Un agricultor ha comprado un terreno de 5,3ha y 4,6 a. Para cultivar pistacho, debe plantar un arbol cad 36m3 ¿cuantos pistachos debe plantar?

Answer: 28x+8y+2

Step-by-step explanation:

Remove parentheses: -2(14x-4y-1)

Solution: 28x+8y+2

Answer:

28x+8y+2

Step-by-step explanation:

Answer:

Step-by-step explanation:

3x^16

Answer:

3x^16

Step-by-step explanation:

Solution: x + 12

Step-by-step explanation:

first simplify then distribute!

Answer:x+12

Step-by-step explanation:

we have to multiply number outside of brackets to each of the numbers inside it:

2(2x+5)=4x+10

-(3x-2)=-1(3x-2)=-3x+2

4x+10-3x+2=x+12

Equations are used in many fields, including physics, engineering, economics, and finance, to model and analyse relationships between variables.

The expressions on either side of the equal sign are referred to as the left-hand side (LHS) and the right-hand side (RHS) of the equation. Solving an equation involves finding the values of the variables that make the equation true.

Two amounts or values are equal when they are stated as so in a mathematical equation, such as 6 x 4 = 12 x 2. two. Noun with a count.

A situation that requires the consideration of two or more elements in order to perceive or comprehend the total situation is referred to as an equation.

[tex]2+2+2+2+2+3-3-3-3-21-456--42 =[/tex]

[tex]2+2+2+2+2+3 = 13[/tex]

[tex]-3-3-3 = -9[/tex]

[tex]-456--42 = -456+42 = -414[/tex]

Therefore,

[tex]2+2+2+2+2+3-3-3-3-21-456--42 = 13 - 9 - 414 = -410[/tex]

Therefore, the value of the expression is  [tex]2+2+2+2+2+3-3-3-3-21-456--42[/tex]  [tex]= 13 - 9 - 414 = -410[/tex]     . As per the given equation.

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Answer:

Step-by-step explanation:

(g * h)(24)

gh*24

24gh

The angle of x in the given circle is 51 degrees.

In the case when a tangent line and a secant line originate from the same external point, the mean of the exterior angle is equal to half the difference between the major arc and the minor arc that are created by the tangent and secant lines.

From the given figure we observe that the secant line and tangent line start from the same exterior angle.

Thus, according to the theorem of secant and tangent line of circle we have:

angle x = (DA - DB) /2

angle x = (178 - 76) / 2

angle x = 102 /2

angle x = 51 degrees.

Hence, the angle of x in the given circle is 51 degrees.

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The value of x that makes the given image to be a rectangle is: x = 2.5

The length of the diagonals of a rectangle are equal and congruent to each other. It is also pertinent to note that the diagonals of a rectangle bisect each other at the center. Thus:

From the given parameters, we see that half of the diagonals have been labelled and as such we have:

8x - 3 = 4x + 7

Rearrange to have like terms on same side and so we have:

8x - 4x = 7 + 3

4x = 10

x = 10/4

x = 2.5

Thus, the value of x from the calculation above of the rectangle is 2.5

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The probability P(Xˉ−2<114<Xˉ+2) is approximately 0.95. (Option 3)

We know that the population mean is 114 and the standard deviation is 6. We also know that the sample size is 36, which is large enough to apply the central limit theorem. Thus, the distribution of sample means follows a normal distribution with a mean of 114 and a standard deviation of 6/√36=1.

Now, we need to calculate the z-score for the lower and upper bound of the inequality.

z1=(114-2-114)/1=-2

z2=(114+2-114)/1=2

Using a standard normal distribution table, we can find that the area to the left of -2 is 0.0228 and the area to the left of 2 is 0.9772. Thus, the area between -2 and 2 is approximately 0.9772-0.0228=0.9544, which is approximately 0.95.

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Complete Question:

In 2022, NBA teams score 114 points per game on average with a standard deviation of 6 points per game. Suppose that we randomly sample 36 NBA scores from this season and consider the sample mean (

Xˉ ); the average points per game scored in this sample.

Calculate P( Xˉ 2<114< Xˉ +2)

In response to the stated question, we may state that therefore, the expressions value of vector will be n = 25.73.

