Do you think that practicing
mathematics will bring Peace
and Harmony in your life?.
Explain your thinking.

Answer:

381 ft

Step-by-step explanation:

first get the area of the kitchen

22 x 18= 396 ft

now get the area of the island which is 3x 5= 15 ft

to know how much she would need to cover (not including the island), subtract 396 to 15

that would be 381 ft

Using the P-value method of testing hypotheses with a significance level of 0.05, the sample provides strong evidence to support the claim that the mean score of job applicants from the university is greater than 160.

To test the claim that the mean score of job applicants from the university is greater than 160, we will perform a one-sample t-test using the P-value method. The null hypothesis (H0) assumes that the mean score is equal to 160, while the alternative hypothesis (Ha) assumes that the mean score is greater than 160.

First, we calculate the test statistic, which is the t-value. The formula for the t-value is:

t = (sample mean - hypothesized mean) / (sample standard deviation / sqrt(sample size))

Plugging in the given values, we have:

t = (183 - 160) / (12 /   √(25))

  = 23 / (12 / 5)

  = 23 * (5 / 12)

  ≈ 9.58

Next, we find the P-value associated with the test statistic. The P-value represents the probability of obtaining a test statistic as extreme as the observed value, assuming the null hypothesis is true. Since the alternative hypothesis is one-sided (greater than 160), we calculate the P-value by finding the probability of the t-distribution with 24 degrees of freedom being greater than the calculated t-value.

Consulting statistical tables or using software, we find that the P-value is very small (less than 0.0001).

Since the P-value (less than 0.0001) is less than the significance level (0.05), we reject the null hypothesis. This provides strong evidence to support the claim that the mean score of job applicants from the university is greater than 160.

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The given mathematical expression is evaluated for the input values (2, 4). The result of the expression is calculated using various operations such as addition, multiplication, square root, natural logarithm, minimum, maximum, and function composition.

The expression f(x1, x2) involves several mathematical operations. Let's evaluate each part of the expression step by step:

1. The first term is 421 + 222, which equals 643.

2. The second term is 3x² + 213. Plugging in x1 = 2 and x2 = 4, we get 3(2)² + 213 = 3(4) + 213 = 12 + 213 = 225.

3. The third term is 5x11². Substituting x1 = 2 and x2 = 4, we have 5(2)(11)² = 5(2)(121) = 1210.

4. The fourth term is (√₁+√₂)². Replacing x1 = 2 and x2 = 4, we obtain (√2 + √4)² = (1 + 2)² = 3² = 9.

5. The fifth term is 10ln(₁). Plugging in x1 = 2, we have 10ln(2) = 10 * 0.69314718 ≈ 6.9314718.

6. The sixth term is (x₁+x₂)(x² + x3). Substituting x1 = 2 and x2 = 4, we get (2 + 4)(2² + 4³) = 6(4 + 64) = 6(68) = 408.

7. The seventh term is min(3r1, 10√2). As we don't have the value of r1, we cannot determine the minimum between 3r1 and 10√2.

8. The eighth term is max{5x1,2r2}. Since we don't know the value of r2, we cannot find the maximum between 5x1 and 2r2.

9. Finally, we have MP1(x1, x2), MP2(X1, X2), and TRS(x1, x2), which are not defined or given.

Considering the given expression, the evaluated terms for the input values (2, 4) are as follows:

- 421 + 222 = 643

- 3x² + 213 = 225

- 5x11² = 1210

- (√₁+√₂)² = 9

- 10ln(₁) ≈ 6.9314718

- (x₁+x₂)(x² + x3) = 408

The terms involving min() and max() cannot be calculated without knowing the values of r1 and r2, respectively. Additionally, MP1(x1, x2), MP2(X1, X2), and TRS(x1, x2) are not defined.

