Explain why∣2x+5∣=−7 has no solutions.​

Answer:

Discrete set

Step-by-step explanation:

A discrete consists of a fixed number of elements, such that each element or point in the set is a standalone point or isolated. This differs from a continous set which still consists of points in between each set of values. Therefore the property of isolated point is a characteristic of a discrete set. The can be written as a list. The values are isolated on a discrete set as opposed to that of a continuous set whereby values can take up any value within any two intervals in the set.

Answer:

(4x - 3)(4x + 3)

Step-by-step explanation:

16x² - 9 ← is a difference of squares and factors in general as

a² - b² = (a - b)(a + b) , then

16x² - 9

= (4x)² - 3²

= (4x - 3)(4x + 3)

Answer:

1 Cards = 36%

2 Cards = 28%  

3 Cards = 24%

4 Cards = 12.5%

Step-by-step explanation:

The resulting units are kilograms grams milligrams, mole grams liter, ft^2g mL ft gmL

To evaluate the given unit expressions, we can apply the rules of multiplying units similar to multiplying fractions.

(kilogram)(kilogram grams)(gram milligrams):

The units cancel out as follows:

(kilogram)(kilogram grams)(gram milligrams) = (kilogram)(grams)(milligrams)

Therefore, the resulting units are kilograms grams milligrams.

(mole)(mole grams)(gram liter):

The units cancel out as follows:

(mole)(mole grams)(gram liter) = (mole)(grams)(liter)

Therefore, the resulting units are mole grams liter.

(mL)(cmmg)(mLcm3):

The units cancel out as follows:

(mL)(cmmg)(mLcm3) = (mL)(cm mg)(cm^3)

Therefore, the resulting units are mL cm mg cm^3.

(f^2g)(mLft)(gmL):

The units cancel out as follows:

(f^2g)(mLft)(gmL) = (ft^2g)(mLft)(gmL)

Therefore, the resulting units are ft^2g mL ft gmL.

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To calculate the probabilities in this scenario, we need to understand the concept of combinations. A combination is the number of ways to choose a specific number of objects from a larger set, without regard to the order in which the objects are chosen. In this case, we can use the formula for combinations to determine the probabilities.

(a) To find the probability that the number 3 appears at least once, we need to calculate the probability of drawing at least one 3 in four draws without replacement. We can calculate this by finding the probability of drawing no 3's and subtracting that from 1. The probability of not drawing a 3 in the first draw is 7/10, and this decreases by 1/9 in each subsequent draw. So the probability of not drawing any 3's in four draws is (7/10) x (6/9) x (5/8) x (4/7) = 0.252. Subtracting this from 1 gives us the probability of drawing at least one 3, which is 0.748.

(b) To find the probability of drawing four numbers in a strictly increasing order, we need to consider the number of ways this can be done. There is only one way to choose four numbers in a strictly increasing order, so the probability is 1/10 x 1/9 x 1/8 x 1/7 = 0.00018.

(c) To find the probability of drawing four numbers with a sum of 13, we need to consider the combinations of numbers that could add up to 13. These are: 1+2+5+5, 1+3+4+5, 2+3+4+4. For each of these combinations, we can calculate the probability of drawing them by multiplying the probabilities of each individual draw. For example, the probability of drawing 1+2+5+5 is (1/10) x (2/9) x (1/8) x (1/7) = 0.000028. The probability of drawing 1+3+4+5 is (1/10) x (2/9) x (3/8) x (1/7) = 0.000054. The probability of drawing 2+3+4+4 is (1/10) x (2/9) x (2/8) x (1/7) = 0.000042. Adding these probabilities together gives us the total probability of drawing numbers with a sum of 13, which is 0.000124.

In summary, the probabilities in this scenario can be calculated using the concept of combinations. The probability of drawing at least one 3 is 0.748, the probability of drawing four numbers in a strictly increasing order is 0.00018, and the probability of drawing numbers with a sum of 13 is 0.000124.

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The ions indicated by the given information are as follows: (i) \(Se^2-\) (selenium ion), (ii) \(Al^3\)+ (aluminum ion), (iii)\(Ni^2\)+ (nickel ion), (iv) \(Ba^2\)+ (barium ion), and (v)\(F^-\)(fluoride ion).

The number of protons and electrons in an atom determines its atomic number and the element it represents. However, ions have a different number of electrons compared to their neutral atoms, resulting in a different charge. The given information provides the number of protons and electrons for each ion, allowing us to identify them:

(i) 34 protons and 36 selectron: An atom with 34 protons corresponds to selenium (Se). Since it has 36 electrons (2 more than the neutral atom), it becomes \(Se^2\)-, an ion with a charge of -2.

