y is less than -2 in inequality notation

the answer for the equation is 0

The sample size required to achieve a 90% confidence level and 5% accuracy is 67. The z-score for the 90% confidence level is 1.645.

The sample size is calculated using the following formula: N = (er × zα/2 × s)²

where:

In this case, we are given that the desired accuracy is 5%, the confidence level is 90%, and the standard deviation of the population is unknown.

The z-score for the 90% confidence level is 1.645.

Therefore, the sample size is:

N = (0.05 × 1.645 × s)²

We do not know the standard deviation of the population, so we must estimate it. We can use the sample standard deviation from the 50 readings that were taken.

The sample standard deviation is 1.5.

Therefore, the sample size is:

N = (0.05 × 1.645 × 1.5)² = 67

Therefore, the sample size required to achieve a 90% confidence level and 5% accuracy is 67.

Here are the steps involved in calculating the sample size:

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The type of sample that a store prints a request on each receipt asking customers to fill out a satisfaction survey online if they are willing is a voluntary response sample.

In the statistics world, a voluntary response sample is a research sample that relies on individual self-selection. This is in opposition to random sampling, in which the samples are gathered in a way that ensures everyone in a particular population has an equal chance of being chosen. This method of self-selection is also known as self-selection bias or sampling bias.Voluntary response samples are subject to a variety of flaws. The responses may not be representative of the population as a whole, and they may be skewed by numerous factors, such as which individuals are most likely to volunteer, how these volunteers differ from the population as a whole, and how the survey questions are phrased.The results of a voluntary response survey may be useful in generating hypotheses or gaining a general sense of how individuals feel about a particular subject. However, they are not reliable enough to draw any conclusions or to be taken seriously in the scientific sense.

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Answer:

30

Step-by-step explanation:

These two triangles are similar, thus they have equal angle measures.

Answer:

30

Step-by-step explanation:

all the interior angles in the triangle adds up to 180

85 + 65=150

180-150=30

The simplest form of the expression can be shown as;

(y - 4) (y + 1)/(y + 4) (y - 3)

Simplifying algebraic expressions involves reducing or combining like terms, applying the distributive property, and performing operations such as addition, subtraction, multiplication, and division.

Step 1;

We know that we can write the expression as shown in the form;

(y - 1) (y + 2)/ (y + 3) ( y + 4) ÷ (y + 2) (y - 5)/(y + 3) ( y - 4) * (y + 1) (y - 5)/ (y -1) (y - 3)

Step 2;

(y - 1) (y + 2)/ (y + 3) ( y + 4) * (y + 3) ( y - 4)/(y + 2) (y - 5) * (y + 1) (y - 5)/ (y -1) (y - 3)

Step 3;

The simplest form then becomes;

(y - 4) (y + 1)/(y + 4) (y - 3)

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The measure of the angle m∠STU is 42 degrees

Find the diagram to the image attached below:

Since the side RS = RU, this shows that:

m∠RTS = m∠RTU

Given the following parameters

m∠RTS = 5w - 69°

m∠RTU = w + 3°

Equating both expressions

5w - 69° = w + 3

5w - w = 3 + 69

4w = 72

w = 72/4

w = 18

Get the measure of m∠STU

m∠STU = m∠RTS + m∠RTU

m∠STU = 5w - 69 + w + 3

m∠STU = 6w - 66

m∠STU = 6(18) - 66

m∠STU = 42 degrees

Hence the measure of the angle m∠STU is 42 degrees

The coordinates of the fourth corner are (x, y) = (5, 11)

The coordinates of a square are (1,7), (5,3), and (9,7)

Let's assume the four vertices of the square be A(1,7), B(5,3), C(9,7), and D(x,y).

Now coordinates of vertices D need to be found. For that use the following property of a square:

Diagonal squares bisect each other.

