Graph the line. Y = -3×

Getting the correct answer in an online school environment can be achieved through a few methods:

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Remember, the key to success in online school is to stay organized, be proactive, and actively engage with the course materials and your classmates. Good luck!

Getting every answer in an online school course depends on several factors, including the format of the course, the resources available, and the instructor's teaching style. Here are some tips that may help:

Utilize course resources: Most online courses provide students with a variety of resources such as textbooks, lectures, discussion forums, and quizzes. Utilize these resources to gain a deeper understanding of the material.

Participate in discussions: Participating in online discussions can be an excellent way to get answers to questions you have about the course material. You can also collaborate with other students to gain a better understanding of the course content.

Ask the instructor: If you're still unsure about a particular topic, don't be afraid to reach out to your instructor for clarification. You can send an email, participate in virtual office hours, or use a discussion forum to ask questions.

Research outside sources: You can also conduct research outside of the course to supplement your understanding of the material. Search for reputable sources, such as academic journals, to gain a better understanding of the course content.

Form a study group: If you're taking the course with other students, consider forming a study group. Studying with others can be an effective way to get answers to questions you have and to collaborate on course assignments.

Remember, the key to getting every answer in an online course is to be an active and engaged learner. Utilize all the resources available to you and don't be afraid to ask for help when needed.

The value of mathematical expression is,

⇒ 10 + 3P

Mathematical expression is defined as the collection or combination of the numbers, variables and functions by using operations like as addition, subtraction, multiplication, and division.

Given that;

The expression is,

⇒ 10 more than triple P.

Now, We can simplify mathematical expression as;

Triple P = 3P

Hence, We get the expression as;

⇒ 10 more than triple P.

⇒ 10 + 3 × P

⇒ 10 + 3P

Therefore, We get;

The value of mathematical expression is,

⇒ 10 + 3P

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The percentage of students with more than 1 brother or sister is 41%.

To find the percentage, first complete the table:

118 - 62 = 56 female students with 1 brother and sister.
8 + 14 = 22 7th years with 3 brothers and sisters.
118 + x + 22 + 6 = 200, x = 200 - 146, x = 54 7th years with 2 brothers and sisters.
54 - 24 = 30 male students with 2 brothers and sisters.
62 + 30 + 8 + y = 104, y = 104 - 100, y = 4 male students with 4 brothers and sisters.
6 - 4 = 2 female students with 4 brothers and sisters.
56 + 24 + 14 + 2 = 96 total female students of 7th year.

The percentage of students with more than i brother and sister is:

54 + 22 + 6 = 82 students or 200 - 118 = 82 students.

% = 82 / 200 x 100 = 41%.

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The number of possible outcomes is given as follows:

18 outcomes.

The Fundamental Counting Theorem states that if there are n independent trials, each with [tex]n_1, n_2, \cdots, n_n[/tex] possible results, the total number of outcomes is calculated by the multiplication of the number of outcomes for each trial as presented as follows:

[tex]N = n_1 \times n_2 \times \cdots \times n_n[/tex]

For this problem, we have that:

Hence the number of outcomes is given as follows:

3 x 6 = 18 outcomes.

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Answer:

B.10

Step-by-step explanation:

√2.√10.√5

√2.10.5

√100

√2².5²

2.5

10

Answer:

√64 a⁶

or, √2⁶a⁶

or, √(2a)⁶

or, (2a)³

=8a³

A 90% confidence interval estimate of the population standard deviation is 1596.45-4466.73

We can use the chi-square distribution to construct a confidence interval for the population standard deviation, where the lower and upper limits are given by:

[tex]\sqrt{(n-1)*s^2/chi2(alpha/2,n-1)) }[/tex]         and [tex]\sqrt{(n-1)*s^2/chi2(1-alpha/2,n-1)}[/tex]

Standard Deviation is a measure that shows how much variation (such as spread, dispersion, spread,) from the mean exists.

Where n is the sample size, s is the sample standard deviation, alpha is the significance level, and chi2 is the chi-square distribution function.

Substituting the given values, we get:

Lower limit

= [tex]\sqrt{(31-1)*50^2/chi2(0.05/2,31-1)}[/tex]

= 1596.45

Upper limit

= [tex]\sqrt{(31-1)*50^2/chi2(1-0.05/2,31-1)}[/tex]

= 4466.73

Therefore, a 90% confidence interval estimate of the population standard deviation is 1596.45-4466.73.

