Given this equation...

4x - 1 6
--------- = -----
3x + 7 5

What is the value of x in this proportion?

The complete statement is (d) Any rational root of f(x) is a factor of 9 divided by a factor of 12.

The polynomial is given as:

12x^3 - 5x^2 + 6x + 9

The first term in the above equation is 12 i.e. 12x^3, while the last term is 9

To determine the possible rational roots, we divide the factors of the last term i.e. 12 by the factors of the first term i.e. 12

Hence, the complete statement is (d) Any rational root of f(x) is a factor of 9 divided by a factor of 12.

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Answer:

D

Step-by-step explanation:

I took the test

The time needed for the investment to become $10,413 is 54.7 years

Number of years = (In FV / PV) / r  

r = 4.252/12 = 0.354%

(In 10, 413 / 1,018.40) / 0.00354 = 656.73 months = 54.7 years

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Answer:

Please see picture below.

Step-by-step explanation:

Answer:

Step-by-step explanation:

Use the mean and the standard deviation obtained from the last discussion and test the claim that the mean age of all books in the library is greater than 2005.

This is the last discussion:

The science portion of my library has roughly 400 books.

They are arranged, on shelves, in order of their Library of Congress

code and, within equal codes, by alphabetical order of author.

Sections Q and QA have a total of 108 books.

The bulk of the library was established in the early 1990s.

I used a deck of cards, removing the jokers and face cards, keeping

only aces (1) and "non-paints" (2 to 10). Starting from the start of

the first shelf, I would turn a card (revealing its number N) and I

would go to the Nth book. This uses ordinal numbers (1 would means the

"first" book, not a gap of 1).

The cards were shuffled sufficiently to assume that the cards have a

random order.

I sampled only the LC subsections Q and QA (therefore, not a true

sample of the entire library, as this classification is not purely

random).

Thus, I picked 21 books (the expected number being 108/5.5 = 19.6 --> 20 books)

The sampled dates of publication were (presented as an ordered set):

1967, 1968, 1969, 1975, 1979, 1983, 1984,

1984, 1985, 1989, 1990, 1990, 1991, 1991,

1991, 1991, 1992, 1992, 1992, 1997, 1999

The median date is 1990

The mean date is 1985.67

Variance = 84.93 ( Sum of (date-mean)^2 )

SQRT of variance = 9.2 (sample standard deviation)

The "sigma-one" confidence interval (containing 68% of the books), if

the sample were NOT skewed, and if the distribution were "normal"

would contain dates from "mean - 9.2" to "mean + 9.2"

Sigma-2 (95%) would have mean - 2(9.2) to mean + 2(9.2)

Sigma-3(99.7%) from "1985.67 - 3(9.2)" to "1985.67 + 3(9.2)"

1958.07 to 2012.6

There, you see one reason why the distribution is skewed (it is

possible to have books older than 1958, but it is impossible to have

book newer than 2012), but it still represents a usable model for the

library. If it applies to the entire library of 400 books, you would

expect one book (0.3% of 400) to fall outside the 3-sigma interval.

Answer:

2.4536 years

Step-by-step explanation:

compound formula:

A=P*(1+r/n)^nt

A = 6000

P = 5000

r = rate

n = 4 times per year compounded

t = time in years

6000 = 5000(1 + .075/4)^4t

divide both sides by 5000

1.2 = (1.01875)^4t

log both sides

log(1.2) = 4t * log(1.01875)

solve for t

t = 2.4537 years

115.2 pounds of type A coffee were used.

A mathematical statement that has an "equal to" symbol between two expressions with equal values is called an equation.

Equations are mathematical statements with two algebraic expressions flanking the equals (=) sign on either side. It demonstrates the equality of the relationship between the expressions printed on the left and right sides. We have LHS = RHS (left hand side = right hand side) in every mathematical equation. To determine the value of an unknown variable that represents an unknown quantity, equations can be solved.

Let the number of pounds of type A coffee be x and the number of pounds of Type B coffee be y.

