Solve the following problems.
Given: AABC, DE AC,
BD DC, mZ1=m22,
mZBDC= 100°
Find: m< A, m< b , m

With four decimal places added, we have P(2.04 Z 3.04) 0.0189.

To round a decimal value to two decimal places, use the hundredths place, which is the second place to the right of the decimal point.

Subtracting the area to the left of 1.25 from the area to the left of 2.15 will give us the standard normal area between 1.25 and 2.15.

The area to the left of 1.25 is 0.8944, and the area to the left of 2.15 is 0.9842, according to a conventional normal distribution table or calculator.

So, the standard normal area between 1.25 and 2.15 is:

P(1.25 < Z < 2.15) = 0.9842 - 0.8944 = 0.0898

Rounding to four decimal places, we get:

P(1.25 < Z < 2.15) ≈ 0.0898

We follow the same procedure as before to determine the standard normal region between 2.04 and 3.04:

P(2.04 < Z < 3.04) = P(Z < 3.04) - P(Z < 2.04)\

The area to the left of 2.04 is 0.9798, and the area to the left of 3.04 is 0.9987, according to a conventional normal distribution table or calculator.

So, the standard normal area between 2.04 and 3.04 is:

P(2.04 < Z < 3.04) = 0.9987 - 0.9798 = 0.0189

Rounding to four decimal places, we get:

P(2.04 < Z < 3.04) ≈ 0.0189

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The given statements are true or false are shown below, about linear system has n variables and m equations, then the augmented matrix has n rows.

First, let's write how A and C look like.

A = [C|b], where b is the constant matrix.

(a) False.

Example

[tex]\left[\begin{array}{ccc}1&0&2\\0&1&3\\\end{array}\right][/tex]

We can see that z = t and so we have infinitely many solutions but there's no row of zeros.

(b) False.

Example

[tex]\left[\begin{array}{cc}1&0&0\1&1&0&0\\\end{array}\right][/tex]

Here; x1 = 1 and x2 = 1 is a unique solution and we have a row of zeros.

(c) True.

In the row-echelon form, the last row is either a row of zeros or a row that contains a leading 1. If the row has a leading 1, then there is a solution. Since we assume there is no solution, then the row must be a row of zeros.

(d) False.

Example

[tex]\left[\begin{array}{cc}1&3\\0&0\end{array}\right][/tex]

Here; x₁ = 1 − 3t and x2 = t. Thus, the system is consistent.

(e) True.

Suppose we have a typical equation in a system

а1x1 + A2X2 + ··· + anxn = b

Now, if b≠0 and x1 = x2 = ··· = x₂ = 0, then the system is Xn inconsistent. But, if b = 0, then we have a solution.

(f) False.

Example

[tex]\left[\begin{array}{cc}1&2&0&0\end{array}\right][/tex]

If a = 0, then it's consistent(infinitely many solutions) but if a 0, then it's inconsistent.

(g) Ture.

Since the rank would be at most 3 and this will lead to a free variable (4 columns in C and the rank is 3, so there is at leat 1 free variable). Thus, the system has more thatn one solution.

(h) True.

Because the rank is the number of leading 1's lying in different rows and A has 3 rows. Thus, the rank ≤ 3.

(i) False.

Because we could have a row of zeros in C and a leading 1 in A. In other words, a31 = a32 = A33 = A34 = 0 and c3 1. This makes the system inconsistent.

(j) True.

If the rank of C = 3, then there will be a free variable and this means the system is consistent.

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Complete question:

In each case either show that the statement is true, or give an example showing it is false. (a) If a linear system has n variables and m equations, then the augmented matrix has n rows. quations • ( *b) A consistent linear system must have infinitely many solutions. . (c) If a row operation is done to a consistent linear system, the resulting system must be consistent. (d) If a series of row operations on a linear system results in an inconsistent system, the original system is inconsistent.

Random sampling is crucial when surveying as it ensures that the sample selected is representative of the population.

By randomly selecting participants from the population, the sample is likely to be unbiased on average, which means that the results of the study can be generalized to the entire population. Without random sampling, the results of the study may be skewed or biased towards a certain group, which can lead to incorrect conclusions and poor decision-making. Therefore, it is essential to use random sampling when surveying to obtain accurate and reliable results.

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By answering the presented questiοn, we may cοnclude that Since a line equatiοn parallel tο this οne will have the same slοpe, the slοpe οf the parallel line is alsο 5/6.