Since diagonal AC and BD bisect each other at point O(p,q). Therefore O is the mid-point of both AC and BD.

midpoint=(x1+x2/2,y1+y2/2)

Now since O is the mid-point of AC. Therefore,

\(O(p,q)=(\frac{1+9}{2} ,\frac{7+7}{2} )\)

Also since O is the mid-point of BD. Therefore,

\(O(p,q)=(\frac{5+x}{2} ,\frac{3+y}{2} )\\\\= > (5,7)=(\frac{5+x}{2} ,\frac{3+y}{2} )\\\)

\(\frac{5+x}{2} =5,\frac{3+y}{2}=7\\\)

5+x=10,3+y=14

x=5,y=11

Thus coordinates of the fourth corner are (5,11)

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The fourth quadrant of the square has the following coordinates: (5,11).

The three points of a square have the following coordinates: (1, 7), (5, 3), & (9, 7).

Rectangle ABCD:

A: (1,7) (left corner)

B: (5,3) (bottom corner)

C: (9,7) (right corner)

AB is 4 up and 4 left

BC is 4 up and 4 right.

The result is that the center of the square is at a 45 ° angle (5,7).

The four locations are situated on lines that are perpendicular to the x & y axes & pass through the center, four units away (5,7).

The omitted point, shown by the letter D, is the top corner.

It will be at X = 5 since it ought to have the same X position as point B. Given that it is 4 units higher than (5,7), its y coordinate is 7 + 4 = 11.

Consequently, the omitted position is at (5,11).

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The parameterization of the ellipse in a counterclockwise manner is x = 3cos(t) y = 2sin(t) where t varies from 0 to 2π.

One common way to parameterize an ellipse is to use trigonometric functions such as sine and cosine. We can write the equation of the ellipse as:

4x² + 9y² = 36

We can then use the following parameterization:

x = 3cos(t) y = 2sin(t)

where t is the parameter that varies between 0 and 2π, traversing the ellipse once in a counterclockwise manner.

To see why this parameterization works, let's substitute x and y into the equation of the ellipse:

4(3cos(t))² + 9(2sin(t))² = 36

Simplifying this equation gives:

36cos²(t) + 36sin²(t) = 36

Which is true for any value of t. This shows that our parameterization does indeed describe the ellipse 4x² + 9y² = 36.

Furthermore, we can see that when t=0, we get x=3 and y=0, which is the starting point (3,0). As t varies from 0 to 2π, x and y will trace out the ellipse exactly once in a counterclockwise manner.

Therefore, the parameterization of the ellipse 4x² + 9y² = 36 that begins at the point (3,0) and traverses once in a counterclockwise manner is:

x = 3cos(t) y = 2sin(t)

where t varies from 0 to 2π.

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The factor of the symmetric matrix into  A = LDL^T is equal to

\(\left|\begin{array}{ccc}\sqrt{29}-3/2 &-\sqrt{29}-3/2\\1&1\\\end{array}\right|\).

Factor the matrices into A=LDL^T$ form, follow these steps,

Find the eigenvalues and eigenvectors of the matrix A.

Construct the diagonal matrix D using the eigenvalues.

Construct the matrix L using the eigenvectors.

Let us apply these steps to each matrix,

\($A = \begin{bmatrix} 1 & 3 \ 3 & 2 \end{bmatrix}$\)

The eigenvalues can be found by solving the characteristic equation

det(A−λI)

= \(\left|\begin{array}{ccc}1-\lambda&3\\3&2-\lambda\\\end{array}\right|\)

⇒ λ²-3 λ -7 = 0

Solving for λ using the quadratic formula gives,

\($\lambda = \frac{3 \pm \sqrt{29}}{2}$.\)

The eigenvectors can be found by solving for each eigenvalue,

\($(A-\lambda I)x = 0$\)

\($\lambda_1 = \frac{3 + \sqrt{29}}{2}$\)

(A−λ₁ I)x = 0

⇒ \(\left|\begin{array}{ccc}1-\lambda_{1} &3\\3&2-\lambda_{1} \\\end{array}\right|x\) = 0

⇒ \(\left|\begin{array}{ccc}-\sqrt{29} -1/2 &3\\3&-\sqrt{29} -1/2\\\end{array}\right|x\) = 0

Solving this system of equations gives the eigenvector.

