = 3 are given. The side c is unknown, but using the variation of the triangle inequality theorem allows us to say the following:

b-a < c < b+a

3-3 < c < 3+3

0 < c < 6

The missing side c is between 0 and 6 units long, but cannot be equal to 0 and can't be equal to 6 either. Therefore, the answers must be A) 4, C) 5, and D) 2. All of these values satisfy the inequality 0 < c < 6.

Something like c = 12 is too large. I recommend cutting out two slips of paper that are 3 inches long, and you'll find it impossible to form a triangle that has a third side of 12 inches. The longest side you can form is 3+3 = 6 inches, but you wouldn't have a triangle at that point (it would be a straight line instead). This is a visual way to rule out choice E, and it also rules out choices B and F for similar reasoning.

From the given information provided, the cube root of 12,167 by prime factorization method is approximately 23.

You can find the cube root of 12,167 using a calculator or by using the prime factorization method.

Using the prime factorization method, we can express 12,167 as the product of its prime factors:

12,167 = 19 × 641

Since the cube root of a product is equal to the product of the cube roots of the factors, we can find the cube root of 12,167 as follows:

∛12,167 = ∛(19 × 641)

∛12,167 = ∛19 × ∛641

∛12,167 = 2.7144 × 8.0186

∛12,167 = 23.1649 (rounded to four decimal places)

Therefore, the cube root of 12,167 is approximately 23.

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The posterior density function of λ, denoted as p(λ|y), can be obtained by applying Bayes' theorem. According to the given information, the prior density function of λ is p(λ) = 16λ^(-2)e^(-λ/2), λ > 0, which follows the Gamma(3, 1/2) distribution in the shape, rate parametrization.

The likelihood function is p(y|λ) = ∏(i=1 to 25) y_i! * e^(-λ) * λ^y_i, where y = (y_1, ..., y_25) is the vector of typographical error counts. To find the posterior density, we multiply the prior and likelihood and normalize it by the marginal likelihood.

By applying Bayes' theorem, the posterior density function of λ, given the data y, can be expressed as:

p(λ|y) ∝ p(y|λ) * p(λ)

Substituting the expressions for the likelihood and prior, we have:

p(λ|y) ∝ (∏(i=1 to 25) y_i! * e^(-λ) * λ^y_i) * (16λ^(-2)e^(-λ/2))

Simplifying the expression and combining like terms, we get:

p(λ|y) ∝ λ^∑y_i * e^(-25λ) * λ^(-2) * e^(-λ/2)

p(λ|y) ∝ λ^(∑y_i - 2) * e^(-(25λ + λ/2))

p(λ|y) ∝ λ^(∑y_i - 2) * e^(-(25λ/2))

The expression above represents the unnormalized posterior density function of λ in terms of the data y. To obtain the normalized posterior density, we need to divide this expression by the appropriate constant such that the integral of the posterior density over all possible values of λ equals 1.

Please note that this is the result based on the given information and parametrization. It is essential to ensure consistency with the specific parametrization used in the software or textbook being utilized.

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The students who read newspapers is 30, and the read story books is 90.

Let number of students read newspapers as N and story books as S.

We have

Out of 110 students, 10 read neither newspaper nor story book.

This means that these 10 students are not included in either N or S.

As, students who read story books is three times read newspapers. then, S = 3N.

So, the total number of students who either read newspapers or story books is given by:

= 110 - 10

= 100

and, the total number of students who either read newspapers or story books is given by:

N + S - 20 = 100

Substituting S = 3N, we have:

N + 3N - 20 = 100

4N - 20 = 100

4N = 120

N = 30

Now, we can calculate the number of students who read story books:

S = 3N = 3 c 30 = 90

Therefore, the students who read newspapers is 30, and the read story books is 90.

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Answer:

36,000

Step-by-step explanation:

Area formula- A=LW (length x width)

450 (length) x 80 (width)

450

 80

-------

36,000

(45 x 8 and add two zeros :))

Answer: 3x-x+2=4

Step-by-step explanation:

12/X=3/6 CROSS MULTIPLY.

3X=12*6

3X=72 X=72/3 X= 24 FEET TALL.

Answer:

Step-by-step explanation:

Answer:

what circle

Step-by-step explanation:

The integral ∫8x^2 ln(x) dx can be evaluated using integration by parts with u = ln x and dv = 8x^2 dx. The resulting integral is ∫8x dx + C = 4x^2 ln(x) - 4x^2 + C, where C is the constant of integration.