When twο expressiοns are equal, a mathematical equatiοn is a statement stating that equality. Twο sides are jοined by the algebraic symbοl (=), and tοgether they make up an equatiοn. Fοr instance, the claim that "2x + 3 = 9" means that "2x plus 3" equals the number "9" is made in this argument. Finding the value(s) οf the variable(s) necessary fοr the equatiοn tο be true is the gοal οf sοlving equatiοns.

There are variοus types οf equatiοns, including regular and nοnlinear οnes with οne οr mοre elements. "x² + 2x - 3 = 0" is an equatiοn that raises the variable x tο the secοnd pοwer. Mathematical disciplines like algebra, calculus, and geοmetry all make use οf lines.

the given equatiοn:

[tex]$\begin{array}{c}{{5x-6y=30}}\\ {{-6y=-5x+30}}\\ {{y=(5/6)x-5}}\end{array}$[/tex]

Sο the slοpe οf the given line is 5/6.

Since a line parallel tο this οne will have the same slοpe, the slοpe οf the parallel line is alsο 5/6.

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The perimeter of the given figures is: Triangle = 4x - 2. Rectangle = 8x - 8, and square = 12x - 8y.

The whole distance encircling a form is referred to as its perimeter. It is the length of any two-dimensional geometric shape's border or outline. Depending on the size, the perimeter of several figures can be the same. Consider a triangle built of an L-length wire, for instance. If all the sides are the same length, the same wire can be used to create a square.

The perimeter of a figure is the sum of all the segments of the figure.

The perimeter of triangle is:

P = 2x - 5 + x + x + 3 = 4x - 2

The perimeter of rectangle is:

P = 2(l + b)

P = 2(3x + 1 + x - 5)

P = 2(4x - 4)

P = 8x - 8

The perimeter of square is:

P = 4(s)

P = 4(3x - 2y)

P= 12x - 8y

Hence, the perimeter of the given figures is: Triangle = 4x - 2. Rectangle = 8x - 8, and square = 12x - 8y.

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Answer:

6188 ways

Step-by-step explanation:

there ate 5 objects to be choosen and there is no replacement of the object therefore you got

17 choices for the first selection of the object and 16 objects for the selection of the second object and so on until you get 13 objects for the last selection

totally you have 5 selections also arrangement does not matter there fore you have 17!/12!5! which is 6188

note we used 5! cause there are 5 placed objects and 12! are unplaced objects

note

that you have used one so you have to deduct one every time you use one

The probability of waiting at least 26 minutes is 0.316. The average waiting time is 19 minutes.

Step 1:

The probability of waiting at least 26 minutes can be calculated by finding the area under the probability density function from 26 to 38:

P(waiting at least 26 minutes) = ∫26^38 (1/38) dt = [t/38] from 26 to 38

= (38/38) - (26/38) = 12/38 = 0.316

So the probability of waiting at least 26 minutes is 0.316 or approximately 0.316 rounded to 3 decimal places.

Step 2:

The average waiting time can be calculated by finding the expected value of the probability density function:

E(waiting time) = ∫0³⁸ t f(t) dt = ∫0³⁸ (t/38) dt

= [(t²)/(238)] from 0 to 38

= (38²)/(238) = 19

Therefore, the average waiting time is 19 minutes.

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Answer:

invertible

Step-by-step explanation:

If A is invertible then ∣A∣ =0

Answer:

Relating SSE to Other Statistical Data



Answer:

  C. RP = 4.5

Step-by-step explanation:

You want to know what condition is sufficient to show ∆ABC ~ ∆QPR, given  three sides and 2 angles in ∆ABC, and 2 sides in ∆QPR.

Similarity can be shown if all three sides are proportional, or if two angles are congruent.

The offered answer choices only list one angle, so none of those will work. The answer choice that makes the third side of ∆QPR be in the same proportion as the corresponding side of ∆ABC is the condition of interest.

  C. RP = 4.5

__

Additional comment

The side ratios in the two triangles are ...

  AB : BC : CA = 10 : 9 : 6

  QP : PR : RQ = 5 : PR : 3

For these ratios to be the same, PR must be half of BC, just as the other segments in ∆QPR are half their counterparts in ∆ABC.

The probability that the person chosen is a woman is 0.51.

The probability that the person chosen favors pink or purple is 0.67.

The probability that a person that favors turquoise is a man is 0.66.