An expression in mathematics is a collection of numbers, variables, and mathematical (such as addition, reduction, multiplication, division, exponentiation, etc.) that express a quantity or value. Expressions might be as basic as "3 + 4" or as complicated as "(3x2 - 2) / (x + 1)". They may also contain functions like "sin(x)" or "log(y)". Expressions can also be evaluated by substituting values for the variables and performing the mathematical operations in the order specified. If x = 2, for example, the formula "3x + 5" equals 3(2) + 5 = 11. Expressions are commonly used in mathematics to describe real-world situations, construct equations, and simplify complicated mathematical topics.

to find vector and magnitude we have

[tex]n^2 = (22.5)^2 + (12.5)^2\\n^2 = 506.25 + 156.25\\n^2 = 662.5\\n = \sqrt(662.5)\\n = 25.73[/tex]

therefore, the value of vector will be n = 25.73.

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The pay at the Digital Source is a variable amount that depends on the employee's job position and experience level. The pay is represented in dollars per hour, and it ranges from $12 per hour for a new employee in a low-level position to $60 per hour for a senior developer with extensive experience.

For example, if we look at the table created in part a, we can see that a junior developer with less than a year of experience will earn $18 per hour, while a senior developer with more than 10 years of experience will earn $60 per hour. This means that the pay at the Digital Source is directly related to the employee's experience and job position.

Additionally, the table also shows that the pay rate increases as the employee gains more experience and moves up the career ladder. For instance, a junior developer with less than a year of experience will earn $18 per hour, but a mid-level developer with 3-5 years of experience will earn $35 per hour. This shows that employees are rewarded for their skills and experience, and they have the opportunity to advance their careers and earn higher pay rates.

In summary, the pay at the Digital Source is a variable amount that depends on the employee's job position and experience level. The pay rate ranges from $12 per hour to $60 per hour, and it increases as the employee gains more experience and moves up the career ladder. The company values the skills and expertise of their employees and rewards them accordingly.

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The resultant vector is 7 units to the left and 5 units down. As a result, graphs option (c) a vector pointing to the left 7 units and down 5 units is the correct answer.

Mathematicians use graphs to visually display or chart facts or values in order to express them coherently. A graph point usually represents a connection between two or more items. A graph, a non-linear data structure, is made up of nodes (or vertices) and edges. Glue the nodes, also known as vertices, together. This graph includes V=1, 2, 3, 5, and E=1, 2, 1, 3, 2, 4, and (2.5). (3.5). (4.5). Statistical graphs (bar graphs, pie graphs, line graphs, and so on) are graphical representations of exponential development. a logarithmic graph shaped like a triangle

We can add vectors geometrically by setting the starting point of the second vector at the terminal point of the first vector and drawing a vector from the first vector's initial point to the second vector's terminal point.

Next we align the starting point of vector v with the terminal point of vector u and draw a vector from the initial point of vector u to the terminal point of vector v:

The resultant vector is 7 units to the left and 5 units down. As a result, option (c) a vector pointing to the left 7 units and down 5 units is the correct answer.

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The value of the constant C is 3/4.

Probability is a branch of mathematics that deals with the likelihood or chance of an event occurring. It is a measure of the likelihood that a specific event will occur, expressed as a value between 0 and 1

The sum of the probabilities for all possible values of X should equal 1

P(1) + P(2) + P(3) + P(4) = 1

Substituting the given values, we get:

C/4 + 1/2 + 1/8 + C/4 = 1

Simplifying the equation, we get:

C/2 + 5/8 = 1

Move 5/8 to right hand side of the equation

C/2 = 3/8

C = (3/8) x 2

C = 3/4

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Answer:

The probability of selecting a number at random from the sample space {1, 2, 3, 4, 5, 6, 7, 8} that is positive is:

There are 8 possible outcomes in the sample space, and 4 of them are positive (1, 2, 3, 4).

Therefore, P(the number is positive) = 4/8 = 1/2 = 0.5

The probability of selecting a number at random from the sample space {1, 2, 3, 4, 5, 6, 7, 8} that is even is:

There are 8 possible outcomes in the sample space, and 4 of them are even (2, 4, 6, 8).

Therefore, P(the number is even) = 4/8 = 1/2 = 0.5

So the probability of selecting a number at random from the sample space {1, 2, 3, 4, 5, 6, 7, 8} that is positive is 0.5 and the probability of selecting a number at random from the same sample space that is even is also 0.5.

Answer:

1/20

Step-by-step explanation:

because you turn the denominators to 40, then subtract walk from run and you got your answer

the range of the function is at least $400 plus $50 for each month after the purchase. The average monthly cost of owning the cell phone can be represented as:

C(t) = 400 + 50t

Where t is the number of months after the purchase.

Since C(t) is a linear function, its range is all real numbers greater than or equal to 400, since the initial cost of the phone is 400 and the monthly cost is a positive constant. Therefore, the range of C(t) is:

Range of C(t) = {C(t) | C(t) ≥ 400} or [400, ∞)

In other words, the range of the function is all possible average monthly costs of owning the phone, which is at least $400 plus $50 for each month after the purchase.

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The probability of getting 3 black cards out of 4 drawn from a standard deck of 52 cards is approximately 0.253 or 25.3%.

To determine the probability of getting 3 black cards out of 4 cards drawn from a standard deck of 52 cards, we can use the hypergeometric distribution.

There are 26 black cards in the deck, and the total number of ways to choose 4 cards from 52 is given by the binomial coefficient C(52, 4). The number of ways to choose 3 black cards out of 26 is given by the binomial coefficient C(26, 3).

The number of ways to choose the fourth card from the remaining 26 non-black cards is given by the binomial coefficient C(26, 1).

Therefore, the probability of getting 3 black cards out of 4 is:

P(3 black cards) = C(26, 3) * C(26, 1) / C(52, 4)

P(3 black cards) = (2600 * 26) / 270725

P(3 black cards) ≈ 0.253

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The pοssible dimensiοns (length and width) οf the rectangular plοt are:

Length = 21 meters, width = 8.05 meters

Length = 8.05 meters, width = 21 meters

Let the length οf the rectangular plοt be L and the width be W. Then we knοw that the perimeter οf the rectangular plοt, which is equal tο the length οf fencing Baο has, is given by:

2L + W = 39

Alsο, the area οf the rectangular plοt is given by:

L × W = 169

We can use these twο equatiοns tο sοlve fοr L and W. First, we can sοlve the equatiοn 2L + W = 39 fοr W:

W = 39 - 2L

Substituting this into the equation for the area, we get:

L × (39 - 2L) = 169

Expanding and rearranging this equation, we get a quadratic equation in terms of L:

[tex]-2L^2 + 39L - 169 = 0[/tex]

We can solve for L using the quadratic formula:

[tex]L = [ -39 \± sqrt(39^2 - 4(-2)(-169)) ] / (2(-2))\\L = [ -39 \± sqrt(1681) ] / 4\\L = [ -39 \± 41 ] / 4[/tex]

L = 1/2 or L = 21

If L = 1/2, then W = 38, which gives a negative area and is therefore not possible.

If L = 21, then W = 169/21, which simplifies to W = 8.05 (rounded to two decimal places).

Therefore, the possible dimensions (length and width) of the rectangular plot are:

Length = 21 meters, width = 8.05 meters

Length = 8.05 meters, width = 21 meters

Note that these are the only possible dimensions, as the sum of the length and width must be less than or equal to 19.5 meters (half of the available fencing).

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Suppose that IQ scores have a bell-shaped distribution with a mean of 96 and a standard deviation of 17. Using the empirical rule, 15.87 percentage of IQ scores are less than 79.

The empirical rule says that for a normal distribution, almost all of the data will fall within three standard deviations of the mean.

Specifically:
Approximately 68% of the data falls within one standard deviation of the mean.
Approximately 95% of the data falls within two standard deviations of the mean.
Almost all (99.7%) of the data falls within three standard deviations of the mean.
Given that the mean of IQ scores is 96 and the standard deviation is 17.

Hence the Z score can be calculated as
z=(x−μ)/σ=(79−96)/17=−1
From the standard normal distribution table, the percentage of the data that is less than z = -1 is 15.87%.
Hence the percentage of IQ scores that are less than 79 is approximately 15.87%.

