Find the length of the third side. If necessary, round to the nearest tenth. With 8 and 6

From the given table it is visible that the value of s(5)=4.2

When the variables and constants in a mathematical expression are given values, the outcome of the computation it describes is the expression's value. The amount that the function assumes for these argument values is the value of a function, given the value(s) assigned to its argument(s).

functions with return values and arguments. Functions with arguments and no return values.... This function takes arguments and returns a value.

functions with return values and no arguments.

without return values and without arguments

Function value can mean: The result of applying a function to an argument in mathematics. a closure in computer science.

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The number of solutions of the system y = 3x + 5 is one because it is a linear equation.

An equation in which is  the highest power of to the  variables 1 is knowns as the  linear equation. Mathematically: it is an  algebraic equations that can be also  written in the form of  ax + b = 0 or ax + by + c = 0, where a, b and c are definitely real numbers and x and y are variables with the highest power one.

Linear equations are equations in which the variables are raised to the power of 1, and the graph of the equation is a straight line. As a result, there is only one solution to the equation, which is when the line intersects the x-axis. This means that the equation has only one set of x- and y-coordinates, where x is the independent variable and y is the dependent variable, that satisfy the equation. The x-coordinate of the solution is the same as the value of x that makes the equation true and the y-coordinate is the value of y for this same x-value.

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The fairness of a passenger survey could be improved by oversampling data from the group of passengers who travel during peak ridership hours between 7:00 am and 5:00 pm.

This will ensure that the survey adequately represents the population of interest and provides accurate results.

Oversampling is the method of selecting respondents so that some groups make up a larger share of the survey sample than they do in the population. Oversampling small groups can be difficult and costly, but it lets polls to shed light on groups that would otherwise be too small to report on.

Oversampling and undersampling are methods used in data mining and data analytics to change unequal data classes to create balanced data sets. Oversampling and undersampling are also known as resampling.

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Answer:

te te te te te te te rog vreau să mă iai la un cont decent în privat de la școală dar trebuie să se poate mai bine tu nu mi-ai spus sa te sun se va întâmpla la un cont decent și a ride la un moment de la card memorie micro

Step-by-step explanation:

trebuie sa faci așa 4629+(703)-105

Answer:

cos(A + B) = - cosC

Step-by-step explanation:

using the addition identity

cos(x ± y) = cosxcosy ∓ sinxsiny

given

A + B + C = π ( subtract C from both sides )

A+ B = π - C

then

cos(A + B)

= cos(π - C)

= cosπ cosC + sinπ sinC

= - 1 × cosC + 0 × sinC

= - cosC + 0

= - cosC

Answer:

Step-by-step explanation:

(1) Yes, the table is proportional

4.5/25  = 5.4/30 = 6.3/35 = 7.2/40  

= 0.18

For a $50 bill the donation will be

50 X 0.18 = $9

Hence, 0.18 times of the dinner bill is donated to charity

(2) Yes, this is also proportional

a1 = 0.47

a2 = 0.68

a3 = 0.89

a4 = 1.10

d = a2 -a1 = a3 -a2 = a4 - a3

= 0.21

For a 6 ounce letter the A.P will be a6 so

a6 = a + 5d

= 0.47 + 5(0.21)

= 0.47 + 1.05

= 1.52

Hence , For every ounce of weight the cost gets increased by 0.21 the weight

The dimensions of the largest banner are 54 inches by 87.372 inches.

A dividable rectangle is referred to as a "golden rectangle." into shapes that are similar to the original square and rectangle. Repeated golden rectangles form a pattern. displayed to the right. Each produced golden rectangle is duplicated and divided once more. All of the golden rectangles are like the original rectangle in shape.

Given a golden rectangle,

width of rectangle = 54 inch

let length be l,

according to the golden ratio,

l/b = 1.618/1

l = 1.618b

l = 1.618*54

l = 87.372 inches

Hence the dimensions are 54 inches by 87.372 inches.

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Yes, the sampling distribution of the difference in sample mean ages is approximately normal.