Probability is the odds that a random event would occur. The odds that the event occurs has a probability value that lies between 0 and 1. The more likely it is that the event would happen, the closer the probability value would be to 1.

The probability that the person chosen is a woman = number of women / total number of people = 152 / 300 =  0.51.

The probability that the person chosen favors pink or purple = (number of people who favor pink / total number of people) + (total number of people that favor purple / total number of people) = (50 /300) + (50 / 300) =  0.67.

The probability that a person that favors turquoise is a man = men who favor turquoise / total number of people who favor turquoise = 79/120 = 0.66

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Answer:

2.

a) log base 10 of 100

b) The expression means that 10 to the 2nd power equals 100.

c) 2

3.

4. It would make sense that the value is between 1 and 2 because 10 to the 1st power is 10, and 10 to the 2nd power is 100. 50 is between 10 and 100 so the value would have to be between 1 and 2. This works because logs are the inverse function to exponentiation.

a) 2.6021

b) 3 and the value is exact because the base 10 in the log expression is technically the exponent when you convert it to exponent form. This works because logs are the inverse function to exponentiation. So 10 to the 3rd power would give you exactly 1000.

The length of the top of the bookcase should be approximately 25 inches to display the soap carving collection with an area of 300 in².

To find the length of the top of the bookcase (which we'll call "b"), we need to know the area of the collection of soap carvings and the formula for the area of a rectangle:

Area = length x width

We're given the area of the soap carving collection (300 square inches), and we know that the soap carvings will be displayed on top of the bookcase, which is a rectangle.

Let's assume that the width of the bookcase is 1 unit (we can choose any unit we want, as long as we're consistent). Then we can write:

300 = b x 1

Simplifying this equation, we get:

b = 300/1

b = 300

So the length of the top of the bookcase should be 300 inches. However, this assumes that the width of the bookcase is only 1 inch, which is quite narrow.

If we assume a more reasonable width of, say, 12 inches, then we can write:

300 = b x 12

Simplifying this equation, we get:

b = 300/12

b = 25

So the length of the top of the bookcase should be 25 inches (if the width of the bookcase is 12 inches).

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Initially, while maintaining the very same [tex]x-[/tex]coordinates, the negative reflect across the [tex]x-axis[/tex] modifies the signs of all locations' [tex]y-[/tex]coordinates.

A negative value is one that always has a value lower than zero and is denoted by the minus (-) symbol. In a number line, negative values are displayed just left of zero.

If a number is higher than zero, it is considered positive. Or less zero numbers are referred to as negative numbers. If a number is bigger than that or equal to zero, it is not negative. If a number falls below or equal to zero, it is non-positive.

This transforms Figure [tex]1[/tex] into a new figure with points [tex](-2, -5), (-2, -4), (-3, -4), (-3, -2), (-5, -2), (-5, -5)[/tex].

Next, the [tex]180^{0}[/tex] rotation about the origin swaps the signs of both [tex]x[/tex] and [tex]y[/tex] coordinates of each point. This results in the final figure with points [tex](2, -5), (2, -4), (3, -4), (3, -2), (5, -2), (5, -5)[/tex].

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The probability of obtaining a reading between -0.08°C and 1.68°C is approximately 0.4854 or 48.54%.

Probability is the branch of mathematics concerned with the occurrence of a random event, and there are four types of probability: classical, empirical, subjective, and axiomatic.

The readings at freezing on a set of thermometers are normally distributed, with a mean () of 0°C and a standard deviation () of 1.00°C. We want to know how likely it is that we will get a reading between -0.08°C and 1.68°C.

To solve this problem, we must use the z-score formula to standardise the values:

z = (x - μ) / σ

where x is the value for which we want to calculate the probability, is the mean, and is the standard deviation.

The lower bound is -0.08°C:

z1 = (-0.08 - 0) / 1.00 = -0.08

1.68°C is the upper bound:

z2 = (1.68 - 0) / 1.00 = 1.68

We can now use a standard normal distribution table or calculator to calculate the probabilities for each z-score.

The probability of obtaining a z-score of -0.08 or less is 0.4681, and the probability of obtaining a z-score of 1.68 or less is 0.9535, according to the table. We subtract the probability associated with the lower bound from the probability associated with the upper bound to find the probability of obtaining a reading between -0.08°C and 1.68°C:

P(-0.08°C x 1.68°C) = P(z1 z z2) = P(z 1.68) minus P(z -0.08) = 0.9535 - 0.4681 = 0.4854

As a result, the chance of getting a reading between -0.08°C and 1.68°C is approximately 0.4854 or 48.54%.