\(x_1 = \begin{bmatrix} \frac{\sqrt{29}-3}{2} \ 1 \end{bmatrix}$.\)

Similarly for,

\($\lambda_2 = \frac{3 - \sqrt{29}}{2}$\)

(A−λ₂I)x = 0

⇒ \(\left|\begin{array}{ccc}1-\lambda_{2} &3\\3&2-\lambda_{2} \\\end{array}\right|x\) = 0

⇒  \(\left|\begin{array}{ccc}\sqrt{29}+1/2 &3\\3&\sqrt{29} -1/2\\\end{array}\right|x\) =0

Solving this system of equations gives the eigenvector

\($x_2 = \begin{bmatrix} \frac{-\sqrt{29}-3}{2} \ 1 \end{bmatrix}$\)

Now we can construct the diagonal matrix D using the eigenvalues,

D = \(\left|\begin{array}{ccc}\lambda_{1} &0\\0&\lambda_{2} \\\end{array}\right|\)

   = \(\left|\begin{array}{ccc}\frac{3+\sqrt{29} }{2} &0\\0&\frac{3-\sqrt{29} }{2} \\\end{array}\right|\)

And we can construct the matrix L using the eigenvectors,

L = \(\begin{bmatrix} x_{1} \ x_{2} \end{bmatrix}$\)

  = \(\left|\begin{array}{ccc}\sqrt{29}-3/2 &-\sqrt{29}-3/2\\1&1\\\end{array}\right|\)

So we have A = LDL^T$ with

L = \(\left|\begin{array}{ccc}\sqrt{29}-3/2 &-\sqrt{29}-3/2\\1&1\\\end{array}\right|\)

Therefore, the value of the matrix is equal to \(\left|\begin{array}{ccc}\sqrt{29}-3/2 &-\sqrt{29}-3/2\\1&1\\\end{array}\right|\).

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The  pounds of chocolate-covered almonds needed is 27 pounds while the pounds of chocolate-covered peanuts is 18 pounds.

The first step is to set up linear equations that represent the information in the question

10.5a + 8b  = (9 x 45)

10.5a + 8b = 405 equation 1

a + b = 45 equation 2

Where:

The elimination method would be used to solve the questions

Multiply equation 2 by 8

8a + 8b = 360 equation 3

Subtract equation 3 from equation 2:

2.5a = 45

Divide both sides of the equation by 2.5

a = 45 / 2.5

a = 18

Substitute for a in equation 2:

18 + b = 45

b = 45 - 18

b = 27

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ANSWER:

A. 337.5 miles

STEP-BY-STEP EXPLANATION:

What we must do is calculate the distance in each section, when the speed and time vary, the sum of both section is the total distance:

Section 1:

Section 2:

The total distance is:

The volume of the cube expressed as a fraction will be 27x³y³ / 64.

Given that:

Volume, V = (0.75xy)³

It is frequently mathematically quantified using SI-derived units or different imperial or US traditional units. The concept of length is linked to the notion of capacity.

The volume of the cube expressed as a fraction is calculated as,

V = (0.75xy)³

V = (3xy/4)³

V = 27x³y³ / 64

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The time shift property tells us that if we shift the function f(t) = u(t - a) by 'a' units to the right, the Laplace transform F(s) will be multiplied by \(e^{(-as)}\), which represents the time delay.

a) To find F(s) for the given function \(f(t) = ate^{(-bt)} sin(mt)u(t)\), where u(t) is the unit step function, we can use the Laplace transform.

- The Laplace transform of a is A/s, where A is the value of a.

