Find the surface area.​

Answer:

2.2 pounds

Step-by-step explanation:

Your welcome

Answer: B. statistic

Step-by-step explanation: A characteristic, usually a numerical value, which describes a sample, is called a statistic.

Answer:

1.8291

Step-by-step explanation:

The probability that Juan loses his next tennis match is 48%

From the question, we have the following parameters

The proportion of tennis matches won

p = 52%

This proportion implies that the probability that the player will win his next tennis match

So, we have the following representations

Probability of winning, p = 52%

Express the percentage as decimal

So, we have the following representations

Probability of winning, p = 0.52

The probability of losing the next match is the complement of the probability of winning

This is represented as

Probability of losing = 1 - Probability of winning,

Substitute the known values in the above equation

So, we have the following equation

Probability of losing = 1 - 0.52

Evaluate

Probability of losing = 0.48

Hence, the probability is 0.48 or 48%

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Answer:

Step-by-step explanation:

yes it is

Answer:D

The answer is D.

Step-by-step explanation:

Like terms are only related to the variable at the end, and the variable also has to have the same exponent or it is not a like term.

The pasture is 68.73 acres. Total fence length: 7000 ft + 32 ft (drive-through gates) + 16 ft (walk-through gates). You need approximately 22 rolls of woven wire fencing.



To find the area of the pasture in acres, we need to convert the given measurements from feet to acres.

1 acre = 43,560 square feet

Area of the pasture = 1500 ft * 2000 ft = 3,000,000 square feet

Area in acres = 3,000,000 square feet / 43,560 square feet per acre ≈ 68.73 acres

So, the area being fenced is approximately 68.73 acres.

Now, let's calculate the total length of fencing required, taking into account the gates.

Length of fence needed = perimeter of the pasture + length of drive-through gates + length of walk-through gates

Perimeter of the pasture = 2 * (length + width)

Perimeter = 2 * (1500 ft + 2000 ft) = 2 * 3500 ft = 7000 ft

Length of drive-through gates = 2 * 16 ft = 32 ft

Length of walk-through gates = 4 * 4 ft = 16 ft

Total length of fencing needed = 7000 ft + 32 ft + 16 ft = 7048 ft

Now, we can calculate the number of rolls of woven wire fencing needed.

Length of one roll of woven wire = 330 ft

Number of rolls needed = Total length of fencing needed / Length of one roll of woven wire

Number of rolls needed = 7048 ft / 330 ft ≈ 21.39 rolls

Since we can't purchase a fraction of a roll, we'll need to round up to the nearest whole number.

Therefore, you would need to purchase approximately 22 rolls of woven wire fencing.

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Step-by-step explanation:

2+3=120

5=120

=120

5

=24

apple=2 multiply 24 is 48

orange=3 multiply 24 is 72

In testing for differences between the means of 2 independent populations the null hypothesis is zero.

This is defined as a statistical hypothesis which has no statistical significance in a set of given observations.

In testing for differences between the means of 2 independent populations the null hypothesis is the difference between the two population means and is not significantly different from zero which is denoted below:

H₀: µ₁ - µ₂ = 0

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The radius of the cone is 783.84/√h

A cone is the surface traced by a moving straight line (the generatrix) that always passes through a fixed point (the vertex).

Volume is defined as the space occupied within the boundaries of an object in three-dimensional space.

The volume of a cone is expressed as;

V = 1/3πr²h

643072 × 3 = 3.14 × r²h

r²h = 614400

r² = 614400/h

r = 783.84/√h

therefore the radius of the cone is 783.84/√h

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Answer:

length • width = area

expression:

area • height = (result) • 1/3 = volume

Example:

x (area) • 6 (height) = 6x

6x • 1/3 = 2x(inches)³ or 6x ÷ 3 = 2x(inches)³

Modifications are just a starting point, and you can explore endless possibilities by experimenting with different mathematical functions, operations, and parameters.

To create even more fanciful curves, you can modify the equation from part (a) by introducing additional terms or manipulating the existing terms. Here are a few examples of modifications you can make to create interesting shapes:

Adding or multiplying terms: You can introduce additional terms to the equation, such as adding or multiplying variables or constants. For example, if the original equation was y = x², you can modify it to y = x³ + 2x² + x, which would create a more intricate curve with higher-order terms.

Trigonometric functions: You can incorporate trigonometric functions like sine, cosine, or tangent into the equation. This can produce curves with periodic patterns or wave-like shapes. For instance, modifying the equation y = x² to y = x² + sin(x) would introduce oscillations to the curve.

Exponential functions: Exponential functions can also be used to create interesting curves. By including terms like eˣ or e⁻ˣ, you can achieve curves that grow or decay exponentially. Modifying the equation y = x²to y = x² + e⁻ˣ would create a curve that decreases exponentially as x increases.