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Answer:

1. I don't know

2. 10

Step-by-step explanation:

[tex]c=\sqrt{a^{2}+b^{2} } = \sqrt{6^{2}+8^{2} } = \sqrt{36+64} = \sqrt{100} = 10[/tex]

To maximize the area of a rectangular field with one side along a stream using 1100 feet of fencing, the dimensions should be 183.33 feet by 550 feet.

Let's call the length of the field that runs parallel to the stream "x" and the width of the field that runs perpendicular to the stream "y".

The total amount of fencing that the farmer has available is 1100 feet, and we know that three sides of the rectangular field will be fenced. Since one side of the field is already bordered by the stream and does not require fencing, we can set up the following equation:

2x + y = 1100 - x

Simplifying this equation, we get:

3x + y = 1100

We want to maximize the area of the field, which is given by the formula:

A = xy

We can use the equation we derived earlier to solve for y in terms of x:

y = 1100 - 3x

Substituting this into the area formula, we get:

A = x(1100 - 3x)

Expanding the parentheses, we get:

A = 1100x - 3x^2

To find the maximum area, we need to find the value of x that maximizes this equation. We can do this by taking the derivative of A with respect to x and setting it equal to zero:

dA/dx = 1100 - 6x = 0

Solving for x, we get:

x = 183.33 feet

We can use this value of x to find the corresponding value of y:

y = 1100 - 3x = 550 feet

Therefore, the dimensions of the rectangular field that maximize the area are 183.33 feet by 550 feet.

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Destiny sold 23 cups of espresso and 53 cups of cappuccino on Friday, for a total sales amount of $178.50.

Let's start by defining our variables. Let x be the number of cups of espresso sold and y be the number of cups of cappuccino sold. We know that the price of espresso is $2 per cup and the price of cappuccino is $2.50 per cup. We also know that Destiny sold 30 more cups of cappuccino than espresso and the total sales amount was $178.50.

Using this information, we can create a system of equations to solve for x and y. The first equation comes from the fact that Destiny sold 30 more cups of cappuccino than espresso:

y = x + 30

The second equation comes from the total sales amount:

2x + 2.5y = 178.5

Now we can substitute the first equation into the second equation and solve for x:

2x + 2.5(x + 30) = 178.5

2x + 2.5x + 75 = 178.5

4.5x = 103.5

x = 23        

So Destiny sold 23 cups of espresso. Using the first equation, we can find that she sold 53 cups of cappuccino:

y = x + 30

y = 23 + 30    

y = 53

Therefore, Destiny sold 23 cups of espresso and 53 cups of cappuccino on Friday, for a total sales amount of $178.50.

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The final equation of motion for the system is: x(t) = [(8v0 - 2x0)/6]*[tex]e^{(-2t)} + [(2x0 - 8v0)/6]*e^{(-8t).[/tex]

To determine the initial conditions and equations of motion, we can use the following steps:

Write down the equation of motion for the system. The equation of motion for a mass-spring-damper system is given by:

m * [tex]d^2x/dt^2[/tex] + c * dx/dt + k * x = 0

where m is the mass, c is the damping coefficient, k is the spring constant, x is the displacement from the equilibrium position, and t is time.

In this case, the mass is 1 kg, the spring constant is 16 N/m, and the damping coefficient is 10 times the instantaneous velocity, or 10 * dx/dt.

Substituting these values into the equation of motion, we get:

1 * [tex]d^2x/dt^2[/tex] + 10 * dx/dt + 16 * x = 0

Determine the initial conditions. The initial conditions refer to the position and velocity of the mass at time t = 0.

Let x(0) be the initial displacement of the mass from the equilibrium position, and let v(0) be the initial velocity of the mass. Then the initial conditions are:

x(0) = ?

v(0) = ?

These values are typically given in the problem statement.

Solve the differential equation. To solve the differential equation, we can use the characteristic equation method. First, we assume that the displacement of the mass is of the form:

x(t) = A[tex]e^{(rt)[/tex]

where A and r are constants to be determined.