Hence, the correct answer is B) 1596.45-4466.73

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The surface area Sam needs to paint on the cylinder is  πx² + 2πxy i.e. B.

What exactly is a cylinder?

In geometry, a cylinder is one of the fundamental 3d forms with two parallel circular bases at a distance. A curving surface connects the two circular bases at a predetermined distance from the centre. The axis of the cylinder is the line segment connecting the centres of two circular bases. The height of the cylinder is defined as the distance between the two circular bases. One real-world example of a cylinder is an LPG gas-cylinder.

Because the cylinder is a three-dimensional form, it has two primary properties: surface area and volume. The cylinder's total surface area (TSA) is equal to the sum of its curved surface area and the area of its two circular bases.

The volume of a three-dimensional cylinder is the space it occupies (V).

TSA=2πr²+2πrh, where r is the radius and h is the height.

Now,

Given that the radius of the roof is x feet and the silo is y feet high

Area of roof=πx² and

curved surface area=2πxy

then the surface area Sam needs to paint=πx²+2πxy

Hence,

        The surface area Sam needs to paint on the cylinder is

πx² + 2πxy.

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Using the law of cosines, as we have two sides and an angle, the length b is given as follows:

b = 17.92.

The law of cosines states that we can find the side c of a triangle as follows:

c² = a² + b² - 2abcos(C)

In which:

For this problem, the length b is opposite to the angle of 47º, hence it is obtained as follows:

b² = 20² + 24² - 2 x 20 x 24 x cosine of 47 degrees

b² = 321.28

b = sqrt(321.28)

b = 17.92.

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(a) P = [tex](56035e^(0.3283t)) / (560 + 35(e^(0.3283t)-1))[/tex] (b) 345 panthers (c) 1.59 years (d) dP/dt = kP(1 - P/560), P(5) ≈ 373 panthers (e) 1.59 years.

(a) The logistic equation that models the population of panthers in the preserve is:

P = [tex](560P0e^{(kt)}) / (560 + P0(e^{(kt)-1}))[/tex]

where P0 is the initial population (35), k is the growth rate constant, and t is the time in years.

To solve for k, we can use the fact that after 2 years the population is 200:

200 =[tex](56035e^(2k)) / (560 + 35(e^{(2k)-1}))[/tex]

Solving for k, we get:

k = 0.3283 (rounded to 4 decimal places)

Therefore, the logistic equation is:

P = (56035e^(0.3283t)) / (560 + 35(e^(0.3283t)-1))

(b) To find the population after 5 years, we substitute t=5 into the logistic equation:

P = (56035e^(0.32835)) / (560 + 35(e^(0.32835)-1))

P ≈ 345 panthers (rounded to the nearest whole number)

(c) To find when the population reaches 280, we can solve the logistic equation for t when P=280:

280 = [tex](56035e^{(0.3283t)}) / (560 + 35(e^{(0.3283t)-1}))[/tex]

Solving for t, we get:

t ≈ 1.59 years (rounded to 2 decimal places)

(d) The logistic differential equation that models the growth rate of the panther population is:

dP/dt = kP(1 - P/560)

Using Euler's method with a step size of h=1, we get:

P(5) ≈ P(4) + hdP/dt(4)

P(4) = 345 (from part b)

dP/dt(4) = k345*(1 - 345/560) = 27.522

P(5) ≈ 345 + 27.522 = 373 panthers (rounded to the nearest whole number)

(e) The panther population is growing most rapidly when the growth rate is at its maximum, which occurs at half the carrying capacity (P=280). We can find the time by taking the derivative of the logistic equation and setting it equal to zero:

dP/dt = kP(1 - P/560)

0 = k280(1 - 280/560)

0.5 = 1 - 280/560

280 = 560/2

So, the panther population is growing most rapidly when P=280, which occurs at half the carrying capacity. Therefore, the time when the population is growing most rapidly is at t=1.59 years (from part c).

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The graph of the two linear functions is f(x) = 2x and f(x) = x.

If a system of equations only contains two linear equations with two variables, the system's equation can be graphed, the graph will have two straight lines, and the intersection point(s) of those lines will be the system's solution.

There are only three matching forms of solution for a given system of equations because there are only three different ways that two straight lines in the plane can graph.

Let, The first table be,

x = 2  4  6  8.

y = 4  8  12  16.

We can observe that when x = 2, y = 4, and when x = 4 y = 8.

Therefore, The equation is y = 2x.

The second table is,

x = 1  2  3  4.

y = 1  2  3  4.

Here, y is equal to x, Therefore, The equation is y = x.

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Answer:

y = -1/4x + 2

Step-by-step explanation:

First you have to pick 2 points on the line, then you find the slope of the line from those 2 points. The slope formula is m = (y2-y1) / (x2-x1), and filling the variables in with the points (0,2) and (4,1) gives you m = (1-2) / (4-0), which gives you m = -1/4.

Once you have found the slope, you look for the y-intercept which is the point where the x-value is 0, and that is 2 in this graph.

You can then put this information into slope-intercept form, y= mx + b. m represents the slope and b represents the y-intercept. The answer is y = -1/4x + 2

The answer is option (C) based on the post-treatment median being much lower than the median at baseline, the treatment appears to be effective.

A) To find the approximate 25th percentile, median, and 75th percentile for the PASI scores pre-treatment, we can look at the boxplot. The box in the plot represents the middle 50% of the data, and the line inside the box represents the median. The whiskers show the range of the data, and any points outside of the whiskers are considered outliers.

From the given boxplot, we can see that the median is around 13, the 25th percentile is around 9, and the 75th percentile is around 19 for the PASI scores pre-treatment.

B) To find the approximate 25th percentile, median, and 75th percentile for the PASI scores post-treatment, we can again look at the boxplot.

From the given boxplot, we can see that the median is around 4, the 25th percentile is around 2, and the 75th percentile is around 8 for the PASI scores post-treatment.

C) The effectiveness of the therapy can be determined by comparing the pre-treatment and post-treatment PASI scores. As the goal is to have lower PASI scores post-treatment, a lower median PASI score post-treatment as compared to pre-treatment is an indication of the effectiveness of the therapy. Therefore, based on the post-treatment median being much lower than the median at baseline, the treatment appears to be effective.

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The value of parameter under  null hypothesis which is not plausible , will not be contained in a 95% confidence interval when (a) p value is less than or equal to 0.05  .

The "P Value" is  the probability of observing a test statistic as extreme or more extreme than the one computed from the sample data, assuming the null hypothesis is true.

When the p value is less than equal to 0.05 , it indicates that the observed test statistic is statistically significant at the 5% level of significance, means that the null hypothesis will be rejected.

Therefore , the parameter value that is assumed under the null hypothesis is not plausible and is not likely to be in a 95% confidence interval.

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The given question is incomplete , the complete question is

When conducting a two-sided test of significance, the value of the parameter under the null hypothesis is not plausible and will not be contained in a 95% confidence interval when  

(a) The p-value is less than or equal to 0.05

(b) The p-value is greater than 0.05.

(c) There is no relationship between the p-value and the confidence interval.

Kendra and Allen Tovsky's adjusted gross income is $139,223.

Adjusted Gross Income (AGI) is defined as an individual's gross income minus various adjustments made to that income, such as trade and business deductions.

To calculate Kendra and Allen Tovsky's adjusted gross income (AGI), we need to start with their total income and subtract any deductions or adjustments they are eligible for.

Their total income is:

Combined income: $142,573

Dividends: $650

Rental income: $8,500

Total income before deductions: $151,723

Next, we need to subtract their eligible deductions. In this case, they are able to reduce their income by $12,500, which could include things like contributions to a retirement account or health savings account, self-employed expenses, or student loan interest payments. Let's assume that the full $12,500 deduction applies to their situation.

Adjusted gross income calculation:

Total income before deductions: $151,723

Minus eligible deductions: $12,500

Adjusted gross income: $139,223

Therefore, Kendra and Allen Tovsky's adjusted gross income is $139,223.

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Answer: (F) 6 feet

Step-by-step explanation:

The area of a circle is proportional to the square of its radius. So if we increase the diameter of the flower bed by a factor of x, its area will increase by a factor of x^2. We want to find the diameter of the new bed that will have nine times the planting area of the old bed.

Let d be the diameter of the new bed. Then its radius is r = d/2, and its area is A = πr^2. We want:

9A = π(D/2)^2

where D is the diameter of the old bed. Substituting r = D/2, we get:

9A = π(D/2)^2

= πr^2

So we can solve for d by equating the right-hand sides and taking the square root:

9πr^2 = πd^2

d^2 = 9r^2

d = 3r

Therefore, the diameter of the new bed must be three times the diameter of the old bed. The diameter of the old bed is 2 feet, so the diameter of the new bed is 3 times that, or 6 feet.

The graph given is of the function f(x) = –5x – 19

A function is a relation from a set of inputs to a set of possible outputs where each input is related to exactly one output.

Given is a graph, we need to determine the graph belongs to which function,

From the figure it is noticed that the f(x) is a straight line, so it must be in the form of,

f(x) = ax+b

Where, a coefficient of x and b are constants.

From the graph it is noticed that as the value of x increases the value f(x) decreases. It means the function have negative slope and the coefficient of x must be negative.

The function intersect the y-axis at below the origin. It means the value of b must be negative.

The sign of both constant and coefficient are negative. According to this statement the correct option is D.

In function f(x) = –5x – 19 both constant and coefficient are negative.

Put x=0, we get f(x)=-19, so the y-intercept is (0,-19).

Put f(x)=0, we get x = -19/5, so the x-intercept is (-19/5, 0).

Since both x and y intercepts are negative,

Therefore, option D is correct.

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The area of the rectangle is 322 square centimeters.

Given data :- The perimeter of a rectangle P =74 meter. If the length is 5 meter less than twice it's width.

The dimensions of the rectangle =?

Let assume that the length of the rectangle L = (2x -5)m

And the width of the rectangle W = x meter

Now Perimeter P = 2(L+W)

74 = [2{x+(2x-5)}]

3x-5=37

3x= 42 , x = (42/3)= 14 meter

So the required width of the Rectangle W = 14 meter

Similarly , L = (2*14 - 5) = 23

Now , area of the rectangle = 14*23

Therefore, area of the rectangle = 322

Hence , the area of the rectangle is 322 square centimeters.

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The measure of the angles are

1. 14°

2. 141°

3. 39°

4. 39°

The 4-letter code is DCAA

From the question, we ae to determine the measure of the angles and then the 4-letter code

From the diagram,

m ∠B + m ∠C = 180° (Sum of angles on a straight line)

Then,

1.

9x + 15° + 3x - 3° = 180°

12x + 12° = 180°

12x = 180° - 12°

12x = 168°

x = 168°/12

x = 14°

Thus, D

2.

m ∠B = 9x + 15°

m ∠B = 9(14) + 15°

m ∠B = 126° + 15°

m ∠B = 141°

Thus, C

3.

m ∠C = 180° - m ∠B

m ∠C = 180° - 141°

m ∠C = 39°

Thus, A

4.

m ∠A = m ∠C (Vertically opposite angle)

m ∠A = 39°

Thus, A

Hence, the code is DCAA

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The coordinates of point Q in the rectangle named QRST is (5, 5).

A pair of numbers called coordinates are used to locate a point or a form in a two-dimensional plane. The x-coordinate and the y-coordinate are two numbers that define a point's location on a 2D plane.

Given:

We have the Graph having coordinate T(5, 2).

and, length of QT = 3 unit

As, the vertical distance shows the y coordinate then the y coordinate of Q is 3.

and, the x coordinate is same as T.

Then, the Coordinate of Q is (5, 2+3) or (5, 5).

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The transformation T1 defined by T1(x1,x2,x3)=(x1,0,x3) is linear transformation among the five given questions.

A linear transformation satisfies two properties: additivity and homogeneity. That is, for any vectors u and v, and any scalars a and b:

T(u + v) = T(u) + T(v)

T(au) = aT(u)

Using these properties, we can determine which of the given transformations are linear transformations:

A. The transformation T1 defined by T1(x1,x2,x3)=(x1,0,x3)

T1(u+v) = (u1+v1, 0, u3+v3) = (u1, 0, u3) + (v1, 0, v3) = T1(u) + T1(v)

T1(au) = (au1, 0, au3) = a(u1, 0, u3) = a*T1(u)

Therefore, T1 is a linear transformation.

The transformation T2 defined by T2(x1,x2)=(2x1−3x2,x1+4,5x2)

T2(u+v) = (2(u1+v1)−3(u2+v2), u1+v1+4, 5(u2+v2))

= (2u1−3u2, u1+4, 5u2) + (2v1−3v2, v1+4, 5v2) = T2(u) + T2(v)

However, T2 does not satisfy homogeneity property. Consider a = -1 and u = (1, 1):

T2(-u) = (-2, 5, -5)

-1*T2(u) = (-2, -5, -5)

Therefore, T2 is not a linear transformation.