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Answer:

H'(0, 3); I'(2, 4); J'(5, 1)

Step-by-step explanation:

H'(-2+2,4-1)

H'(0, 3)

I'(0+2, 5-1)

I'(2, 4)

J'(3+2, 2-1)

J'(5, 1)

In 2.63  hours the prices of the two stocks be the​ same

Two or more expressions with an Equal sign is called as Equation.

Given,

The price of Stock A at 9 A.M. was $12.71.

Price has been increasing at the rate of $0.08 each hour

Price of Stock B was $13.21

The price has been decrease at the rate of $0.11 each hour.

We can write in the form of equation , to find how many hours will the prices of the two stocks be the​ same

12.71+0.08h=13.21-0.11h

0.11h+0.08h=13.21-12.71

0.19h=0.5

Divide both sides by 0.19

h=0.5/0.19

h=2.63

Hence, in 2.63  hours the prices of the two stocks be the​ same

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The parametric representation for the part of the sphere x² + y² + z² = 4 that lies above the cone defined by z = √(x² + y²) is given by x = u, y = v, and z = √(4 - u² - v²), where u and v range over appropriate intervals. The value of z can be greater than, equal to, or less than 2 depending on the values of u and v.

To find the parametric representation, we can assign u and v as parameters. Considering the equations x = u and y = v, we can substitute these values into the equation of the sphere to solve for z. Substituting x = u and y = v into x² + y² + z² = 4 yields u² + v² + z² = 4. Rearranging the equation, we have z² = 4 - u² - v², which gives us z = ±√(4 - u² - v²). However, since we are interested in the upper part of the sphere, we take the positive square root.

Therefore, the parametric representation for the part of the sphere that lies above the given cone is x = u, y = v, and z = √(4 - u² - v²). The values of u and v can vary within appropriate intervals that satisfy the given conditions. As for the value of z, it can be greater than 2, equal to 2, or less than 2 depending on the specific values of u and v.

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When the plane intersects the figure parallel to the circular bases, the cross-sectional area formed is a circle.

A circle is a geometric shape that is defined as a set of points in a plane that are equidistant from a fixed point, known as the center of the circle.

To form a circle, we can start by fixing a point in a plane and then drawing all the points that are at a fixed distance from it. The fixed distance is known as the radius of the circle.

We can use a compass to draw a circle, where the compass point represents the center of the circle, and the distance between the two legs of the compass represents the radius.

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Given:

The function is

\(f(x)=4x+9\)

Domain = {-4, -2, 0, 2}

To find:

The range of the given function for the given domain.

Solution:

We know that domain is the set of input values and range is the set  of output values.

We have, \(f(x)=4x+9\) and domain = {-4, -2, 0, 2}.

Putting x=-4 in the given function, we get

\(f(-4)=4(-4)+9\)

\(f(-4)=-16+9\)

\(f(-4)=-7\)

Putting x=-2 in the given function, we get

\(f(-2)=4(-2)+9\)

\(f(-2)=-8+9\)

\(f(-2)=1\)

Putting x=0 in the given function, we get

\(f(0)=4(0)+9\)

\(f(0)=0+9\)

\(f(0)=9\)

Putting x=2 in the given function, we get

\(f(2)=4(2)+9\)

\(f(2)=8+9\)

\(f(2)=17\)

The output values are -7, 1, 9, 17. So, the range of the function f(x) for the given domain is {-7, 1, 9, 17}.

Therefore, the correct option is D.

Answer: 32

Step-by-step explanation: this is arithmetic sequence

general formula = a+(n-1)d

nth term = 5+(n-1)3

=5+3n-3

=2+3n

10th term = 2+3(10)

=32

Answer:

44 inches

Step-by-step explanation:

Given that :

Scale of drawing = 1 inch : 8 inches

Height of building in drawing = 5 1/2 inches

Using the scale provide for the drawing, the actual height of the building is :

Height of building in drawing * 8 inches

5 1/2 * 8

5.5 * 8 inches

= 44 inches

Answer:

Hello

I think it's 4

Step-by-step explanation:

The integers between -4 and 4 are:

The probability that Lucy selects a cherry Skittle followed by a lime Skittle is 15/380.