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Answer:

[tex]{ \sf{a = \frac{0.012}{0.633 -0.063 } }} \\ \\ { \sf{a = \frac{0.012}{0.57} }} \\ \\ { \sf{a = 0.021 \: (2 \: s.f)}}[/tex]

Answer:

13/8

Step-by-step explanation:

You first find the lcm of 8 and 4 which is 8

Then make a dividing line (/) with the lcm as the numerator.

After which you start diviing the lcm by each denominator of the two terms, and then multiplying with their numerators:

8/8 x 7 + 8/4 x 3 all over the lcm which is 8

(1 x 7 + 2 x 3)/8

(7 + 6)/8

13/8

A defective stamp is likely to be chosen at random 1.4% of the time, or about 0.014 times. The observed percentage probability  faulty stamps is below the 3.5% maximum permitted rate,

Name an event from one outcome. Step 2: Compile a list of all potential outcomes, including any positive ones. Step 3: Subtract the number of favorable outcomes from the total number of possibilities that are feasible.

P(faulty) = quantity of defective stamps divided by total quantity of stamps

P(defective) = 10/700.

0.014 P(defective)

Consequently, there is a 1.4% chance (or about 0.014) that a randomly chosen stamp will be flawed.

B-The observed percentage of flawed stamps is:

10 / 700 0.5 – 0.014

Divide this rate by 100 to get the percentage:

0.014 x 100 approximately 1.4%

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the Fourier series solution is not directly used to find the general solution but is used as a part of it, along with the radial solution. The Fourier series solution helps in finding the solution to the given boundary value problem, which, when combined with the radial solution, gives the complete solution to the Laplace equation.

The Fourier series approach that you have used helps in finding the solution to the boundary value problem, i.e., finding u(a,ϕ) for the given boundary conditions. However, to find a general solution to the Laplace equation, we need to use the superposition principle, which states that the sum of any two solutions to the Laplace equation is also a solution.

Therefore, we can use the previously obtained Fourier series solution for u(a,ϕ) as a building block to construct the general solution. We know that the Laplace equation has radial symmetry, which means that the temperature distribution is only a function of radius (rho) and not of angle (ϕ). Hence, we can write the general solution as:

u(rho,ϕ) = f(rho) + u(a,ϕ)

where f(rho) is the radial component of the solution and u(a,ϕ) is the previously obtained Fourier series solution.

To find f(rho), we need to solve the radial ODE using the boundary conditions at rho=0 and rho=a. Once we have obtained f(rho), we can add it to u(a,ϕ) to get the general solution.

Therefore, the Fourier series solution is not directly used to find the general solution but is used as a part of it, along with the radial solution. The Fourier series solution helps in finding the solution to the given boundary value problem, which, when combined with the radial solution, gives the complete solution to the Laplace equation.

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Answer:

going from left to right:

AA

BD

CB

DC

Answer:

m = 0

Step-by-step explanation:

Slope = rise/run or (y2 - y1) / (x2 - x1)

Pick 2 points (0,2) (1,2)

We see the y stay the same and the x increase by 1, so the slope is

m = 0/1 = 0

So, the slope is 0

The total amount of money Carol originally invested is £22,000 in the bank account.

Compound interest is interest that is calculated not only on the initial amount of money invested or borrowed, but also on any accumulated interest from previous periods.

This results in exponential growth or accumulation of interest over time.

Let X be the amount that Carol originally invested in the account.

After 1 year, the amount of money in the account will be X(1+0.025) = X(1.025).

After Carol withdrew £1000, the amount of money in the account will be X(1.025) - £1000.

After 2 years (i.e. on 1st January 2016), the amount of money in the account will be (X(1.025) - £1000)(1+0.025) = (X(1.025) - £1000)(1.025).

We know that the amount of money in the account on 1st January 2016 was £23,517.60, so we can write the equation:

(X(1.025) - £1000)(1.025) = £23,517.60

Expanding the left-hand side and simplifying, we get:

X(1.025)² - £1000(1.025) = £23,517.60

X(1.025)² = £24,567.63

Dividing both sides by (1.025)², we get:

X = £22,000 (rounded to the nearest pound)

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The complete question is -

On the 1st of January 2014, Carol invested some money in a bank account. The account pays 2.5% compound interest per year. On the 1st of January 2015, Carol withdrew £1000 from the account. On the 1st of January 2016, she had £23 517.60 in the account. Work out how much Carol originally invested in the account?