The transformation T3 defined by T3(x1,x2,x3)=(x1,x2,−x3)

T3(u+v) = (u1+v1, u2+v2, -(u3+v3)) = (u1, u2, -u3) + (v1, v2, -v3) = T3(u) + T3(v)

T3(au) = (au1, au2, -au3) = a*(u1, u2, -u3) = a*T3(u)

Therefore, T3 is a linear transformation.

The transformation T4 defined by T4(x1,x2,x3)=(1,x2,x3)

T4(u+v) = (1, u2+v2, u3+v3) ≠ (1, u2, u3) + (1, v2, v3) = T4(u) + T4(v)

T4(au) = (1, au2, au3) ≠ a(1, u2, u3) = a*T4(u)

Therefore, T4 is not a linear transformation.

The transformation T5 defined by T5(x1,x2)=(4x1−2x2,3|x2|)

T5(u+v) = (4(u1+v1)−2(u2+v2), 3|u2+v2|)

= (4u1−2u2, 3|u2|) + (4v1−2v2, 3|v2|) ≠ T5(u) + T5(v)

T5(au) = (4au1−2au2, 3|au2|) ≠ a*T5(u)

Therefore, T5 is not a linear transformation.

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The value of V = 2π/3. To find the volume of the solid, we need to use the method of cylindrical shells.

First, we need to solve for the x and y intercepts of the curve. Setting x = 0 gives y = 0, and setting y = 0 gives x = ±1. So the curve intersects the x-axis at (-1, 0) and (1, 0).

Next, we need to find the height of each cylindrical shell. The height is given by the difference between the y-values of the curve and the line x = 1. Solving for y in 2xy = 1 + x^2, we get y = (1 + x^2)/(2x). So the height of the cylindrical shell at x is given by (1 + x^2)/(2x).

Finally, we need to find the radius of each cylindrical shell. The radius is simply x.

So the volume of the solid is given by the integral from x = 0 to x = 1 of 2πx(1 + x^2)/(2x) dx. Simplifying and integrating, we get V = 2π/3.

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Complete Question:

Find the volume of the solid formed by revolving the region bounded by the curve 2x y = the line 1 + x2 y = 0, and the line x = 1 about the y-axis. V=?

The  cosine of angle R is 3/5 and tangent of angle R is 4/3.

Trigonometry is a branch of mathematics that studies relationships between side lengths and angles of triangles.

The given right triangle is RTS.

We have to find the cosine of angle R

We know that cos theta = Adjacent side/hypotenuse.

CosR=6/10

=3/5

Now let us find tangent of angle R

Tan theta = opposite side/adjacent side

=8/6

=4/3

Hence, the  cosine of angle R is 3/5 and tangent of angle R is 4/3.

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0.1038 is the probability that a randomly selected full carton of Grade A eggs will weigh more than 850 grams.

0.413 is the mean of X?

0.643 is the standard deviation of X

As per the data given:

Each full carton of Grade A eggs consists of 1 randomly selected empty cardboard container and 12 randomly selected eggs.

The weights of such full cartons are approximately normally distributed with a mean of 840 grams and a standard deviation of 7.9 grams.

[tex]\mu=840[/tex]

[tex]\sigma = 7.9[/tex]

(a) Here we have to determine the probability that a randomly selected full carton of Grade A eggs will weigh more than 850 grams

Here, a full carton [tex]$y \sim N(840,7.9)$[/tex]

[tex]& \mu=840 \\[/tex]

As, z = [tex]$ \frac{y-840}{7.9} \sim N(0.1) \\[/tex]

[tex]$P(y > 850)= & P\left(z > \frac{850-840}{7.9}\right)=1-P(z < 1.26) \\[/tex]

[tex]= & 1-\phi(1.26) \\[/tex]

= 1 - 0.8962

= 0.1038

(b) Here, B is the weight of the carton [tex]$\sim N(20,1.7)$[/tex]

Weight of each egg [tex]$x=\frac{y-B}{12} \\[/tex]

[tex]& x \sim N\left(\frac{1}{12}(\epsilon(y)-\epsilon(B)), a\right)^{\prime}[/tex]

Here we have to determine the mean of x

Here [tex]$ \epsilon(x)=\epsilon \frac{\epsilon(y)-\epsilon(B)}{12} \\[/tex]