To determine the probability that Lucy selects a cherry Skittle followed by a lime Skittle, we need to consider the total number of Skittles available and the number of cherry and lime Skittles remaining.

Let's calculate the probability step by step:

Step 1: Calculate the probability of selecting a cherry Skittle first.

Lucy has a total of 3 cherry Skittles remaining out of a total of 3 + 5 + 4 + 8 = 20 Skittles remaining.

The probability of selecting a cherry Skittle first is 3/20.

Step 2: Calculate the probability of selecting a lime Skittle second.

After Lucy has eaten the cherry Skittle, she has 2 cherry Skittles remaining, along with 5 lime Skittles out of a total of 19 Skittles remaining.

The probability of selecting a lime Skittle second is 5/19.

Step 3: Calculate the probability of selecting cherry and then lime.

To calculate the probability of two independent events occurring in sequence, we multiply their individual probabilities.

Therefore, the probability of selecting a cherry Skittle first and then a lime Skittle is (3/20) * (5/19) = 15/380.

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Answer:

(4,0) and (1,-3) are the solutions sets.

Step-by-step explanation:

Given that:

y = x²-4x       Eqn 1

y = x - 4        Eqn 2

In substitution method, the equation is solved for one variable and then put in the other.

Putting value of y from Eqn 2 in Eqn 1

x - 4 = x²- 4x

Adding -x+4 on both sides

\(x-4-x+4=x^2-4x-x+4\\0=x^2-5x+4\\x^2-5x+4=0\)

Factorizing the equation;

\(x^2-4x-x+4=0\\x(x-4)-1(x-4)=0\\(x-4)(x-1)=0\)

Either,

x-4 = 0,     => x=4

Or,

x-1=0,       => x=1

Putting x=4 in Eqn 2

y = 4-4

y = 0

Putting x=1 in Eqn 2

y = 1-4

y = -3

Hence,

(4,0) and (1,-3) are the solutions sets.

The standard error of the mean for a sample of size 8 and population mean 436 is equals to the 36.42.

The standard error of the mean, or usually say standard error, shows how different the population mean from a sample mean. We have a population with population mean, \(\mu\) = 436

Standard deviations for population, \( \sigma\) = 103

Sample size, n = 8

We have to determine the value of standard error of the mean. SEM is calculated simply as dividing the standard deviation by the square root of the sample size, n. Mathematically,

\(SE = \frac{ \sigma}{\sqrt n}\)

Substitues all known values in above formula, \(= \frac{ 103}{\sqrt 8}\)

= 36.42

Hence, required standard error value is 36.42.

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The expected value of the discrete distribution, for each case, is given as follows:

a) Amount won by one entry: $0.075.

b) If the cost is the cost of a postage stamp:  -$0.525.

The expected value of a discrete distribution is obtained with the sum of each outcome multiplied by it's probabillity.

In this problem, the distribution for the prizes is given as follows:

Hence the expected value for the prizes is:

E(X) = 10,000,000 x 1/300,000,000 + 350,000 x 1/150,000,000 + 50,000 x 1/10,000,000 + 20,000 x 1/1,000,000 + 400 x 1/200,000 + 75 x 1/6,000 = $0.075.

The cost of a postage stamp is of $0.6, hence the expected value considering this cost would be of:

0.075 - 0.6 = -$0.525.

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For Helen's canvas is 12 in, wide, length of canvas is, 18 inch

Any relationship that is always in the same ratio and quantity which vary directly with each other is called the proportional.

Given that;

Helen is copying a printed reproduction of the Mona Lisa.

And, The print is 24 in. wide and 36 in tall.

Now,  If Helen's canvas is 12 in, wide, length of canvas = x

Hence, We get;

24 / 36 = 12 / x

x = 12 x 36 / 24

x = 18 inch

Thus, Lenght of canvas is, 18 inch.