Answer:

To solve the quadratic equation 9×^2-15×-6=0, we can use the quadratic formula, which is given by:

x = (-b ± sqrt(b^2 - 4ac)) / 2a

where a, b, and c are the coefficients of the quadratic equation ax^2 + bx + c = 0.

In this case, a = 9, b = -15, and c = -6, so we can substitute these values into the quadratic formula:

x = (-(-15) ± sqrt((-15)^2 - 4(9)(-6))) / 2(9)

Simplifying this expression gives:

x = (15 ± sqrt(225 + 216)) / 18

x = (15 ± sqrt(441)) / 18

x = (15 ± 21) / 18

So the two solutions to the quadratic equation are:

x = (15 + 21) / 18 = 2

x = (15 - 21) / 18 = -1/3

Therefore, the solutions to the quadratic equation 9×^2-15×-6=0 are x = 2 and x = -1/3.

Descriptive statistics are the most useful in life. Descriptive statistics provide information about a data set and can help to summarize and interpret data. Specifically, I find the mean and median to be the most useful.

Descriptive statistics involves the use of measures such as the mean, median, mode, and range, as well as graphical representations of the data, such as histograms, box plots, and scatter plots.

The mean is the average of a set of data and is useful for summarizing and interpreting data. For example, when I am studying for a test, I often use the mean of my practice test scores to understand my overall performance.

The median is the middle value of a set of data and is useful for understanding the spread of the data. For example, when I am tracking my monthly expenses, I often use the median to understand how much I am spending each month. By taking the median of my monthly expenses, I can get an idea of which expenses are taking up the most of my budget.

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The solution to the compound inequality -3+b> 7 OR b +9 <17 is given as follows:

b < 8 or b > 10.

The inequality for this problem is defined as follows:

-3+b> 7 OR b +9 <17.

The or operation means that the we must solve each operation separately, and then the solution set to the compound inequality is given by the union of the solution set of each of the inequalities.

The solution set for the first inequality is given as follows:

-3+b> 7

b > 10.

The solution set for the second inequality is given as follows:

b + 9 < 17

b < 8.

Hence the solution for the entire inequality is of:

b < 8 or b > 10.

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Answer:

54°

Step-by-step explanation:

AB = BC = CD = AD (Because all sides of a rhombus are equal)

Let's consider the triangle DBC:

BC = DC => ∠DBC = 36°

The angle DCB is equal to:

180° - 36*2 = 180° - 72° = 108°

In a parallelogram, opposite angles are equal.

Then the angle DAB = DCB = 108°

Also, we know that the diagonals of a rhombus are the bisectors of the angles from which they come.

So, the angle DAE = EAB = 108° / 2 = 54°

The line segment AB and CB are perpendicular to each other.

Check whether the slopes of the lines are the negative reciprocals of one another to see if they are perpendicular to one another. The steps are as follows:

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The process to show that if 16 people are seated in a row of 20 chairs, then some group of 4 consecutive chairs must be occupied is shown below.

We prove this using the Pigeonhole Principle, which states that if n items are placed into m containers, and n > m, then at least one container must contain more than one item.

Let us consider the 16 people seated in a row of 20 chairs. Each person occupies one chair, so there are 20 - 16 = 4 empty chairs in the row.

We assume that empty chairs as containers, and people as items that need to be placed into containers.

Since there are more items (people) than containers (empty chairs), there must be at least one group of 2 or more consecutive empty chairs.

Now, let's consider the complement of this statement: Suppose there are no groups of 4 consecutive chairs that are occupied. Then, each group of 4 consecutive chairs contains at most 3 people.

We partition the row of chairs into groups of 4 consecutive chairs.

So, there are 20 - 3 = 17 such groups. By the statement above, each of these groups contains at most 3 people. Therefore, the total number of people seated in the row is at most 17×3 = 51.

But, we know that there are actually 16 people seated in the row. This is a contradiction, since 51 < 16. Therefore, our assumption that there are no groups of 4 consecutive chairs that are occupied must be false, and we have proved that some group of 4 consecutive chairs must be occupied.

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Answer:

a

Step-by-step explanation: