Find the maximum/minimum value of the quadratic function q² + 22q = y - 85 by
completing the square method.
O-36
O-48
C
-24
-64

In both cases, the sum of x and y is not divisible by 3, we have demonstrated that the proposition is true. and the proposition is false, and we have shown a counterexample where the sum of two integers, one of which is not divisible by 3 and the other is divisible by 3, can be divisible by 3.

(a) To prove that the sum of two integers, x and y, neither of which is divisible by 3, cannot be divisible by 3, we can use a case-based proof.

Case 1: x and y leave a remainder of 1 when divided by 3.

Let x = 3m + 1 and y = 3n + 1, where m and n are integers. Then, the sum of x and y is 3m + 3n + 2, which leaves a remainder of 2 when divided by 3. Therefore, x + y is not divisible by 3.

Case 2: x and y leave a remainder of 2 when divided by 3.

Let x = 3m + 2 and y = 3n + 2, where m and n are integers. Then, the sum of x and y is 3m + 3n + 4, which leaves a remainder of 1 when divided by 3. Therefore, x + y is not divisible by 3.

Since in both cases, the sum of x and y is not divisible by 3, we have  demonstrated that the proposition is true.

(b) To prove that the sum of two integers, x and y, where x is not divisible by 3 and y is divisible by 3, cannot be divisible by 3, we can use a counterexample.

Let x = 2 and y = 6. Then, x is not divisible by 3 and y is divisible by 3. However, x + y = 8, which is not divisible by 3.

Therefore, the proposition is false, and we have shown a counterexample where the sum of two integers, one of which is not divisible by 3 and the other is divisible by 3, can be divisible by 3.

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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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Answer:

Relating SSE to Other Statistical Data



Answer:

B

Step-by-step explanation:

It's like a piece of paper, fold it in half

Answer:

1.11

2.4

3.5

pls correct me if I'm wrong

Answer:

38. (b) 11

39. (c) 4

40. (c) 5

Step-by-step explanation:

38.)

[tex]\implies \: \sf \sqrt{3xx - 8} = 5 \\ \\ \implies \: \sf 3xx - 8 = {(5)}^{2} \\ \\ \implies \: \sf 3xx - 8 = 25 \\ \\ \implies \: \sf 3xx = 25 + 8 \\ \\ \implies \: \sf 3xx = 33 \\ \\ \implies \: \sf xx = \dfrac{33}{3} \\ \\ \implies \: \sf xx = 11\\ [/tex]

Hence, Required answer is option (b) 11.

39.)

[tex]\implies \: \sf \sqrt{4xx -7 } - 3 = 0 \\ \\ \implies \: \sf \sqrt{4xx - 7} = 3 \\ \\ \implies \: \sf 4xx - 7 = {(3)}^{2} \\ \\ \implies \: \sf 4xx - 7 = 9 \\ \\ \implies \: \sf 4xx = 9 + 7 \\ \\ \implies \: \sf 4xx = 16 \\ \\ \implies \: \sf xx = \dfrac{16}{4} \\ \\ \implies \: \sf xx = 4 \\ [/tex]

Hence, Required answer is option (c) 4.

40.)

[tex]\implies \: \sf \sqrt{6xx + 6} - 6 = 0 \\ \\ \implies \: \sf \sqrt{6xx + 6} = 6 \\ \\ \implies \: \sf 6xx + 6 = {(6)}^{2} \\ \\ \implies \: \sf 6xx + 6 = 36 \\ \\ \implies \: \sf 6xx = 36 - 6 \\ \\ \implies \: \sf 6xx = 30 \\ \\ \implies \: \sf xx = \dfrac{30}{6} \\ \\ \implies \: \sf xx = 5 \\ [/tex]

Hence, Required answer is option (c) 5.

Answer:

I HOPE ITS HELPFUL

Step-by-step explanation:

Cost price of 300 articles = ₹1500Hence cost of each item = 1500/300 = ₹5With 20% profit, selling price of each item = 5 * 120/100 = ₹6Hence, total selling price of 260 items = 260 * 6 = ₹1560Selling price of balance items = 6/2 = ₹3Total selling price for balance items = 40 * 3 = ₹120Total selling price = 1560 + 120 = ₹1680Profit gained = 1680 - 1500 = ₹180Profit percentage = 180/1500 * 100 = 12

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: x=16, y=9

Step-by-step explanation:  Find x first. The unmarked angle underneath the x + 19 is 9x + 1 (corresponding angles so congruent so same measure as the angle below)

So the two angles add up to 180°

x+19 + 9x + 1 = 180°

combine like terms

10x + 20 = 180

subtract 20

10x = 160

divide by 10

x = 16

Now you can find the measure of the angle marked 9x+1.

9(16) + 1

= 144 + 1

= 145

Now find y. The angle marked 9x+1 is now known to be a 145° angle. So that angle with the angle marked 3y+8 must make 180°

3y + 8 + 145 = 180

combine like terms

3y + 153 = 180

subtract 153

3y = 27

divide by 3

y = 9

-Hope this helps! Thanks, have a good day :-)

a) The value of Hy = 1.15

b) The proportion of cartons with less than 1.15 broken eggs is 0.60

c) The answer is B)This computation of the mean is incorrect because it does not take into account that the number of broken eggs are not all equally like.

(a) Hy = 0.67, which means that on average, there are 0.67 broken eggs per carton.

Hy = (0*0.60 + 1*0.20 + 2*0.15 + 3*0.03 + 4*0.02)/1

= 0.67

(b) The proportion of cartons with less than 0.67 broken eggs is 0.60 which is probability of 0 broken eggs.. This may be surprising as the probability of having less than 0.67 broken eggs is only 0.60, but this is due to the fact that the distribution is skewed towards higher values.

(c) Hy is not equal to the simple average of 0, 1, 2, 3, and 4 because the distribution is not symmetrical. The probabilities are weighted towards higher values, which increases the mean. Additionally, the value of zero should be included in the calculation, as it represents a possible outcome. Therefore, the correct formula for Hy is (0*0.60 + 1*0.20 + 2*0.15 + 3*0.03 + 4*0.02)/1 = 0.67.

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--The question is incomplete, answering to the question below--

"Let y denote the number of broken eggs in a randomly selected carton of one dozen eggs.

y 0 1 2 3 4

p(y) 0.60 0.20 0.15 0.03 0.02

(a) Calculate and interpret Hy  

(b) In the long run, for what proportion of cartons is the number of broken eggs less than Hy? (Round your answer to two decimal places).

Does this surprise you?

Yes

No

(c) Explain why Hy is not equal to (0 + 1 + 2 + 3 + 4)/5 = 2.0.  

A. This computation of the mean is incorrect because the value in the denominator should equal the maximum y value.

B. This computation of the mean is incorrect because it does not take into account that the number of broken eggs are not all equally like

C. This computation of the mean is incorrect because it includes zero in the numerator which should not be taken into account when

D. This computation of the mean is incorrect because it does not take into account the number of partially broken eggs."

Answer: the distribution to the x and the 3 makes the solution 4

Step-by-step explanation: i got you broski

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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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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Answer:

1

Step-by-step explanation:

slope = rise/run = 1/1 = 1

To rewrite a number with a radical in the denominator without the radical, we can use the conjugate of the denominator. The conjugate of a binomial expression a + b is a - b, and the conjugate of a - b is a + b.

Suppose we have a number with a radical in the denominator, such as 1/√2. To rewrite this without the radical, we multiply the numerator and denominator by the conjugate of the denominator, which is

√2: 1/√2 = (1/√2) * (√2/√2) = √2/2

Note that the product of the denominator and its conjugate is always a difference of squares, which can be simplified to an integer without a radical.

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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.

Answer:

Matt would walk 8 miles in to hours.

Step-by-step explanation:

15 minutes is 1/4 of an hour, meaning it would be 1/8 of 2 hours.

1 mile times 8 = 8 miles

Hope that helps!

Answer:

[tex](x-4)^2+(y+1)^2=162[/tex]

Step-by-step explanation:

Determine r² by using the equation of a circle and plugging in the center (h,k)->(4,-1) as well as (x,y)->(13,8):

[tex](x-h)^2+(y-k)^2=r^2\\(x-4)^2+(y-(-1))^2=r^2\\(13-4)^2+(8+1)^2=r^2\\9^2+9^2=r^2\\81+81=r^2\\162=r^2\\[/tex]

Hence, the equation of the circle that meets these criteria is[tex](x-4)^2+(y+1)^2=162[/tex]

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.

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:

Step-by-step explanation:

long leg = 78  (means that 26√3*√3 = 26√9 = 26*3 = 78

for x: Short leg= 26√3

Hypotenuse= 2*26√3 = 52√3 for y

The two transformations applied to quadrilateral JKLM are rigid transformations, quadrilateral PQRS is congruent to quadrilateral JKLM, and the corresponding sides are proportional but not equal.

The dilation is a transformation in which all points are moved by the same scale factor, but the scale factor is different for each point. In this case, the dilation was centered at the origin and had a scale factor of . This means that all points were moved by the same scale factor of .

Quadrilateral PQRS is the image of quadrilateral JKLM after translating it 6 units down and 6 units to the right, followed by a dilation centered at the origin with a scale factor of . This means that quadrilateral PQRS is congruent to quadrilateral JKLM, and the corresponding sides are proportional but not equal.

In this case, the translation of quadrilateral JKLM was 6 units down and 6 units to the right.

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Answer:

a

Step-by-step explanation:

The factorization of the polynomial x³ + 9x² + 3x + 27 by grouping is:

x³ + 9x² + 3x + 27 = (x + 9)(x² + 3)

In algebra, "grouping" refers to a method of factoring polynomials that involves grouping together pairs of terms within the polynomial, in order to factor out a common factor.

In mathematics, a polynomial is a mathematical expression consisting of variables (also called indeterminates) and coefficients, that involves only the operations of addition, subtraction, multiplication, and non-negative integer exponents of variables.

In the given question,

To factor the four-term polynomial by grouping, we can follow these steps:

Step 1: Group the first two terms and the last two terms together.

x³ + 9x² + 3x + 27

= (x³ + 9x²) + (3x + 27)

Step 2: Factor out the common factor from each group.

= x²(x + 9) + 3(x + 9)

Step 3: Notice that the expression (x + 9) is a common factor of both terms, and factor it out.

= (x + 9)(x² + 3)

Therefore, the factorization of the polynomial x³ + 9x² + 3x + 27 by grouping is:

x³ + 9x² + 3x + 27 = (x + 9)(x² + 3).

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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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The volume of the composite figure is 837.33 cubic units.

The area or capacity of a cone is determined by its volume. A cone's circular base tapers from a flat base to a point known as the apex or vertex in three dimensions. A cone is made up of a collection of line segments, half-lines, or lines that link the apex—the common point—to each point on the base, which is on a plane without the apex. A cone may be thought of as a collection of irregularly shaped circular discs placed on top of one another with the ratio of their radii remaining constant.

The volume of any composite figure is the addition of the volume of all the shapes present in the figure.

Here, the composite solid is made of 2 cones and a cylinder.

The volume of cone is:

V = 1/3(πr²h)

For cone with height 12 and r= 4:

V = 1/3(π)(4)²(12)

V = 200.96 cubic units.

For cone with h = 8 and r = 4:

V = 1/3(π)(4)²(8)

V = 133.97

The volume of cylinder is:

V = πr²h

V = (3.14)(4)²(10)

V = 502.4 cubic units.

The volume of the composite solid is:

Total volume = 200.96 + 133.97 + 502.4

Total volume = 837.33 cubic units.

Hence, the volume of the composite figure is 837.33 cubic units.

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Answer:

7 miles

Step-by-step explanation:

Note there are 63,360 inches in a mile.

So, rewrite scale as 1 inch: 1 mile, where inch replaces the unit.

Therefore 7 inches: 7 miles.

The two requirements for a discrete probability distribution are that the probabilities assigned to each possible outcome must be between 0 and 1 inclusive, and the sum of probabilities of all possible outcomes must be equal to 1.

There are two main requirements for a discrete probability distribution,

The probabilities assigned to each possible outcome must be between 0 and 1 inclusive. That is, the probability of each outcome must be a non-negative number, and the sum of probabilities of all possible outcomes must be equal to 1.

Each possible outcome must be mutually exclusive. That is, only one of the possible outcomes can occur on any given trial of the random experiment.

These two requirements ensure that the probabilities assigned to each outcome are valid and that the total probability space is complete, allowing us to make valid inferences and predictions about the outcomes of random events.

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I have solved the question in general, as the given question is incomplete.

The complete question is:

What are the two requirements for a discrete probability distribution?

Answer:

7. 23      

8. (3 - 8) x 5

Step-by-step explanation:

I think the second one is right but I know the first one is.

There are 3 seniors who take Math and Spanish, but not Physics.

What is Venn Diagram?

A diagram representing mathematical or logical sets pictorially as circles or closed curves within an enclosing rectangle (the universal set), common elements of the sets being represented by the areas of overlap among the circles.

To solve this problem, we can use a Venn diagram to organize the information given about the students.

First, we know that there are 150 seniors in total, so we can label the outer rectangle of the Venn diagram as having a total of 150 students.

Next, we can label the three circles as Math, Spanish, and Physics, with the numbers given in the problem:

80 seniors take Math, so we label the Math circle with 80.

41 seniors take Spanish, so we label the Spanish circle with 41.

54 seniors take Physics, so we label the Physics circle with 54.

We also know that 10 seniors take Math and Spanish, 19 take Math and Physics, 12 take Physics and Spanish, and 7 take all three. We can label these overlaps in the Venn diagram as follows:

The overlap of Math and Spanish is 10.

The overlap of Math and Physics is 19.

The overlap of Physics and Spanish is 12.

The overlap of all three is 7.

To find the number of seniors who take Math and Spanish, but not Physics, we need to subtract the number of seniors who take all three (because they are counted in all three circles) and the number of seniors who take Math, Spanish, and Physics (because they are counted in the overlap of all three circles). Then we add the number of seniors who take only Math and Spanish (because they are not counted in the Physics circle).

Using the formula for the number of elements in the union of three sets, we have:

Math or Spanish or Physics = Math + Spanish + Physics - (Math and Spanish) - (Math and Physics) - (Spanish and Physics) + (Math and Spanish and Physics)

So, the number of seniors who take Math and Spanish, but not Physics is:

Math and Spanish - (Math and Spanish and Physics) + (Math or Spanish or Physics - (Math + Spanish + Physics) + (Math and Spanish and Physics))

= 10 - 7 + (80 + 41 + 54 - 2(10) - 2(19) - 2(12) + 7)

= 3

Therefore, there are 3 seniors who take Math and Spanish, but not Physics.

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