This is because the Central Limit Theorem states that the sampling distribution of the mean of any sample of size n will be approximately normal, provided that the samples are drawn from a population with a finite variance. Furthermore, the sample sizes in this case (32 and 41) are both greater than 30, which is the minimum sample size needed for the CLT to be valid.

The Central Limit Theorem is a fundamental theorem of statistics that states that the sampling distribution of the means of any sample of size n drawn from a population with a finite variance will be approximately normal, regardless of the shape of the population’s probability distribution.

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After the second month, the charge will remain the same. 14 + 28 (2) = $70 would be the final price.

The first club offers a lower membership cost but a higher monthly fee in this instance, while the second club offers the polar opposite.

The computation would then be as follows to determine how long it would take for the total cost of each health club to equalize:

Example:

1 club fee equals 2 clubs fees

14+ 28x= 24+ 23x

28x-23x= 24-14

5x= 10 \sx=2

After the second month, the charge will remain the same. 14 + 28 (2) = $70 would be the final price.

The Complete Question.

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The relation between mode, mean and median is:

mode=3median-2mean

Now, to find mean:

2mean=3median-mode

mean=7/2 = 3.5

Let we wish to evaluate the mean  μ of a population. In real practice we would typically take just one sample. Imagine still that we take sample after sample, all of the same size n, and compute the sample mean x¯ every time.

The sample mean x is a random variable: it differs from sample to sample in a way that cannot be predicted with sureness. We will write X¯ when the sample mean is idea of as a random variable and write x for the values that it takes.

The random variable X¯ has a mean, denoted  μX¯, and a standard deviation, denoted  σX¯. Here is an example with such a small population and small sample size that we can indeed write down every single sample.

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The remaining fractions and percentages can be represented

as

0             1/4            2/4           3/4              1

|--------------|--------------|--------------|--------------|

0%          25%         50%           75%       100%

Real numbers are represented by a straight line known as a number line. The distance between the dots on the line, which correlates to the magnitude of the numbers, is how the numbers are typically represented. Positive or negative numerals can be used, and they are normally marked at regular intervals.

Given that,

The number line is divided into equal parts,

The missing parts are

0             1/4            2/4           3/4              1

|--------------|--------------|--------------|--------------|

0%          25%         50%           75%       100%

This is how we can fill in the remaining fractions and percentages.

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(a) P(X < 1470) = 0.8849, P(X >= 950) = 0.7487, (b) the probability that the size of a single droplet is between 950 and 1470 μm is 0.6336, (c) the smallest 2% of all droplets are those smaller than 827 μm in size, and (d) the probability that at least one exceeds 1470 μm is 0.3995.

(a) To find the probability that the size of a single droplet is less than 1470 μm or at least 950 μm, we need to use the standard normal distribution table. To do this, we first need to convert the droplet size to a standard normal variable.

Let X be the droplet size. Then we have:

Z = (X - 1050)/150

The probability that the size of a single droplet is less than 1470 μm is equal to the cumulative distribution function of Z evaluated at Z = (1470 - 1050)/150 = 1.2:

P(X < 1470) = P(Z < 1.2) = 0.8849 (rounded to four decimal places)

The probability that the size of a single droplet is at least 950 μm is equal to 1 minus the cumulative distribution function of Z evaluated at Z = (950 - 1050)/150 = -0.6:

P(X >= 950) = 1 - P(Z < -0.6) = 0.7487 (rounded to four decimal places)

(b) To find the probability that the size of a single droplet is between 950 and 1470 μm, we subtract the cumulative distribution function of Z evaluated at Z = (950 - 1050)/150 = -0.6 from the cumulative distribution function of Z evaluated at Z = (1470 - 1050)/150 = 1.2:

P(950 <= X <= 1470) = P(Z <= 1.2) - P(Z <= -0.6) = 0.8849 - 0.2513 = 0.6336 (rounded to four decimal places)

(c) To find the smallest 2% of all droplets, we find the value of Z such that P(Z <= Z) = 0.02. Using the standard normal distribution table, we find that Z = -2.33. Then, we convert Z back to the original variable X:

X = Z * 150 + 1050 = -2.33 * 150 + 1050 = 827 (rounded to two decimal places)

So, the smallest 2% of all droplets are those smaller than 827 μm in size.