According to the question, the equations are,

5.50x + 4.40y = 762.30

x = 4y

So, the solution of the equation is obtained as follows:

5.50(4y) + 4.40y = 762.30

26.40y = 762.30

y = 762.30/26.40

y = 28.8 pounds

x = 4*28.8 = 115.2 pounds

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Considering the function for the net worth of the function, it is found that it will not be in debt for the first time after 6 months in month 14, at point (14,0), which is plotted on the graph.

A company is said to be in debt if it's net worth is negative. On a graph, it is represented by the curve being below the x-axis.

From month 2 until the end of month 13, the company is in debt, hence it will not be in debt for the first time after 6 months in month 14, at point (14,0), which is plotted on the graph.

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Answer:

13

Step-by-step explanation:

7+2x=3x-6

knowing this then you solve it

Step-by-step explanation:

7 more = + 7

twice a number = 2*x = 2x

6 less = -6

3 times the same number = 3 * x = 3x

Putting this together we get:

2x + 7 = 3x - 6

subtract two from both sides to isolate x

(-2x) + 2x + 7 = 3x - 6 (-2x)

7 = x - 6

add 6 to both sides to get your answer

7 (+6) = x - 6 (+6)

13 = x

or

x = 13

From the calculation, we can see that /BC/ is 8 sin 35. Option D

We can see the image of the triangular piece of cloth as shown in the image. This shows that we have to approach the problem by the use of the trigonometric ratios.

Hence;

/AC/ = 8 in

/BC/ = x

<A = 35 degrees

Sin 35 = /BC//8

/BC/ = 8 sin 35

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The number of female contestants who signed up to compete is; 43.

Let number of females = x

Let number of males = y.

Hence, it follows from the task content that;

x+y = 50

x-y = 36.

Hence, by adding both equations together in a bid to determine the number of females; we have;

2x = 86

x = 43.

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Answer:

0.2135

Step-by-step explanation:

When something asks to find the product, it indicates multiplication.

We are given two decimals which are: 3.05 and 0.07.

We are asked to multiply and find the product of these decimals.

Let's use the algorithm method.

                   3  .    0   5

x                 0   .    0   7

_____________________

                  2        3

          0 .    2      1    3    5

Therefore, 3.05 × 0.07 = 0.2135.

Answer:

hello :

f(x)=x/4-3 and g(x)= 4x²+2x-4,

(f+g)(x).=f(x)+g(x)=x/4-3  + 4x²+2x-4,

(f+g)(x).= 4x²+(2x+x/4)-7

(f+g)(x).= 4x²+x(3+1/4) - 7......continu

Answer:

The answer to your question is p = 10.50b

Step-by-step explanation:

Since you want to find p, you're looking for an equation that has

p = <something>

The last selection has this form

p = 10.50b

I hope this helps and have a good day!

Answer:

a = 4 first term, r = ?, T10 = 100.Tn = ar^n - 1 formula for g.pT10 = 4r^ 10 - 1100 = 4r^9Divide both side by 4100/4 = 4/4r^925 = r^9 take the 9th root of both side 9√25 = 9√r^9r = 9√25To find the nth term Since Tn = ar^n - 1

Please see the graph below

Answer:

y = - 7x + 7

Step-by-step explanation:

the equation of a line in slope- intercept form is

y = mx + c ( m is the slope and c the y- intercept )

given

7x + y = 7 ( subtract 7x from both sides )

y = - 7x + 7 ← in slope- intercept form

The equivalent expression of [tex]1 - \frac{1}{x} \div 2[/tex] is [tex]\frac{2x - 1}{2x}[/tex]

The expression is given as:

1 minus StartFraction 1 Over x EndFraction divided by 2

Rewrite properly as:

[tex]1 - \frac{1}{x} \div 2[/tex]

Express the division as product

[tex]1 - \frac{1}{x} \times \frac 12[/tex]

Evaluate the product

[tex]1 - \frac{1}{2x}[/tex]

Take the LCM

[tex]\frac{2x - 1}{2x}[/tex]

Hence, the equivalent expression of [tex]1 - \frac{1}{x} \div 2[/tex] is [tex]\frac{2x - 1}{2x}[/tex]