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The value of 60.32 percent of 675.64 is 407.55.

It is important to note that percentage is a value that may be stated as a fraction of 100. When we need to calculate a percentage of a number, we divide it's entirety and then multiply it by 100.

In this case, the value of 60.32% of 675.64 will be:

= 60.32/100 × 675.64

= 0.6032 × 675.64

= 407.55

Therefore, the value is 407.55

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Answer:

1. the best teacher

teach from the heart ❤️

not from the book

2. a good education can change anyone.

a good teacher can change everything.

3.teaching is a work of a Heart

The volume of the resulting solid, after drilling a round hole of radius 5 through the center of a ball with a radius of 10, is 5243.7 cubic units.

To find the volume of the resulting solid, we can subtract the volume of the drilled hole from the volume of the original ball. The volume of a sphere is given by the formula V = (4/3)πr^3, where r is the radius.

The volume of the original ball can be calculated as:

V_original = (4/3)π(10^3) = 4188.8 cubic units.

The volume of the drilled hole can be calculated as:

V_hole = (4/3)π(5^3) = 523.6 cubic units.

Subtracting the volume of the hole from the volume of the original ball, we get:

V_resulting_solid = V_original - V_hole

                 = 4188.8 - 523.6

                 = 4665.2 cubic units.

Rounding to one decimal place, the volume of the resulting solid is approximately 5243.7 cubic units.

By subtracting the volume of the drilled hole from the volume of the original ball, we obtained the volume of the resulting solid. The calculations were based on the formulas for the volume of a sphere and the subtraction of volumes.

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Answer:

620 cm²

Step-by-step explanation:

First I'll describe the net. There are 3 rectangles and 2 triangles. The triangles are identical and 2 out 3 of the rectangles are the same.

So, we know that the two similar rectangles have measurements of 13cm x 10cm. So, that's 130 cm² for one such rectangle. Add the other one and we get 260 cm².

The bottom rectangle has dimensions of 10cm x 24 cm, or 240 cm². Add that to our previous total and we'll get 500 cm².

On to the triangles. Now, we know that 1. these are isosceles triangles, and 2. they have a height of 5 cm. We can pretend to cut them in half at their highest point, which would give us 4 triangles with height 5 and length 12. Now, if you put two identical triangles together, it's a four-sided shape, and you can just multiply length by width. Let's do that.

Now we have 2 rectangles, each with width 5 and length 12. Calculate the area for each, and now we have 2 rectangles with area 60 cm². Add them together....120cm².

Add that to our previous total, and we'll get 620 cm². That should be the surface area of your prism :D

Answer:

G) AC = 12

Step-by-step explanation:

5/10 = (x + 5)/(3x + 3)

15x + 15 = 10x + 50

5x = 35

x = 7

AC = 7 + 5 = 12

By using  Fundamental Theorem of Calculus, we find the derivative of the function g(x) = In { sqrt( t + t^3)dt } limit from x to 0 is  ln(sqrt(x + x^3)).  The derivative of the function g(x) = { In (1+t^2) dt} where limit are from x to 1 is  ln(1 + x^2).

The Fundamental Theorem of Calculus, which states that if a function is defined as the definite integral of another function, then its derivative is equal to the integrand evaluated at the upper limit of integration.

So, applying this theorem, we have:

g'(x) = d/dx [∫x_0 ln(sqrt(t + t^3)) dt]

= ln(sqrt(x + x^3)) * d/dx (x) - ln(sqrt(0 + 0^3)) * d/dx (0)

= ln(sqrt(x + x^3))

Therefore, g'(x) = ln(sqrt(x + x^3)).

Using the Fundamental Theorem of Calculus, we have:

g'(x) = d/dx [∫1_x ln(1 + t^2) dt]

= ln(1 + x^2) * d/dx (x) - ln(1 + 1^2) * d/dx (1)

= ln(1 + x^2)

Therefore, g'(x) = ln(1 + x^2).