(d) To find the probability that at least one of five independently selected droplets exceeds 1470 μm, we use the cumulative distribution function of a binomial distribution with n = 5 and p = P(X > 1470). The cumulative distribution function of a binomial distribution with parameters n and p is given by:

1 - (1 - p)^n

Using the standard normal distribution table, we find that P(X > 1470) = 1 - P(Z < (1470 - 1050)/150) = 1 - 0.8849 = 0.1151. Then, we plug in n = 5 and p = 0.1151 into the cumulative distribution function formula:

P(at least one exceeds 1470 μm) = 1 - (1 - 0.1151)^5 = 0.3995 (rounded to four decimal places).

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Answer: 630 rabbits

Step-by-step explanation: it shows that the population of rabbits grows by 70 each 4 years so we will have to add 70 to 560 which is 630 rabbits.

The population of rabbits in the year 2014 will be growing exponentially.

So, we can write a function

P = A · [tex]\alpha ^{x}[/tex]

where,

A is the original rabbits

[tex]\alpha[/tex] is the ratio of the increase

[tex]x[/tex] is the time (the unit is year)

In 2004, the population of the rabbits was 490

∴ P = 490 · [tex]\alpha ^{x}[/tex]

In 2008, the population increases to 560

∴ 560 = 490 · [tex]\alpha ^{2008-2004}[/tex]

   560 = 490 · [tex]\alpha ^{4}[/tex]

  [tex]\alpha ^{4\\[/tex] = [tex]\frac{560}{490}[/tex] = [tex]\frac{8}{7}[/tex]

  [tex]\alpha[/tex]   = [tex]\sqrt{\frac{2\sqrt{14} }{7} }[/tex]

In 2014, x = 2014- 2004 = 10

P = 490 · [tex]\sqrt{\frac{2\sqrt{14}}{7} } ^{10}[/tex]

  ≅ 684

Hence, the population of rabbits in the year 2014 will be approximately 684.

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The equation of a plane in 3-space with equation c1x + c2y + c3z + c4 = 0 can be determined by passing through three non-collinear points (x1, y1, z1), (x2, y2, z2), and (x3, y3, z3). This is given by the determinant equation:

Copy code

| x  y  z  1 |

| x1 y1 z1 1 | = 0.

| x2 y2 z2 1 |

| x3 y3 z3 1 |

However, if the three distinct points are collinear, meaning they lie on the same line, then the determinant equation becomes undefined as the determinant of a 3x3 or smaller matrix with linearly dependent rows or columns is equal to zero. This means that there is no plane that passes through three collinear points as a plane requires at least three non-collinear points to be defined.

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Step-by-step explanation:

"∠B is 2 more than 3 times the measure of ∠C" translates into B = 3C + 2....(1)

"...∠B and ∠C are complementary angles" translates to B + C = 90...............(2)

Now, all that's left to do is to solve for B and C. So, we'll take equation (1) and substitute it into equation (2) as follows:

(3C + 2) + C = 90

4C + 2 = 90

4C = 88

C = 22

Plugging C = 22 back into either equation (1) or (2) gives us

B = 90 - 22 = 68 (or B = 3(22) + 2 = 68)

Thus, ∠B = 68º; ∠C = 22º

The tight asymptotic bound for the function f(n) = 4f(n/2) + c is T(n) = O(n²).

A recursion tree is a graphical representation of a recursive function that shows the repeated calls of the function. The tree is constructed by expanding each call of the function into a separate node.

For the given function f(n) = 4f(n/2) + c, the recursion tree is attaced below:

f(n/4) f(n/4) f(n/4) f(n/4)

The height of the tree is log2(n), and at each level, there are 4 calls of the function. Hence, the total number of nodes in the tree would be [tex]4^{log2(n)} = n^2[/tex].