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The equation of the asymptote is y = 3x - 5. The correct answer is option C

The distance from a point and the distance to a line in hyperbola is known as asymptote. The general equation is [tex]x^{2}/ a^{2} - y^{2}/b^{2} = 1[/tex]

From the given equation of hyperbola, which is

[tex]\frac{(x - 2)^{2}}{4}[/tex]  - [tex]\frac{(y - 1)^{2} }{36}[/tex] = 1

The center (h , k) of the hyperbola = C(2, 1)

a = 2

b = 6

Where C = [tex]\sqrt{a^{2} + b^{2} }[/tex]

C = [tex]\sqrt{4 + 36}[/tex]

C = [tex]\sqrt{40}[/tex]

C = [tex]2\sqrt{10}[/tex]

The equation of the asymptote will be y - K = +/-(b/a)(x - h)

That is,

y - 1 = +/-(6/2)(x - 2)

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

y - 1 = +/-3x - 6

y = +/-3x - 6 + 1

y = +/- 3x - 5

Therefore, the equation of the asymptote is y = 3x - 5.

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Answer:

Vertex (2 , -25)

Axis of symmetry: x = 2

Step-by-step explanation:

 The vertex is the highest point if the parabola open downwards and the lowest point if the parabola opens upward.

 f(x) = (x - 2)² - 25

The given quadratic function  is in vertex form.

 f(x)  = a(x - h)² + k

Here,  (h , k) is the vertex of the parabola.

 h = 2  ; k = -25

[tex]\sf \boxed{\bf Vertex (2 , -25)}[/tex]

 The axis of symmetry is the vertical line that divides the parabola into two equal halves and it passes through the vertex of the parabola.

Axis of symmetry: x = h

  [tex]\sf \boxed{\bf x = 2}[/tex]

Answer:

For an exact answer the width would be the square root of 1011 and the length would be 1.5x the square root of 1011.

Step-by-step explanation:

A = lw

1516.5 = w(1.5)w

1516.5 = 1.5w^2  Divide both sides by 1.5

1011 = w^2  Take the square root of each side

Square root of 1011 = w.

A piece-wise function that has domain of all real numbers and range of [tex][-3, \infty)[/tex] is f(x) = |x| - 3, as:

[tex]|x| - 3 = x - 3, x \geq 0[/tex]

[tex]|x| - 3 = -x - 3, x < 0[/tex]

A piece-wise function is a function that has different definitions, depending on the input.

One example is the absolute value function, defined as follows:

[tex]|x| = x, x \geq 0[/tex]

[tex]|x| = -x, x < 0[/tex]

The function is defined for all real values, with a range of [tex][0, \infty)[/tex]. Shifting it down 3 units, it will have the desired range, as follows:

[tex]|x| - 3 = x - 3, x \geq 0[/tex]

[tex]|x| - 3 = -x - 3, x < 0[/tex]

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The quantity of rainfall that fell in the second half of the month is; ¹/₆ of an inch.

We are given;

Rainfall during first half of the month = 1/2 inch

Total rainfall for the month = 2/3 inch

Now, to find the rainfall for the second half of the month, we just subtract the one for the first half from the one for the month to get;

Rainfall in second half of the month = ²/₃ - ¹/₂ = ¹/₆ inches

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A statement which could be true for g is that: A. g(-13) = 20.

A domain can be defined as the set of all real numbers for which a particular function is completely defined.

Since the domain of this function is given by -20 ≤ x ≤ 5, it simply means that the value of x must between -20 and 5. Also, with a range of -5 ≤ g(x) ≤ 45, the value of x must between -5 and 45.

By extrapolating the function, we can deduce that:

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Complete Question:

Given that a function, g, has a domain of -20 ≤ x ≤ 5 and a range of -5 ≤ g(x) ≤ 45 and that g(0) = -2 and g(-9) = 6, select the statement that could be true for g.