- The Laplace transform of \(e^{(-bt)}\) is 1/(s + b).

- The Laplace transform of sin(mt) is \(m/(s^2 + m^2)\), using the property of the Laplace transform for sine functions.

- The Laplace transform of u(t) is 1/s.

Now, using the linearity property of the Laplace transform, we can combine these transforms:

\(F(s) = (A/s) \times (1/(s + b)) \times (m/(s^2 + m^2)) \times (1/s)\)

    \(= Am/(s^2(s + b)(s^2 + m^2))\)

b) The time shift property in the Laplace transform states that if the function f(t) has a Laplace transform F(s), then the Laplace transform of the function f(t - a) is \(e^{(-as)}F(s)\).

This property allows us to shift the function in the time domain and see the corresponding effect on its Laplace transform in the frequency domain. It is particularly useful when dealing with time-delay systems or when we need to express a function in terms of a different time reference.

For example, let's consider the function f(t) = u(t - a), where u(t) is the unit step function and 'a' is a positive constant. This function represents a step function that starts at t = a. The Laplace transform of this function is F(s) = \(e^{(-as)}/s.\)

The time shift property tells us that if we shift the function f(t) = u(t - a) by 'a' units to the right, the Laplace transform F(s) will be multiplied by \(e^{(-as)}\), which represents the time delay. This property allows us to analyze and solve problems involving time-delay systems in the Laplace domain.

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Answer: linear combo of terms involving x & y, with respective numbers determining their contribution to the expression

Answer:

(7) - Option (1)

(8) -  Option (4)

Refer to the step-by-step.

Step-by-step explanation:

(7) - Rewrite the following expression as fractional exponent.

\(\dfrac{\sqrt{x} \cdot x}{x^{-2}}\)

We can use the rules for exponents to solve this problem.  

\(\boxed{\left\begin{array}{ccc}\text{\underline{Exponent rules:}}\\\\1.\ a^0=1\\\\2.\ a^m \times a^n=a^{m+n}\\\\3.\ a^m \div a^n=a^{m-n}\\\\4.\ (ab)^m=a^mb^m\\\\5.\ (a/b)^m=a^m/b^m\\\\6.\ (a^m)^n=a^{mn}\\\\7.\ a^{-m}=1/a^m\\\\8.\ a^{m/n}=(\sqrt[n]{a} )^m\end{array}\right}\)

\(\dfrac{\sqrt{x} \cdot x}{x^{-2}}\\\\\\\Longrightarrow \dfrac{x^{\frac{1}{2} } \cdot x}{x^{-2}} \ \text{(Used rule 8)} \\\\\\\Longrightarrow \dfrac{x^{\frac{1}{2}+1 } }{x^{-2}}\\\\\\\Longrightarrow \dfrac{x^{\frac{3}{2} } }{x^{-2}} \ \text{(Used rule 2)}\\\\\\ \Longrightarrow x^{\frac{3}{2}-(-2) } \\ \\ \\\Longrightarrow \boxed{x^{\frac{7}{2} }} \ \text{(Used rule 3)}\)

Thus, option (1) is correct.

\(\hrulefill\)

(8) - Express the following expression as a radical

\(\dfrac{x^{\frac{1}{3} }x^{\frac{1}{2} }}{x^{-1}}\)

Once again, we can use the rules for exponents to solve this problem.

\(\dfrac{x^{\frac{1}{3} }x^{\frac{1}{2} }}{x^{-1}}\\\\\\\Longrightarrow \dfrac{x^{\frac{1}{3} +\frac{1}{2} }}{x^{-1}} \\\\\\\Longrightarrow \dfrac{x^{\frac{5}{6} }}{x^{-1}} \ \text{(Used rule 2)} \\ \\ \\ \Longrightarrow x^{\frac{5}{6}-(-1) }\\\\\\\Longrightarrow x^{\frac{11}{6} }} \ \text{(Used rule 3)} \\ \\ \\ \Longrightarrow \boxed{\sqrt[6]{x^{11}}} \ \text{(Used rule 8)}\)

Thus, option (4) is correct.