The tight asymptotic bound for the function can be found using the Master Theorem, which states that for a recurrence relation of the form T(n) = aT(n/b) + f(n), if f(n) = O([tex]n^c[/tex]) for some constant c, then T(n) = O([tex]n^d[/tex]), where d = max(c, log b(a)).

For this function, a = 4, b = 2, and c = 1. Hence, d = max(1, log2(4)) = max(1, 2) = 2. Thus, the tight asymptotic bound for the function is T(n) = O([tex]n^2[/tex]).

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Using property 8 and the midpoint rule, we have estimated the value of the integral ∫1147xdx to be approximately 14.

Property 8 of the definite integral states that if a < b and 0 < c < d, then:

∫baf(x)dx = ∫dcf(x)dx - ∫caf(x)dx

We can use this property to estimate the value of the integral ∫1147xdx. Let's choose c = 1 and d = 4, so that 0 < c < d and a < b. Then:

∫1147xdx = ∫447xdx - ∫11x dx

Next, we can use a numerical method, such as the midpoint rule or trapezoidal rule, to estimate the value of each of the integrals on the right side of the equation.

For example, using the midpoint rule with n = 4 subintervals for each integral:

∫447xdx ≈ (4 - 4/2) + (4 - 3/2) + (4 - 2/2) + (4 - 1/2) = 4 + 4 + 4 + 4 = 16

∫11x dx ≈ (1 - 1/2) + (1 - 2/2) = 1 + 1 = 2

So, our estimate of ∫1147xdx becomes:

∫1147xdx ≈ 16 - 2 = 14

Therefore, using property 8 and the midpoint rule, we have estimated the value of the integral ∫1147xdx to be approximately 14.

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Answer:

The answer would be 70.4

Answer:

Why not and what??

Step-by-step explanation:

wait why is that and also that not even a question

Answer: 31.4

Step-by-step explanation: the circumference is 31.4 because the radius is 5, you multiply by 2 to get a diameter that is 10, then multiply that by 3.14 because in this question pi=3.14.

The area of the semicircle is 3.14 square feet.

The formula for  area of a semicircle is A = ([tex]\pi[/tex]* r^2) / 2

Here A = Area

         r = Radius

Given:

Given that the radius of the semicircle is 2 feet.

We need to Find out the area of the semicircle.

The value of [tex]\pi[/tex] is 3.14

A = ([tex]\pi[/tex] * 2^2) / 2

Substituting π = 3.14 and r = 2, we get;

A = (3.14 * 4) / 2

A = 6.28 / 2

A = 3.14 square feet

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The resultant image of point A(-12,3) , after a scale factor 1/3 and a rotation of 270 degrees is (1, 4).

In Geometry, dilation simply refers to a type of transformation which typically changes the size of a geometric object with respect to a specific scale factor, but not its shape.

Next, we would have to dilate the coordinates of the pre-image by using a scale factor of 1/3 centered at the origin as follows:

Ordered pair A (-12, 3) → Ordered pair A' (-12 × 1/3, 3 × 1/3) = Ordered pair A' (-4, 1).

By applying a rotation of 270° about the origin to the given figure, the location of A" is given by:

(x, y)                                   →         (y, -x)

Ordered pair A' (-12, 3) → Ordered pair A'' = (1, -(-4) = (1, 4).

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Taylor's Line Best Fit does not accurately anticipate the relationship between crocodile body mass and bite force and has a tendency to overestimate the bite force.

An infinite sum of terms that are expressed in terms of the function's derivatives at a single point is the Taylor series, also known as the Taylor expansion, of a function. Near this point, the function and the sum of its Taylor series are equivalent for the majority of common functions.

P1(x) = (a)+ f'(x a) The function close to x = an is best approximated linearly by the tangent line. The curve and tangent line have the same first derivative, or slope, at the point where x = a. This close estimate is a Taylor polynomial of degree 1 is referred to as P1 (x).