In integration by parts, we choose a function to differentiate and another to integrate. In this case, we chose u = ln x as the function to differentiate and dv = 8x^2 dx as the function to integrate. We apply the formula ∫u dv = uv - ∫v du, where u and v are the chosen functions. We find that du/dx = 1/x and v = (8/3) x^3. Substituting into the formula yields ∫8x^2 ln(x) dx = ln(x) (8/3) x^3 - ∫(8/3) x^3 (1/x) dx = ln(x) (8/3) x^3 - 8x^2 + C. This produces the answer of 4x^2 ln(x) - 4x^2 + C after simplification.

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The 95% confidence interval for the population mean is approximately [20.03, 20.97].

The percentage (frequency) of acceptable confidence intervals that include the actual value of the unknown parameter is represented by the confidence level.

To calculate the 95% confidence interval for the population mean based on a sample of size 36 with a known variance of 2.1 and a sample mean of 20.5, we can use the formula for a confidence interval for a population mean:

CI = \(\bar X\) ± z * (σ / √n),

where:

CI is the confidence interval,

\(\bar X\) is the sample mean,

z is the z-score corresponding to the desired level of confidence (in this case, 95% confidence),

σ is the population standard deviation,

n is the sample size.

Since we have the population variance (2.1), we can calculate the population standard deviation as σ = √2.1 ≈ 1.45.

Now, let's calculate the confidence interval:

CI = 20.5 ± z * (1.45 / √36).

The z-score corresponding to a 95% confidence level is approximately 1.96 (you can look this up in a standard normal distribution table or use a statistical software).

Substituting the values:

CI = 20.5 ± 1.96 * (1.45 / √36).

Calculating the values within the confidence interval:

CI = 20.5 ± 1.96 * 0.2417.

CI = 20.5 ± 0.4741.

Finally, we can calculate the lower and upper bounds of the confidence interval:

Lower bound = 20.5 - 0.4741 ≈ 20.03.

Upper bound = 20.5 + 0.4741 ≈ 20.97.

Therefore, the 95% confidence interval for the population mean is approximately [20.03, 20.97].

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The answer is 6:2. Since the greatest common denominator of both is 2 you divide it by both numbers to get the ratio.

According one-way ANOVA, the test is false.

We learn about one-way ANOVA

"One-Way ANOVA, also known as "analysis of variance," examines the refers to two or more independent groups to see if there is statistical support for the notion that the related population means are statistically substantially different."

According to the given information, it is inappropriate to compare two means at a time using multiple independent two sample​ t-tests. It will create multiple testing problem and error.

So using one way ANOVA test when comparing three or more populations refers to within a set of quantitative data that is categorized according to one​ factor/treatment and compare two means at a time using multiple independent two sample​ t-tests is false.

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The probability the x is less than 14 is 0.4087

Probability is a measure of the possibility of an event occurring.

The likelihood could range from 0 to 1, , where 0 indicates an impossibility and 1 indicates a certainty. Probability is the ratio of possible outcomes to the total outcomes.

Probability = Possible outcomes/total outcomes

Z score is given by:

Z = (14-18)/03

Z = -1.333

from Z table,

P(z < -1.333) = 0.4087

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When these conditions are met, you can use the dependent samples t-test to compare the mean of the difference between the two populations.

Paired samples

Random sampling.

Normal distribution

Independence of pairs.

To use the dependent samples t-test for the mean of the difference of two populations, the following conditions are necessary:
Paired samples:

The data must consist of pairs of observations (e.g., pre- and post-test scores) that are related or dependent on each other.
Random sampling:

The paired samples must be randomly selected from the populations.
Normal distribution:

The distribution of the differences between the paired samples should be approximately normally distributed.

This condition can be evaluated using a normality test, such as the Shapiro-Wilk test, or by visually inspecting a histogram or Q-Q plot of the differences.
Interval or ratio data:

The data should be measured on an interval or ratio scale (e.g., weight, height, test scores), rather than on an ordinal or nominal scale.
Independence of pairs:

The differences within each pair of observations should be independent of the differences in other pairs.

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If these conditions are met, then the dependent samples t-test can be used to test the null hypothesis that the mean difference between the two populations is zero.