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____The given question is incomplete, the complete question is given below:

Use Part 1 of the Fundamental Theorem of Calculus to find the derivative of the function. 7 g(x) = In { sqrt( t + t^3)dt } limit from x to 0.  8. g(x) = { In (1+t^2) dt} where limit are from x to 1.

The number of milligrams (mg) of silver (Ag) in 250 mL of a saturated solution of Ag₂CO₃ (Ksp = 8.1 × 10⁻¹²) is approximately 1.5 mg.

To calculate the amount of silver in the solution, we need to consider the solubility product constant (Ksp) and the stoichiometry of the compound. The balanced equation for the dissolution of Ag₂CO₃ is:

Ag₂CO₃(s) ⇌ 2Ag⁺(aq) + CO₃²⁻(aq)

From the equation, we can see that each mole of Ag₂CO₃ dissociates into 2 moles of Ag⁺ ions. Given the Ksp value, we can assume that the concentration of Ag⁺ in the saturated solution is equal to the solubility of Ag₂CO₃.

Using the equation for Ksp:

Ksp = [Ag⁺]²[CO₃²⁻]

We can rearrange the equation to solve for [Ag⁺]:

[Ag⁺]² = Ksp / [CO₃²⁻]

Since Ag₂CO₃ is sparingly soluble, we can assume that the concentration of CO₃²⁻ is negligible compared to Ag⁺, and thus [CO₃²⁻] ≈ 0.

Therefore, [Ag⁺] ≈ √Ksp = √(8.1 × 10⁻¹²) ≈ 9.0 × 10⁻⁶ M.

Now, to find the amount of silver in the solution, we multiply the concentration by the volume:

Amount of Ag = [Ag⁺] × Volume = (9.0 × 10⁻⁶ M) × (250 mL) ≈ 2.25 × 10⁻³ mol.

To convert this to milligrams, we need to multiply by the molar mass of silver:

Amount of Ag in mg = (2.25 × 10⁻³ mol) × (107.87 g/mol) × (1000 mg/g) ≈ 1.5 mg.

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Answer:

Step-by-step explanation:

Equation for the linear function has been given as,

y = 26 - 4x

For x = 1,

y = 26 - 4(1)

  = 26 - 4

  = 22

For x = 2,

y = 26 - 4(2)

  = 26 - 8

  = 18

For x = 3,

y = 26 - 4(3)

  = 26 - 12

  = 14

For x = 6,

y = 26 - 4(6)

  = 26 - 24

  = 2

Therefore, table for the function will be,

x               y

1              22  

2              18

3              14

6               2

Answer:

3 different ways

Step-by-step explanation:

there are 6 classes and 2 twins, so 6 divided by 2 = 3

The given statement is true. The outside bank of a meander that is prone to erosion is known as a cut bank.

As it flows across the land, a river erodes the soil and forms banks. A cut bank is a nearly vertical bank that is also known as a river cliff or river-cut cliff. Cut banks can be found on the river's rim. The river's moving water wears away the earth, resulting in cut banks.

A cut bank, also known as a river cliff or river-cut cliff, is the outside bank of a water channel's constantly eroding curve or meander.

Hence, the given statement is true.

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It is plausible to defend their assertion that more than 25% of the pumpkins picked weighed 25 pounds or more. By carefully considering the data representation, the characteristics of pumpkins, and the farming practices involved.

What is a histogram?

A histogram is a graphical representation of data that organizes and displays the frequency or count of observations within different intervals or bins. It is commonly used to depict the distribution of a continuous or discrete variable. The horizontal axis of a histogram represents the variable being measured, divided into intervals or bins, while the vertical axis represents the frequency or count of observations falling within each interval.

Introduction:

In this argument, we will defend the claim made by a student on a pumpkin farm who asserted that more than 25% of the pumpkins picked weighed 25 pounds or more. We will examine the histogram representing the words of pumpkins picked by students on the farm to evaluate the accuracy of this claim. By analyzing the data and considering the characteristics of pumpkins and the farming process, we will present a compelling argument supporting the student's assertion.