(ii) 13 protons and 10 electrons: An atom with 13 protons corresponds to aluminum (Al). Since it has 10 electrons (3 fewer than the neutral atom), it becomes \(Al^3\)+, an ion with a charge of +3.

(iii) 28 protons and 26 electrons: An atom with 28 protons corresponds to nickel (Ni). Since it has 26 electrons (2 fewer than the neutral atom), it becomes\(Ni^2\)+, an ion with a charge of +2.

(iv) 56 protons and 54 electrons: An atom with 56 protons corresponds to barium (Ba). Since it has 54 electrons (2 fewer than the neutral atom), it becomes \(Ba^2\)+, an ion with a charge of +2.

(v) 9 protons and 10 electrons: An atom with 9 protons corresponds to fluorine (F). Since it has 10 electrons (1 more than the neutral atom), it becomes \(F^-\), an ion with a charge of -1.

Therefore, the ions indicated by the given information are \(Se^2-\),\(Al^3\)+, \(Ni^2\)+, \(Ba^2\)+, and\(F^-\).

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Answer:

200 cm²

Step-by-step explanation:

The combined area of the two triangular faces is \((15)(8)=120\) cm².

The area of the back rectangular face is \((8)(2)=16\) cm².

The area of the bottom rectangular face is \((15)(2)=30\) cm².

The area of the slanted rectangular face is \((17)(2)=34\) cm².

Taking the sum of these areas, the surface area is \(120+16+30+34=200\) cm².

To find the maximum depth of balanced parentheses in a given string, we can use a stack data structure to keep track of opening and closing parentheses. By iterating through the string and pushing opening parentheses onto the stack and popping them when encountering a closing parenthesis, we can determine the maximum depth. If the parentheses are unbalanced at any point, we return -1.

To solve this problem, we can use the concept of a stack, which follows the Last-In-First-Out (LIFO) principle. We iterate through the given string character by character, and whenever we encounter an opening parenthesis, we push it onto the stack. If we encounter a closing parenthesis, we check if the stack is empty or if the top of the stack is not a matching opening parenthesis. If either of these conditions is true, it means the parentheses are unbalanced, and we return -1. Otherwise, we pop the opening parenthesis from the stack.

To keep track of the maximum depth, we maintain a variable maxDepth and a variable currentDepth. Whenever we push an opening parenthesis onto the stack, we increment currentDepth by 1. If we pop an opening parenthesis from the stack, we update currentDepth by subtracting 1. At each step, we compare currentDepth with maxDepth and update maxDepth if currentDepth is greater.

After iterating through the entire string, we check if the stack is empty. If it is not empty, it means the parentheses are unbalanced, and we return -1. Otherwise, we return the maxDepth, which represents the maximum depth of balanced parentheses in the given string.

In the example "(((X)) (((Y))))", the maximum depth of balanced parentheses is 4, as 'Y' is surrounded by 4 balanced parentheses. We can achieve this by pushing each opening parenthesis onto the stack and popping them when encountering the corresponding closing parenthesis.

By using the stack data structure and following the described steps, we can efficiently find the maximum depth of balanced parentheses in a given string.

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The amount of interest that the customer will pay is given as follows:

$31.88.

The balance of an account after t periods is given as follows:

A(t) = P(1 + rt).

In which the parameters of the equation are explained as follows:

Hence the interest accrued is given as follows:

I(t) = Prt.

The parameters for this problem are given as follows:

P = 2530.16, r = 0.00042, t = 30.

Hence the interest is given as follows:

I(30) = 2530.16 x 0.00042 x 30

I(30) = $31.88.

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The 16th term of the sequence is `a16 = 141`.

Given that the first term, `a = 6` and common difference, `d = 9`

The nth term of an arithmetic sequence is given by the formula

`an = a + (n - 1)d`.

Substitute `a = 6` and `d = 9` in the above equation

`an = a + (n - 1)d6 + 9(n - 1)`

Simplify

`an = 9n - 3`

Therefore, the nth term of the given sequence is

`9n - 3`.

Now, we have to find the 16th term, that is, `a16`.

`a16 = a + (16 - 1)d`

Substitute `a = 6` and `d = 9` in the above equation.

`a16 = 6 + 15 × 9`

Simplify`a16 = 141`

Therefore, `a16 = 141`.

Thus, the nth term of the arithmetic sequence is given by `an = 9n - 3`.