Answer:

no its 1749

Step-by-step explanation:

Answer:

Step-by-step explanation:

The slope of a perpendicular line is the negative inverse of the reference line.  If the reference line is y=mx + b, with slope of m, the perpendicular line will have slope of -(1/m).  The perpendicular line would take the form:

y = -(1/m)+b

Example:  Reference Line:  y = 2x + 2

Perpendicular Line:  y = -(1/2)x + 2

See attachment.

In the given question, none of the function exhibits an end behavior

The end behavior of a polynomial is the way the graph of the polynomial behaves as the input values become very large or very small. Generally, the end behavior of a polynomial is either up, down, or a combination of both. If the highest degree term has a positive coefficient, the end behavior is up; if the highest degree term has a negative coefficient, the end behavior is down. If the highest degree term has a coefficient of zero, the end behavior is a combination of up and down.

In the functions given, we can find the end behavior of each of these functions.

1)

y = √x

The leading coefficient test requires a polynomial and this function is not a polynomial.

2)

y = √(x - 5) - 6

The leading coefficient test requires a polynomial and this function is not a polynomial.

3)

y = -3√(x + 3) + 7

The leading coefficient test requires a polynomial and this function is not a polynomial

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Answer:

D. y=2x^2-1

Step-by-step explanation:

I tried plugging in the other equations with real numbers on the table, but the equations didn't works for more than one value.

|-86| = 86 The Badwater Basin, which is 282 feet (86 metres) below sea level, is the lowest place in North America.

The absolute value is always positive. The elevation of the shoreline, or the ocean-land boundary, is referred to as sea level. Land above this altitude is above sea level, whereas land below it is below sea level. When we visit the shoreline, the small strip of land that separates the land from the water, we experience relative sea level. When we observe the shoreline, we can see that it fluctuates throughout the day. Tides and winds affect how the shoreline narrows and widens, but they do not determine the sea level. To eliminate these effects from the calculations, we use mean sea level (MSL). In a specific location, MSL represents the halfway point between mean high tide and mean low tide.

-86 has an absolute value of 86.

Advice: The absolute value is consistently positive.

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The given expression is:

It is required to simplify the product using the distributive property.

Apply the distributive property to the product:

Apply the distributive property again to simplify the expression on the right:

The answer is x²+2x-8.

Therefore, the values of constants a, b, and c are a = -2, b = 14, and c = 10 to satisfy the given conditions.

To find the values of constants a, b, and c so that the graph of y = ax^3 + bx^2 + cx has a local maximum at x = -2, a local minimum at x = 4, and an inflection point at (1, -26), we can use the given conditions to set up a system of equations.

Condition 1: Local maximum at x = -2

To have a local maximum at x = -2, the derivative of the function at x = -2 should be equal to 0, and the second derivative should be negative.

Taking the derivative of the function, we get:

y' = 3ax^2 + 2bx + c

Evaluating the derivative at x = -2:

0 = 3a(-2)^2 + 2b(-2) + c

0 = 12a - 4b + c ... Equation 1

Taking the second derivative of the function, we get:

y'' = 6ax + 2b

Evaluating the second derivative at x = -2:

y''(-2) < 0

6a(-2) + 2b < 0

-12a + 2b < 0

6a - b > 0 ... Equation 2

Condition 2: Local minimum at x = 4

To have a local minimum at x = 4, the derivative of the function at x = 4 should be equal to 0, and the second derivative should be positive.

Evaluating the derivative at x = 4:

0 = 3a(4)^2 + 2b(4) + c

0 = 48a + 8b + c ... Equation 3

Taking the second derivative of the function:

y'' = 6ax + 2b

Evaluating the second derivative at x = 4:

y''(4) > 0

6a(4) + 2b > 0

24a + 2b > 0

12a + b > 0 ... Equation 4

Condition 3: Inflection point at (1, -26)

To have an inflection point at (1, -26), the function should satisfy the coordinates of the inflection point.