1. Data Representation and Distribution:

The histogram is an effective visual representation of data distribution. It provides an overview of the frequency or count of certain values, in this case, the weight of pumpkins. By examining the histogram, we can identify patterns and draw meaningful conclusions about the data.

2. Interpretation of Histogram:

Upon analyzing the histogram, we observe that the horizontal axis represents the weight ranges of the pumpkins, while the vertical axis represents the frequency or count of pumpkins falling into each weight range. The histogram displays a bar graph, where the height of each bar represents the frequency of pumpkins falling within that specific weight range.

3. Identifying the Weight Range of Pumpkins:

To assess the claim, we need to determine the percentage of pumpkins weighing 25 pounds or more. By examining the histogram, we locate the corresponding bar representing the weight range of 25 pounds or more. The total count of pumpkins falling into this weight range can be estimated by summing up the frequencies of all the bars within that range.

4. Calculating the Percentage:

Next, we calculate the percentage of pumpkins that weigh 25 pounds or more. We divide the estimated count of pumpkins in the 25 pounds or more range by the total count of all pumpkins, and then multiply by 100 to obtain the percentage.

5. Comparing the Percentage:

Now, we compare the calculated percentage with the claim made by the student. If the calculated percentage is higher than 25%, the student's assertion holds true.

6. Factors Supporting the Claim:

Several factors support the likelihood of the student's claim being accurate:

a. Variability in Pumpkin Sizes: Pumpkins can vary significantly in size, ranging from small to very large. Therefore, it is plausible to have a notable portion of pumpkins weighing 25 pounds or more.

b. Farming Practices: Pumpkin farmers often aim to cultivate larger pumpkins as they are generally preferred for carving, decorations, or selling purposes. As a result, farmers may prioritize growing larger pumpkins, leading to a higher proportion of pumpkins weighing 25 pounds or more.

c. Human Perception Bias: Humans tend to remember and focus on outliers and unique occurrences. The student's claim may be a result of observing a few particularly large pumpkins, which left a lasting impression, influencing their perception of the overall pumpkin population.

Therefore, based on the analysis of the histogram and the factors supporting the student's claim, it is plausible to defend their assertion that more than 25% of the pumpkins picked weighed 25 pounds or more. By carefully considering the data representation, the characteristics of pumpkins, and the farming practices involved, we find a convincing argument supporting the student's claim.

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Given data:

The given triangles.

The given trinagles are similar.

Thus, the (D) option is correct.

The number of each type of coins are as follows:

q = 15 quarters.

d = 30 dimes.

n = 25 nickels.

In order to solve this word problem, we would assign a variables to the unknown numbers and then translate the word problem into algebraic equation as follows:

Let d represent the number of dimes.

Let q represent number of quarters.

Let n represent number of nickels.

Let T represent total number of coins.

Note: 1 quarter is equal to 0.25 dollar, 1 nickel is equal to 0.5 dollar, and 1 dime is equal to 0.1 dollar.

Translating the word problem into an algebraic equation, we have;

Dimes; d = 2q     .....equation 1.

Nickels; (70 - (q + 2q)) = (70 - 3q)     .....equation 2.

Total coins; T = n + d + q

0.5(70 - 3q) + 2q(0.1) + q(0.25) = 8.00

Multiplying all through by 100, we have:

5(70 - 3q) + 2q(10) + q(25) = 800

350 - 15q + 20q + 25q = 800

350 + 30q = 800

30q = 800 - 350

30q = 450

q = 450/30

q = 15 quarters.

For the number of dimes, we have:

Dimes, d = 2q

Dimes, d = 2(15)

Dimes, d = 30 dimes.

For the number of nickels, we have:

Nickels, n = (70 - 3q)

Nickels, n = (70 - 3(15))

Nickels, n = (70 - 45)

Nickels, n = 25 nickels.

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