The 16th term of the sequence is `a16 = 141`.
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The solutions for the given triangle with a = 11.4, b = 13.7, and c = 12.2 are valid and the triangle exists.

Given the following data :a = 11.4b = 13.7c = 12.2

By the triangle inequality, it is given that any side of the triangle is shorter than the sum of the other two sides. i.e.,a < b + c; b < a + c; c < a + b

Now, let us use the given data and test it to see if the given triangle can exist or not. a = 11.4b = 13.7c = 12.2

Therefore, to check the validity of the triangle, we will perform the following tests :a < b + c => 11.4 < 13.7 + 12.2 => 11.4 < 25.9 [True]b < a + c => 13.7 < 11.4 + 12.2 => 13.7 < 23.6 [True]c < a + b => 12.2 < 11.4 + 13.7 => 12.2 < 25.1 [True]

Thus, all the tests hold true and hence the given triangle exists.

Similarly, using the cosine rule which states that c^2 = a^2 + b^2 - 2abcosC; one can calculate the value of each of the three angles of the triangle.

Therefore, the solutions for the given triangle with a = 11.4, b = 13.7, and c = 12.2 are valid and the triangle exists.

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It will take 20 minutes to dissipate the queue that resulted from blockage if

the freeway has the capacity of 4000 vph.

Mean service rate(M) = 4000vph

Mean arrival rate(λ)  = 3000vph

Block time duration of queue (r) = 5min

Time taken to dissipate the queue =  t

t = (M * r)/(M-λ)

Substitute  the given values

t  =  (4000*5)/(4000-3000)

t = 20000/1000

t = 20minutes.

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Answer:

  50 lb

Step-by-step explanation:

Before drying, the cranberries have ...

  (100% -99%)(100 lb) = 1 lb

of "meat."

After drying, that represents (100% -98%) = 2% of the total weight. Then the total weight is ...

  1 lb = 2%·x

  x = (1 lb)/0.02 = 50 lb

The dried cranberries will weigh 50 lb.

Answer:

This would be the range, the interquartile range, the variance, and finally the standard deviation.

Step-by-step explanation:

Hope this helps you, please mark as brainlest.

Emma is asking an empirical question so that she she can test her hypothesis

What is a Empirical hypothesis?

Something empirical is something based on real evidence .

Evidence that is verifiable by observation as opposed to something that is true in theory or by some kind of reckoning or logic

A hypothesis is a theory that is put forward for further examination to see if it is correct. It is not a call but it is an explanation that fits the available evidence so , An empirical hypothesis is a theory that is based on real evidence and experience but that has not been proven yet

So, According to the question

Emma wants to know if long distance relationship in college have a greater likelihood of ending than a relationship that are not long distance and for that she creates a questionnaire and the crowds 100 student from our school to complete the questionnaire so she can test her hypothesis

Here, Emma is asking an empirical questions ( question about experience) and will conclude the test based on these 100 students experience and thoughts.

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Answer:

Step-by-step explanation:

Answer:

\(y=\frac{-1}{6}x-\frac{10}{3}\)

Step-by-step explanation:

Slope intercept form of the equation of a line is \(y=mx+b\) where m denotes the slope of line and \(b\) denotes the y-intercept.

Equation of a line u is \(y=6x+6\)

Here, slope of line u = m = 6

As the line v is perpendicular to line u, product of slopes of both the lines is equal to \(-1\).

Slope of line v  \(=\frac{-1}{6}\)

So, slope intercept form of the line v is \(y=\frac{-1}{6}x+b\)

As point \((10,-5)\) lies on the line v, put \((x,y)=(10,-5)\) in equation \(y=\frac{-1}{6}x+b\)

\(-5=\frac{-1}{6}(10)+b\\-5=\frac{-5}{3}+b\\ -5+\frac{5}{3}=b\\\frac{-15+5}{3}=b\\\\\frac{-10}{3}=b\)

Equation of line v is \(y=\frac{-1}{6}x-\frac{10}{3}\)

Using the Newton-Raphson method with an initial guess of X₀ = 0, two iterations are performed to estimate the root of the function.

The Newton-Raphson method is an iterative root-finding algorithm that uses the derivative of a function to approximate its roots. To apply the method, we start with an initial guess, X₀, and use the following equation to calculate the new estimate, X₁:

X₁ = X₀ - f(X₀) / f'(X₀)

In this case, the function f-*-, for which we are estimating the root, is not specified. Therefore, we are unable to provide the exact calculations and results for the iterations. However, by following the process outlined above, we can perform two iterations to refine the estimate of the root.