Combining functions: You can combine multiple functions using addition, subtraction, multiplication, or composition to generate complex shapes. For example, y = sin(x)× cos(2x) would result in a curve that exhibits both sine and cosine patterns simultaneously.

Parameterization: Introducing parameters into the equation allows you to control various aspects of the curve, such as its shape, size, or position. By adjusting these parameters, you can create a wide range of fanciful curves. For example, in the equation y = a× x², you can vary the value of 'a' to stretch or compress the parabola.

Remember, these modifications are just a starting point, and you can explore endless possibilities by experimenting with different mathematical functions, operations, and parameters. Computer algebra systems can help you visualize the resulting curves and explore their properties in detail.

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Answer:

GCF, Greatest Common Factor

Step-by-step explanation:

Hope this helped

By engaging in these exercises, students can develop a deeper understanding of conditional statements and logical reasoning, which are essential skills for further studies in mathematics and logic.

In Unit 2 of a logic and proof course, homework 3 focuses on conditional statements.

Conditional statements are fundamental concepts in logic and mathematics, representing logical implications between two statements.

They are typically expressed in "if-then" format, where the "if" part is the hypothesis and the "then" part is the conclusion.

The homework may involve tasks such as:

Identifying conditional statements: Students are given a set of statements and asked to identify which ones are conditional statements.

They need to recognize the "if-then" structure and correctly identify the hypothesis and conclusion.

Analyzing the truth value of conditional statements:

Students may be given conditional statements and asked to determine whether they are true or false.

They need to evaluate the hypothesis and conclusion to determine if the implication holds in each case.

Writing converse, inverse, and contrapositive statements:

Students may be required to manipulate given conditional statements to form their converse, inverse, and contrapositive statements.

This involves switching the positions of the hypothesis and conclusion or negating both parts.

Applying the laws of logic:

Students may need to apply logical laws, such as the Law of Detachment or the Law of Modus Tollens, to deduce conclusions based on conditional statements.

Constructing counterexamples:

Students may be asked to provide counterexamples to disprove statements that are falsely claimed to be universally true based on a given conditional statement.

They also help students develop critical thinking and problem-solving abilities, as they have to analyze and manipulate logical structures.

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Answer:

See below

Step-by-step explanation:

g (h(6))   :

   h (6) =  3 ( 6^2) + 2 = 110

   then g (110) =   sqrt (110)

h (g(5))

    g(5) = sqrt 5

        then h( sqrt5) = 3 ( sqrt5)^2 + 2 = 17

The approximate annual cost of raising a child born in 2013 at age 17 is 15230.

Scatter plots are used to observe and plot relationships between two numeric variables graphically with the help of dots. The dots in a scatter plot shows the values of individual data points.

a) Plot (2, 12,940), (5, 12,970), (8, 12,800), (11, 13,600) 1nd (14, 14420) on graph.

b) (8, 12800) and (14, 14420)

Here, slope m=(14420-12800)/(14-8)

= 1,620/6

= 270

Substitute m=270 and (x, y)=(8, 12800) in y=mx+c, we get

12800=270(8)+c

12800=2160+c

c=10640

Substitute m=270 and c=10640 in y=mx+c, we get

y=270x+10640

c) Now, substitute x=17 in y=270x+10640, we get

y=270×17+10640

y=15230

Therefore, the approximate annual cost of raising a child born in 2013 at age 17 is 15230.

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In a Runge-Kutta method, the local truncation error measures the error made in one step of the method, while the global truncation error measures the cumulative error over all the steps.

Given that the local truncation error of the Runge-Kutta method is of O(h^3), we can express the local error as:

LTE = C1 h^3 + C2 h^4 + ...

where C1, C2, ... are constants.

To obtain the global truncation error, we need to sum the local truncation errors over all the steps. Suppose we take n steps, each of size h. Then, the total error is of the form:

GTE = C1 h^3 + C2 h^4 + ... + Cn h^(n+2)

where the powers of h increase as we take more steps.

Since the global truncation error is also of O(h^n), we can equate the highest power of h in the above equation to n:

n + 2 = n

Solving for n, we get:

n = 2

Therefore, the value of n is 2.

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Answer:

d

Step-by-step explanation:

8/3

Answer:

63 seat for adults and 27 seat for children.

please mark me as brainliest

The problem can be modeled as a linear programming problem. We need to minimize the total cost of shipping the boots, which can be represented as a linear combination of the shipment quantities.