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Answer:

the answer would be 75% of the journey

The common difference is 7/4 and the required terms are 16a,28a,49a

Any progression in which the ratio of adjacent terms is fixed is a Geometric Progression. The general form of representation of a geometric sequence is a, ar, ar2, ar3, ar4,...

Given here: The fifth, ninth, sixteenth terms of a linear sequence (a.p.) are consecutive terms of an expodencial sequence (g.p.)

5th term of the ap: a+4d

9th term: a+8d

16th term: a+15d

In a gp, the square of any term is equal to the product of the terms before and after it:

(a+8d)²=(a+4d).(a+15d)

[tex](a^2+16ad+64d^2)=a^2+19ad+60d^2\\4d^2-3ad=0\\d(4d-3a)=0[/tex]

(i) ANSWER: d = (3/4)a

(ii) 21st term: a+20d = a+15a = 16a

37th term: a+36d = a+27a = 28a

65th term: a+64d = a+48a = 49a

28a/16a = 7/4; 49a/28a = 7/4; this sequence is geometric with common ratio 7/4

Hence, The common difference is 7/4 and the required terms are 16a,28a,49a

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The translation needed to obtain the graph of g(x) = 9^{x - 2} from the graph of f(x) = 9^x is shift 2 units right.

In Mathematics, a horizontal translation to the right is modeled by this mathematical expression g(x) = f(x + N) while a vertical translation to the negative y-direction (downward) is modeled by this mathematical expression g(x) = f(x) + N.

Where:

By translating the exponential function f(x) = 9^x two (2) units to the right, we have the following:

f(x) = 9^x   →  g(x) = f(x - (2))

g(x) = 9^{x - 2}.

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Jon bought 10.5 boxes of cards and Felipe bought 11 boxes of cards, and Jon and Felipe each spent $215 on business cards.

What is the system of equations?

A system of linear equations is a set of two or more equations that includes common variables. To solve a system of equations, we must find the value of the unknown variables used in the equations that must satisfy both equations.

Let x be the number of boxes of cards that Jon bought and y be the number of boxes of cards that Felipe bought.

The equation for Jon's cost is: 10 + 20x = y * (21 + 9)

The equation for Felipe's cost is: 21y + 9 = 10 + 20x

We will solve the system of equations using substitution.

Starting with the second equation, we can simplify: 21y + 9 = 10 + 20x

21y + 9 - 10 = 20x

21y - 1 = 20x

Dividing both sides by 21: y = (20/21)x + 1

We can now substitute this expression for y into the first equation:

10 + 20x = (20/21)x + 10 + 9

10 + 20x = (20/21)x + 19

10 = 19 - (20/21)x

Multiplying both sides by 21:

210 = 19 * 21 - 20x

Subtracting 19 * 21 from both sides:

210 - 19 * 21 = -20x

Divide both sides by -20:

x = 10.5

So, Jon bought 10.5 boxes of cards and Felipe bought (20/21) * 10.5 + 1 = 11 boxes of cards.

To find out how much each spent, we can plug these values back into either equation:

Jon's cost: 10 + 20 * 10.5 = 215

Felipe's cost: 21 * 11 + 9 = 250

So, Jon and Felipe each spent $215 on business cards, buying 10.5 boxes.

Hence, Jon bought 10.5 boxes of cards and Felipe bought 11 boxes of cards, and Jon and Felipe each spent $215 on business cards.

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The two inequalities that represent the constraints in this situation are

r + c >= 80 and 1.45r + 0.65c <= 200

Let r be the number of roses and c be the number of carnations Victor buys.

So, we have

Total number of flowers: r + c >= 80

Victor has $200 to spend on flowers for the school celebration.

Roses cost $1.45 each and carnations cost $0.65 each.

So, we have

Total spending: 1.45r + 0.65c <= 200

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Answer:

m<4 = 100

Step-by-step explanation:

Because 1 is opposite to 4 they equal the same degree

Answer:

10%

Step-by-step explanation:

11.6/11.1 = 0,1045045