Starting with the initial guess X₀ = 0, we substitute this value into the equation to calculate the new estimate X₁. We repeat this process for the second iteration, using X₁ as the new estimate to find X₂. These iterations continue until the desired level of accuracy is achieved or until a predetermined stopping criterion is met.

By performing two iterations of the Newton-Raphson method, we obtain an improved estimate for the root of the function.

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T test is used to compute the confidence interval, when the mean and standard deviation is given.

What is Standard Deviation?

The standard deviation is a statistic that expresses how much variation or dispersion there is in a set of values. While a high standard deviation suggests that the values are dispersed over a wider range, a low standard deviation suggests that the values tend to be very close to the set mean. The lower case Greek letter (sigma), again for population standard deviation, or even the Latin letter s, again for sample standard deviation, are most frequently used in mathematical texts and equations to represent standard deviation. Standard deviation may be abbreviated as SD. A random variable, specimen, statistical population, data set, as well as probability distribution's standard deviation is equal to the square root of its variance. When the number in a sample, sample mean and sample standard deviation is given, the test statistic that should be used when computing a confidence interval is t test

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To increase usability in a spreadsheet, you can use comments, descriptive labels, and data validation.

To increase usability in a spreadsheet, you can utilize the following techniques:

1. Comments: Comments allow you to add explanatory notes or instructions within cells. They can provide additional context or clarify the purpose of certain data or formulas. Users can view comments by hovering over the respective cell, making it helpful for collaboration and documentation.

2. Descriptive Labels: Instead of relying solely on cell references or generic labels, use descriptive labels that clearly indicate the purpose of each column, row, or range of cells. This approach enhances readability and makes it easier for users to understand the data and navigate through the spreadsheet.

3. Data Validation: Implement data validation rules to ensure the accuracy and integrity of the data entered into the spreadsheet. Data validation allows you to define constraints or conditions for input, such as numeric ranges, predefined lists, or specific text formats. It helps prevent errors and assists users in entering valid data.

4. Formatting: Properly formatting the cells, columns, and rows can greatly enhance usability. Use consistent formatting for similar types of data, apply color schemes or conditional formatting to highlight important information or trends, and adjust the column widths and row heights to optimize readability.

5. Sheet Organization: Organize the spreadsheet by using multiple sheets if needed. Group related information on separate sheets, and provide clear sheet names that reflect the content or purpose of each sheet. This helps users locate and navigate to specific sections or data within the spreadsheet.

6. Clear Instructions or Documentation: Provide clear instructions or documentation either within the spreadsheet or as separate documentation. Explain the purpose of the spreadsheet, any specific procedures or workflows, and how to interpret the data or use any embedded formulas or macros. This helps users understand the spreadsheet's functionality and ensures consistent usage.

7. Error Handling: Implement error handling mechanisms to provide informative error messages or alerts when users input incorrect or incompatible data. Clear error messages guide users in correcting their input and reduce frustration when dealing with potential errors.

By incorporating comments, descriptive labels, data validation, formatting, sheet organization, clear instructions/documentation, and error handling, you can significantly enhance the usability of your spreadsheet and make it more user-friendly for both yourself and others who interact with it.

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The ratio of John's money to Peter's money is 5/4. This means if John has a total amount of 5 then Peter will have a total of 4 as his amount.

Let's assume John has 'x' amount of  money, Peter has 'y' amount of money, The money John saved is 'p' and the money Peter saved is 'q'

So,

p = x - 80x/100                (equation 1)

q = y - 75y/100                (equation 2)

According to the given question, the amount John saved is equal to the amount Peter saved. Hence, we can equate equations 1 and 2.

p = q

x- 80x/100 = y - 75y/100

x - 0.8x = y - 0.75y

0.2x = 0.25y

x =  0.25y/0.2

x/y = 0.25/0.2

x/y = 25/20

x/y = 5/4

Hence, the ratio of John's money to Peter's money is 5/4.

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Answer:

Therefore, June gets £6

Step-by-step explanation:

Total amount = £20

Assume:

June's share = a

Sushil's share = b

Fiona's share = c

a = 2b -----------------------(1)

b = 3c -----------------------(2)

c = b ÷ 3 -------------------(3)

a + b + c = £20  --------(4)

Substitute equations (1) and (3) in equation (4)

2b + b + (b ÷ 3) = £20

10b ÷ 3 = £20

b = £20 * 3 ÷ 10 = £6

a = 2 * £6 = £12

c = £6 ÷ 3 = £3

Therefore, June gets £6

Answer: a

Step-by-step explanation:a

The nearest whole number, we get that we can expect about 419 students on campus to prefer Movie B.