Thus, the objective function to be minimized can be written as:

Minimize Z = 4x11 + 5x12 + 7x13 + 6

32 + y33 ≤ 40 (capacity limit at warehouse 3)

y41 + y42 + y43 ≤ 30 (capacity limit at warehouse 4)

y51 + y52 + y53 ≤ 45 (capacity limit at warehouse 5)

xij, yjk ≥ 0 (non-negativity constraint)

Thus, the complete linear programming problem can be formulated as:

Minimize Z = 4x11 + 5x12 + 7x13 + 6x21 + 4x22 + 5x23 + 4y11 + 3y12 + 6y13 + 5y21 + 5y22 + 4y23 + 6y31 + 7y32 + 5y33 + 4y41 + 6y42 + 5y43 + 5y51 + 4y52 + 6y53

Subject to:

x11 + x12 + x13 ≤ 60

x21 + x22 + x23 ≤ 90

x11 + x21 = y11 + y21 + y31 + y41 + y51

x12 + x22 = y12 + y22 + y32 + y42 + y52

x13 + x23 = y13 + y23 + y33 + y43 + y53

y11 + y12 + y13 ≤ 35

y21 + y22 + y23 ≤ 50

y31 + y32 + y33 ≤ 40

y41 + y42 + y43 ≤ 30

y51 + y52 + y53 ≤ 45

xij, yjk ≥ 0

Solving this linear programming problem using a suitable solver will provide us with the optimal values of xij and yjk, which will help us in determining the minimum cost required for shipping the boots while satisfying all the constraints.

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Answer:

radius, center, chord, diameter

Step-by-step explanation:

a. radius

b. center

c. chord

d. diameter

Answer:

You Have to Find Answer Yourself

Step-by-step explanation:

By This - B+D+C = 130°-20°+E

Answer:

110 degrees

Step-by-step explanation:

when continuing line DC down to line AE, it creates a triangle with the intersection point at AE, C and E.

the angles in that triangle are

E (20 degrees)

the internal angle at C = 180 - 130 = 50 degrees

and the angle at the interception point at AE.

since CD is parallel to AB, that angle is equal to x.

the sum of all angles in a triangle is airways 180 degrees.

therefore this angle is

180 - 20 - 50 = 110 degrees = x

Answer: Length = 28 feet and width = 9 feet

Step-by-step explanation:

Let x = width of the wall.

Then, Length of the wall =  10+2x

Since , Area of rectangle = length x width

Then, as per given,

\(252= (10+2x)x\\\\\Rightarrow\ 252=10x+2x^2\\\\\Rightarrow\ 2x^2+10x-252=0\\\\\Rightarrow\ x^2+5x-126=0\\\\\Rightarrow\ x^2+14x-9x-126=0\\\\\Rightarrow\ (x+14)-9(x+14)=0\\\\\Rightarrow\ (x+14)(x-9)=0\\\\\Rightarrow\ x=-14 \ or \ x= 9\)

Since width cannot be negative , so x = 9

So width = 9 feet and length = 10+2(9)=10+18 = 28 feet

Hence, length is 28 feet and width is 9 feet of the wall of the barn.

Answer:

-2a + 3

Step-by-step explanation:

We can substitute a + 7 for x:

f(a + 7) = 17 - 2(a+7) = 17 - 2a - 14 = -2a + 3

Answer:

To determine the speed, we can use the formula:

speed = distance / time

where distance is measured in feet and time is measured in minutes.

In this case, the distance is 264 feet and the time is 3 minutes. Plugging these values into the formula, we get:

speed = 264 feet / 3 minutes

simplifying, we get:

speed = 88 feet/minute

Therefore, the equipment is traveling at a speed of 88 feet per minute.

Answer:

8/9n - 1 1/3

Step-by-step explanation:

Answer:

Point form: (-2,3)

Equation form: x= -2, y=3

Step-by-step explanation:

Hope it helps!

Answer: What’s the rest of the question?

Step-by-step explanation:

We can prove that there exists a positive constant C such that G(x) > C for all x, where G is a continuous function.

G is a function continuous at a and G(a)>0.

We have to prove that there exists a positive constant C such that G(x)>C for all x.

To prove this statement, we can use the epsilon-delta definition of continuity.

According to the epsilon-delta definition of continuity, if G is continuous at a, then for every ε > 0 there exists a δ > 0 such that for all x with |x - a| < δ, we have |G(x) - G(a)| < ε.

Now since G(a) > 0, let ε = G(a)/2.

So there exists a δ > 0 such that for all x with |x - a| < δ,

we have |G(x) - G(a)| < G(a)/2.

Since G(a) > 0,

we can multiply both sides of this inequality by 2 to get:

2|G(x) - G(a)| < G(a).

Adding G(a) to both sides, we get:

2|G(x) - G(a)| + G(a) < 2G(a).

Let C = G(a)/2.

Then we have:

|G(x) - G(a)| < C for all x with |x - a| < δ.

G(x) - G(a) > -C and G(x) - G(a) < C.

Then G(x) > G(a) - C > 0 for all x with |x - a| < δ, so we can choose δ small enough that |x - a| < δ implies G(x) > G(a) - C > 0.

Thus we have proved that there exists a positive constant C such that G(x) > C for all x.

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