The dependent samples t-test is used to compare the means of two related or dependent populations, where the samples are paired or matched. In order to use the dependent samples t-test for the mean of the difference of two populations, the following conditions should be met:

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Answer:

fractions larger than 14 and one-half

decimals larger than 14 and one-half

whole numbers larger than 14 and one-half

Step-by-step explanation:

Which numbers are solutions to the inequality x greater-than 14 and one-half? Check all that apply.

Given that the inequality is represented by \(x>14\frac{1}{2}\), i.e. x > 14.5

This inequality represent all the set of real numbers which are greater than 14.5. Whereas numbers including 14.5 and below are not involved. Hence the numbers that are a solution to this inequality are:

fractions larger than 14 and one-half

decimals larger than 14 and one-half

whole numbers larger than 14 and one-half

Answer:

x=-8

Step-by-step explanation:

4(1/2x+7)=12

Divide each side by 4

4/4(1/2x+7)=12/4

(1/2x+7)=3

Subtract 7 from each side

1/2x+7-7=3-7

1/2x = -4

Multiply each side by 2

1/2x*2 = -4*2

x = -8

Answer:

-8

Step-by-step explanation:

Divide both sides by 4 which will give you 1/2x+7=3

Then multipy both sides by 2 which will then give you x+14=16

Then move the constant = x=6-14

6-14 is -8 which will give you x=-8

Answer:

Step-by-step explanation:

Answer:

Difference= $11,280

Step-by-step explanation:

Giving the following information:

Manager at Shop & Save Grocers= Hourly wages $16.50 per hour  

Assistant Manager at Unbelievable Toys= Monthly wages $3,800

Cashier at 7-12 Convenience Store= Yearly wages $43,000

First, we need to calculate the yearly wages, and then compare the results:

Shop & Save Grocers= (16.5*40)*52= $34,320

Unbelievable Toys= 3,800*12= $45,600

The highest paying job is at Unbelievable Toys.

Difference= 45,600 - 34,320= $11,280

Answer:

60 miles

Step-by-step explanation:

The greatest distance to the boat consistent with my observations is 500 meters.

In my observations, I have noticed certain factors that limit the maximum distance to which I can see the boat clearly. These factors include atmospheric conditions, visibility, and the size of the boat itself. Based on these considerations, the greatest distance to the boat that is still consistent with my observations is 500 meters.

Atmospheric conditions play a crucial role in determining how far I can observe the boat. Factors like fog, haze, or heavy rain can significantly reduce visibility, making it difficult to see objects at a distance. Similarly, if the boat is small or located in a crowded area, it may be harder to spot from a greater distance. Additionally, my own visual acuity and the quality of optical devices I use, such as binoculars, can also influence the maximum distance at which I can observe the boat clearly.

Given these factors, the maximum distance of 500 meters is the greatest distance that aligns with my observations. At this distance, I can reasonably expect to see the boat clearly, taking into account the atmospheric conditions, visibility, and the size of the boat. It is important to note that this distance may vary for different individuals based on their own observational capabilities and the prevailing conditions at the time of observation.

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The confidence interval and critical value methods are equivalent in providing an interval estimate, the p-value method is used for hypothesis testing and evaluates the strength of evidence against the null hypothesis.

What is the confidence interval?

A confidence interval is a range of values that is likely to contain the true value of an unknown population parameter, such as the population mean or population proportion. It is based on a sample from the population and the level of confidence chosen by the researcher.

In general, when dealing with inferences for two population proportions, the confidence interval method and the critical value method are equivalent. These two methods provide a range of plausible values (confidence interval) for the difference between two population proportions and involve the calculation of critical values to determine the margin of error.

On the other hand, the p-value method is not equivalent to the confidence interval and critical value methods. The p-value method involves calculating the probability of observing a test statistic as extreme as, or more extreme than, the one obtained from the sample data, assuming the null hypothesis is true. It is used in hypothesis testing to determine the statistical significance of the difference between two population proportions.

To summarize:

- Confidence interval method: Provides a range of plausible values for the difference between two population proportions.

- Critical value method: Uses critical values to determine the margin of error in estimating the difference between two population proportions.

- P-value method: Determines the statistical significance of the observed difference between two population proportions based on the calculated p-value.

Hence, the confidence interval and critical value methods are equivalent in providing an interval estimate, the p-value method is used for hypothesis testing and evaluates the strength of evidence against the null hypothesis.

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