How to calculate addition?

Addition is a basic arithmetic operation that involves combining two or more numbers to obtain their sum. Here's how to calculate addition:

Write the numbers to be added one below the other, aligning the digits according to their place value (ones below ones, tens below tens, and so on).

Start by adding the digits in the ones column (the rightmost column). If the sum is less than 10, write it down below the line. If the sum is 10 or greater, carry the 1 to the next column (the tens column).

Move to the next column (the tens column) and add the digits in that column, along with any carried-over 1 from the previous column. Again, if the sum is less than 10, write it down below the line. If the sum is 10 or greater, carry the 1 to the next column (the hundreds column).

Repeat this process for each column, working from right to left, until you have added all the numbers.

The final result is the sum of the numbers.

We can use the proportion of students in the sample who preferred Movie B to estimate the expected number of students on campus who would prefer Movie B.

The proportion of students in the sample who preferred Movie B is:

Proportion of students who prefer Movie B in the sample = Number of students who prefer Movie B in the sample / Sample size

= 66 / (39 + 66 + 2 + 25)

= 0.524

So, we can expect that approximately 0.524 * 800 = 419.2 students on campus would prefer Movie B.

Rounding this to the nearest whole number, we get that we can expect about 419 students on campus to prefer Movie B.

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The correlation between the number of cars passing the school and the hours as time goes is C. no correlation

The strength and direction of a relationship between two or more variables are described by the statistical measure of correlation, which is given as a number. However, a correlation between two variables does not necessarily imply that a change in one variable is the reason for a change in the values of the other.

As time goes on, the number of cars passing the school sometimes increases and then other times it decreases. This shows that there is no correlation between number of hours and cars.

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Answer:

4

Step-by-step explanation:

Answer:2+2 = 4.............

Answer:

1000 grams per pint

Step-by-step explanation:

2 kg / qt * (1 qt / 2 pints) * (1000 grams/ kg ) = 1000 grams /pint

The standard deviation needed to achieve a process capability index of 2.0 is 0.003 feet.

To calculate the required standard deviation to achieve a process capability index of 2.0, we need to use the following formula:

Process Capability Index (Cpk) = (Upper Specification Limit - Lower Specification Limit) / (6 * Standard Deviation)

In this case, the upper specification limit is 1.788 feet, the lower specification limit is 1.752 feet, and the process capability index (Cpk) is 2.0.

Let's plug in the values into the formula and solve for the standard deviation:

2.0 = (1.788 - 1.752) / (6 * Standard Deviation)

Rearranging the equation:

Standard Deviation = (1.788 - 1.752) / (6 * 2.0)

Standard Deviation = 0.036 / 12

Standard Deviation = 0.003

Therefore, the standard deviation needed to achieve a process capability index of 2.0 is 0.003 feet.

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By considering the given information that angles 21 and 23 are supplementary and analyzing the properties of supplementary angles and parallel lines, we have proven that segment HI is parallel to segment JK.

To prove that segment HI is parallel to segment JK based on the given information that angles 21 and 23 are supplementary, we can utilize the properties of supplementary angles and parallel lines.

First, let's examine the given figure and information.

We have a capital letter A with serifs, where segment HI represents one of the serifs, and segment JK represents a horizontal line within the letter A.

To begin the proof, we'll make use of the fact that angles 21 and 23 are supplementary.

Supplementary angles are defined as two angles whose measures sum up to 180 degrees.

We can observe that angle 21 is an interior angle of triangle AHI, and angle 23 is an interior angle of triangle AJK.

Since angles 21 and 23 are supplementary, their sum is equal to 180 degrees.

Now, let's assume that segments HI and JK are not parallel.

In this case, if we extend lines HA and JA, they will eventually intersect at point P.

Since the angles formed at the point of intersection are supplementary (angle 21 + angle 23 = 180 degrees), it would imply that angle 21 and angle PJK, as well as angle 23 and angle PHI, are also supplementary.

However, this leads to a contradiction. In the original figure, we can observe that angle 21 and angle PJK do not form a supplementary pair since angle PJK is a right angle (90 degrees) in the letter A.

Therefore, our assumption that segments HI and JK are not parallel must be incorrect.

Consequently, we can conclude that segment HI is indeed parallel to segment JK.

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Answer:

Step-by-step explanation:

\((5x10^{3})x(9x10^{7})=5*9*x^{1+1+1}*10^{3+7}\\\\=45x^{3}*10^{10}\)