Taking the first and second derivatives of x(t), we get:

dx/dt = Are^(rt)

d^2x/dt^2 = Ar^2*e^(rt)

Substituting these expressions into the equation of motion, we get:

A * ([tex]r^2[/tex] + 10r + 16) * [tex]e^{(rt)[/tex]= 0

Since [tex]e^{(rt)[/tex]is never zero, we can divide both sides by it and simplify to obtain the characteristic equation:

[tex]r^2 + 10r + 16 = 0[/tex]

Solving for r using the quadratic formula, we get:

r1 = -2

r2 = -8

Therefore, the general solution for the displacement of the mass is:

[tex]x(t) = Ae^{(-2t)} + Be^{(-8t)[/tex]

where A and B are constants determined by the initial conditions.

To find A and B, we use the initial conditions x(0) = x0 and v(0) = v0. Taking the derivative of x(t) and setting t = 0, we get:

v(0) = -2A - 8B

Using the second initial condition, we get:

x(0) = A + B = x0

Solving for A and B, we get:

A = (8v0 - 2x0)/6

B = (2x0 - 8v0)/6

Therefore, the final equation of motion for the system is:

[tex]x(t) = [(8v0 - 2x0)/6]*e^{(-2t)} + [(2x0 - 8v0)/6]*e^{(-8t)[/tex]

Complete question: A 1-kilogram mass is attached to a spring whose constant is 16 N/m, and the entire system is then submerged in a liquid that imparts a damping force numerically equal to 10 times the instantaneous velocity. Determine the equations of motion if (a) the mass is initially released from rest from a point 1 meter below the equilibrium position, and then (b) the mass is initially released from a point 1 meter below the equilibrium position with an upward velocity of 12 m/s.

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The two numbers in the system of equation is 5 and 13.

A group of multiple-variable equations that must all be solved concurrently make up a system of equations. Finding values for the variables that satisfy every equation in a system of equations is the objective. Each equation in the system reflects a connection between the variables. In order to find a singular solution or a group of solutions, systems of equations can be solved using a variety of methods, such as substitution or elimination. In many disciplines, including mathematics, physics, engineering, and economics, systems of equations can occur.

Let us suppose the two numbers as x and y.

x + y = 18

x - y = -8

The second equation can be written as:

x + 8 = y

Substitute the value of y in equation 1:

x + x + 8 = 18

2x = 10

x = 5

Substitute the value of x to get y:

x + y = 18

5 + y = 18

y = 13

Hence, the two numbers in the system of equation is 5 and 13.

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The solution of the match is

(1) Orthocenter          (C) The point at which the altitudes of a triangle                                                

                                              meet.

(2) Centroid                 (D) The point at which the medians of a triangle

                                             meet.

(3) Circumcenter         (A) The point of intersection of the perpendicular                  

                                             bisectors of the sides of a triangle.

(4) Incentre                 (B) The point at which the three angle bisectors      

                                             of a triangle meet.

The orthocenter of a triangle is the point where the altitudes of the triangle intersect. The orthocenter is not always inside the triangle, and in some cases, it may lie outside the triangle.

The centroid of a triangle is the point where the three medians of the triangle intersect. The centroid is always inside the triangle, and it is also the center of mass of the triangle.

The circumcenter of a triangle is the point where the perpendicular bisectors of the sides of the triangle intersect.  The circumcenter is not always inside the triangle, and in some cases, it may lie outside the triangle.

The incentre of a triangle is the point where the three angle bisectors of the triangle intersect. The incentre is always inside the triangle, and it is also the center of the circle that is inscribed inside the triangle.

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Complete Question:

Match the following.

(1) Orthocenter         (A) The point of intersection of the perpendicular                  

                                             bisectors of the sides of a triangle.

(2) Centroid                 (B) The point at which the three angle bisectors of

                                             a triangle meet.

(3) Circumcenter         (C) The point at which the altitudes of a triangle

                                              meet.

(4) Incentre                 (D) The point at which the medians of a triangle

                                             meet.

The probability of drawing an ace and a queen from a pack of cards without replacement is 1/78. This can be explained as follows:

In a standard pack of 52 cards, there are 4 aces and 4 queens. When two cards are drawn without replacement, the probability of drawing an ace and then a queen is 4/52 x 3/51 = 12/2652. This can be simplified to 1/78.

Without replacement means that the card that is drawn is not replaced in the deck before the next card is drawn. In this case, when the first card is an ace, there are only 3 queens left in the deck so the probability of the second card being a queen is 3/51.