[tex]$ =\frac{840-20}{12} \\[/tex]

[tex]$& =\frac{820}{12} \\[/tex]

[tex]\epsilon(x) & =68.33 \\[/tex]

[tex]\alpha^2=v(x) & =\left(\frac{1}{12}\right)^2[v(y)-v(B)][/tex] as independent

[tex]$& =\frac{1}{12^2}\left[(7.9)^2-(1.7)^2\right][/tex]

[tex]$& =\frac{1}{144}[62.41-2.89] \\[/tex]

[tex]$ & =\frac{1}{144}[59.52] \\ &[/tex]

= 0.413

(ii) Here we have to determine the standard deviation

[tex]\sigma =\sqrt{0.413} \\ &[/tex]

[tex]\sigma[/tex] = 0.643

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Each full carton of Grade A eggs consists of 1 randomly selected empty cardboard container and 12 randomly selected eggs. The weights of such full cartons are approximately normally distributed with a mean of 840 grams and a standard deviation of 7.9 grams.

(a) What is the probability that a randomly selected full carton of Grade A eggs will weigh more than 850 grams?

(b) The weights of the empty cardboard containers have a mean of 20 grams and a standard deviation of 1.7 grams. It is reasonable to assume independence between the weights of the empty cardboard containers and the weights of the eggs. It is also reasonable to assume independence among the weights of the 12 eggs that are randomly selected for a full carton.

Let the random variable X be the weight of a single randomly selected Grade A egg.

i) What is the mean of X?

ii) What is the standard deviation of X?

Answer: To find the sales tax percentage, we can start by subtracting the cost of the item itself from the total cost with tax included.

In this case, the cost of the item is $3, and the total cost with tax included is $3.24, so:

$3.24 - $3 = $0.24

Next, we can divide the amount of sales tax by the cost of the item, and multiply by 100 to find the percentage:

$0.24 ÷ $3 × 100 = 8%

So the sales tax percentage is 8%.

Step-by-step explanation:

Answer:

Step-by-step explanation:

$3 item

h = 100 *  3.24 - 3/ 3  =  24/3  = 8 1  = 8%

Sales tax percentage is = 8%

The correct way to start finding g(h(x)) is to replace the output of h(x) in

the variables of g(x).

When the output of one function is given to the input of another function they are called composite functions.

In (fog)x which is [f{g(x)}] here the out of the function g(x) is the input of the function f(x).

Given, g(x) = x² - 8 and h(x) = - 6x + 1,

Now, g(h(x)) is the output of h(x) is the input of g(h(x)).

Therefore, g(h(x)) is g(- 6x + 1) = (- 6x + 1)² - 8.

g(h(x)) = - (6x - 1)² - 8.

g(h(x)) = - (36x² - 12x + 1) - 8.

g(h(x)) = - 36x² + 12x - 1 - 8.

g(h(x)) = - 36x² + 12x - 9.

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If a colorimetric indicator is not available, an alternative method for determining the equivalence point in a titration is to use a pH meter.

A pH meter is a device that measures the acidity or basicity of a solution and gives a numerical value for the pH.

In an acid-base titration, the pH of the solution changes as the titrant is added to the analyte. Initially, the pH is determined by the analyte, which is usually acidic. As the titrant is added, it reacts with the analyte to form a neutral or basic solution, and the pH increases. At the equivalence point, all of the analyte has reacted with the titrant.

To use a pH meter to determine the equivalence point, the pH of the solution is monitored as the titrant is added. Initially, the pH is low due to the acidity of the analyte. As the titrant is added, the pH increases, and a sharp increase in pH is observed at the equivalence point.

The advantage of using a pH meter to determine the equivalence point is that it provides a more precise and accurate method than visual indicators, which can be affected by factors such as color blindness or variations in lighting conditions. Additionally, a pH meter can be used for both acid-base and redox titrations.

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Amplitude is the maximum displacement of a function on a graph.

Amplitude is the maximum displacement of a function on a graph. It refers to the maximum height or distance from the baseline of a wave or oscillation.

It is often used in physics and engineering to describe the strength or size of a signal, vibration, or wave. In simple terms, it can be thought of as the "peak-to-peak" distance of a waveform, or the height of the waveform above and below the baseline.

But the minimum displacement of a function on a graph would typically be referred to as the minimum value or the "valley" of the function.

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