Substituting x = 1 and y = -26 into the function:

-26 = a(1)^3 + b(1)^2 + c

-26 = a + b + c ... Equation 5

Now we have a system of equations (Equations 1, 2, 3, 4, and 5) that we can solve simultaneously to find the values of a, b, and c.

Solving this system of equations, we find the values:

a = -2, b = 14, c = 10

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Answer:

We conclude that:

Step-by-step explanation:

Given

Given the polynomial

\(-m^6-10m^3\)

To determine

The degree of the polynomial

In order to determine the degree of the polynomial, we need to check the term with the highest of the degree with non-zero coefficients. The degree of an individual term is basically the exponent of the term.

In our case,

The exponents of the terms of this polynomial are, in order, 6, and 3.

Thus, the highest degree (larger exponent) of the term is 6.

Therefore, we conclude that:

Answer: 1/50

Step-by-step explanation:

Step 1: We need to multiply the numerator and denominator by 100 since there are 2 digits after the decimal.

0.02 = (0.02 × 100) / 100

= 2 / 100 [ since 0.02 × 100 = 2 ]

Step 2: Reduce the obtained fraction to the lowest term

Since 2 is the common factor of 2 and 100 so we divide both the numerator and denominator by 2.

2/100 = (2 ÷ 2) / (100 ÷ 2) = 1/50

The correct answer is C." is either rational or irrational" because it cannot be expressed as the ratio of two integers.

The product of a non-zero rational number and a non-zero irrational number is either rational or irrational, depending on the specific numbers involved.

let's assume that the non-zero rational number is a/b and the irrational number is c. If their product is rational, we can write

a/b×c=d/e

where d and e are integers with no common divisors. This means:

c = (d/b) × (e/a)

Both d/b and e/a are rational numbers, so their product is also a rational number. Therefore, expressing the irrational number c as the product of two rational numbers is a contradiction. 

For example, consider the rational number 2/3 and the irrational number √2.

Multiplying them gives:

(2/3) * √2 = (2√2)/3

It is an irrational number because it cannot be expressed as the ratio of two integers. Therefore answer (C) is either rational or irrational.

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The minimum value of the bacterial population on the interval [2, 6] is 4, and the maximum value is 64.

To find the minimum and maximum values of the bacterial population along the interval [2, 6], we need to evaluate the function b(t) at t = 2 and t = 6, respectively, and then compare the results.

at t = 2

b(2) = 2^2 = 4, which means that after 2 days, the sample has 4 bacteria.

at t = 6

b(6) = 2^6 = 64, which means that after 6 days, the sample has 64 bacteria.

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Answer:

the answer is 6

Step-by-step explanation:

Answer:

A: 2 Feet

Step-by-step explanation:

Got it right

Step-by-step explanation:

Hey there!!!

Here,

The midpoint of AB is M(-3,-3) and coordinates of A is (-8,-2).

Now,

Using midpoint formulae,

\((x,y) =( \frac{x1 + x2}{2} ,\frac{y1 + y2}{2} )\)

Put all values.

\(( - 3,- 3) = ( \frac{ - 8 + x}{2} , \frac{ - 2 + y}{2} )\)

As they are equal, equating with their corresponding elements we get,

\( - 3 = \frac{ - 8 + x}{2} \)

Simplify them.

\( - 6 = - 8 + x\)

\(x = - 6 + 8\)

Therefore, x= 2

Now,

\( - 3 = \frac{ - 2 + y}{2} \)

Simplify them.

\( - 6 = - 2 + y\)

\(y = - 6 + 2\)

Therefore, y = -4.

Therefore, the coordinates of B are (2,-4).

Hope it helps...