To put it another way, the chances of drawing an ace and a queen when the cards are drawn without replacement can be thought of as a ratio of the favorable outcomes to the total number of possible outcomes. There is only one favorable outcome (ace-queen) out of a total of 78 possible outcomes (4 aces and 4 queens combined with the remaining 44 cards). Thus, the probability of drawing an ace and a queen is 1/78.

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The decrease in the time spent exercising is 0.74.

A scatter plot is a type of data visualization that is used to display the relationship between two continuous variables. It consists of a set of data points, each representing an observation, plotted as a point on a two-dimensional graph with one variable on the x-axis and the other variable on the y-axis.

Scatter plots are often used to identify patterns or relationships between two variables, such as correlation or causation. If the points on the plot tend to form a pattern, such as a straight line, then there may be a relationship between the two variables. If the points do not form a pattern, then there may be no relationship between the two variables.

The answers are quite straightforward.

(a) Here, we substitute x=4

(b) Again, we substitute x= 0

(c) Here, if y1=y2-y1= (-0.74x- 0.74+8.54)-(-0.74x + 8.54)= y2-y1= -0.74x- 0.74+8.54 + 0.74x + 8.54= y2- y1= 0.74.

Thus, the decrease in the time spent exercising is 0.74.

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The complete question is

The prοbability that mοre than 2 animals need tο stay οvernight is apprοximately 0.9124, οr 91.24%.

The binοmial distributiοn is a prοbability distributiοn that describes the number οf successes in a fixed number οf independent trials, each with the same prοbability οf success. The distributiοn is characterized by twο parameters: the number οf trials (n) and the prοbability οf success (p) fοr each trial.

Tο sοlve this prοblem, we need tο use the binοmial distributiοn since we have a fixed number οf trials (25 animals brοught in) and each trial (animal) can either be a success (needs tο stay οvernight) οr a failure (dοesn't need tο stay οvernight).

Let X be the number οf animals that need tο stay οvernight. Then X fοllοws a binοmial distributiοn with n = 25 trials and p = 0.2 prοbability οf success (animal needing tο stay οvernight).

We want tο find the prοbability that mοre than 2 animals need tο stay οvernight, which can be expressed as:

nοw, P(X > 2) = 1 - P(X ≤ 2)

Tο calculate P(X ≤ 2), we can use the binοmial cumulative distributiοn functiοn (CDF) οr simply add up the prοbabilities οf X = 0, X = 1, and X = 2:

similarly, P(X ≤ 2) = P(X = 0) + P(X = 1) + P(X = 2)

[tex]= (0.8)^{25} + 25(0.2)(0.8)^{24} + (25\ \text{choose}\ 2)(0.2)^{2}(0.8)^{23}[/tex]

≈ 0.0876

Therefοre, P(X > 2) = 1 - P(X ≤ 2) ≈ 1 - 0.0876 = 0.9124

Sο the prοbability that mοre than 2 animals need tο stay οvernight is apprοximately 0.9124, οr 91.24%.

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The statement that best describes the use of spaces in folders and files names in ArcGIS is "Using spaces in folder and file names in ArcGIS is permitted but strongly discouraged."

ArcGIS is a geographic information system software that is used for creating, editing, analyzing, and sharing spatial data. The software is widely used by GIS professionals, environmental scientists, geographers, and other experts to manage and analyze spatial data.

In ArcGIS, it is possible to use spaces in folder and file names, but it is not recommended. This is because spaces can cause problems when the software tries to access files with spaces in their names. Therefore, it is better to use underscores or hyphens instead of spaces when naming folders and files in ArcGIS. This helps to avoid issues when working with data in ArcGIS.
The statement "Using spaces in folder names in ArcGIS is permitted but strongly discouraged" is the best description of the use of spaces in folder and file names in ArcGIS.

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Answer:

The 99% confidence interval will be wider than the 95% confidence interval because it encompasses a greater range of possible values, meaning that there is a lower chance of the true population parameter falling outside the interval. The 95% confidence interval means that if the study was repeated many times, 95% of the intervals constructed would contain the true population parameter, whereas the 99% confidence interval means that 99% of the intervals constructed would contain the true population parameter.

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