A. g(-13) = 20

B. g(-4) = -11

C. g(7) = -1

D.g(0) = 2

Answer:

Step-by-step explanation:

Okay, so first of all, when a slope is negative, the line goes down from left to right, not up from left to right. Then, plot the Y intercept as -2. Finally, either go up 2 and left 3 or down 2 and right 3.

Answer:

Graph attached.

Step-by-step explanation:

As we have been given the slope and the y-intercept, we can create a linear equation using the slope-intercept form.

Slope-intercept form of a linear equation:

[tex]y=mx+b[/tex]

where:

Substitute the given values into the formula to create an equation for the line:

[tex]\implies y=-\dfrac{2}{3}x-2[/tex]

Now we have an equation for the line, find at least two points on the line by inputting values of x into the found equation:

[tex]x=0 \implies y=-\dfrac{2}{3}(0)-2=-2 \implies (0,-2)[/tex]

[tex]x=3 \implies y=-\dfrac{2}{3}(3)-2=-4 \implies (3,-4)[/tex]

[tex]x=-6 \implies y=-\dfrac{2}{3}(-6)-2=2 \implies (-6,2)[/tex]

Plot the found points and draw a straight line through them (see attachment).

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Answer:

[tex](f \cdot g)(x)[/tex] = [tex]2x^2 -1[/tex]

Step-by-step explanation:

[tex](f \cdot g)(x)[/tex], read as "f of g of x", means we have to use the function [tex]\bf g(x)[/tex] as input for the function [tex]\bf f(x)[/tex]. This means, we have to replace the [tex]x[/tex] in the definition of  [tex]f(x)[/tex] with [tex]g(x)[/tex].

[tex]f(x)=x+3[/tex]

[tex]g(x) = 2x^2 - 4[/tex]

∴ [tex](f \cdot g)(x) = f(g(x))[/tex]

                 ⇒ [tex]g(x) + 3[/tex]

                 ⇒ [tex](2x^2 - 4) + 3[/tex]

                 ⇒ [tex]2x^2 -1[/tex]

Using the Poisson distribution, there is a 0.9978 = 99.78% probability that there will be 4 or more customers at this bank in one hour.

In a Poisson distribution, the probability that X represents the number of successes of a random variable is given by:

[tex]P(X = x) = \frac{e^{-\mu}\mu^{x}}{(x)!}[/tex]

The parameters are:

A bank gets an average of 12 customers per hour, hence the mean is [tex]\mu = 12[/tex].

The probability that there will be 4 or more customers at this bank in one hour is:

[tex]P(X \geq 4) = 1 - P(X < 4)[/tex]

In which:

P(X < 4) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3)

Then:

[tex]P(X = x) = \frac{e^{-\mu}\mu^{x}}{(x)!}[/tex]

[tex]P(X = 0) = \frac{e^{-12}12^{0}}{(0)!} \approx 0[/tex]

[tex]P(X = 1) = \frac{e^{-12}12^{1}}{(1)!} \approx 0[/tex]

[tex]P(X = 2) = \frac{e^{-12}12^{2}}{(2)!} = 0.0004[/tex]

[tex]P(X = 3) = \frac{e^{-12}12^{3}}{(3)!} = 0.0018[/tex]

Then:

P(X < 4) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) = 0 + 0 + 0.0004 + 0.0018 = 0.0022.

[tex]P(X \geq 4) = 1 - P(X < 4) = 1 - 0.0022 = 0.9978[/tex]

0.9978 = 99.78% probability that there will be 4 or more customers at this bank in one hour.

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The point-slope equation of the line is: y - 2 = 8(x - 2).

The point-slope equation is: y - b = m(x - a), where:

Using the coordinates of the point shown in the graph, substitute (a, b) = (2, 2) and m = 8 into y - b = m(x - a):

y - 2 = 8(x - 2)

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Data collected about a population is not a characteristic of both observational studies and experiments.

Through experiments, two variables' cause and effect relationships are examined. This is where they differ from observations and interviews, which can only assume the existence of contexts and cannot provide evidence for them. The environment of the test subjects is managed in an experiment depending on the issue.

Three things characterize statistical experiments in general:

There are various outcomes that the experiment could produce.

In an experiment, the cause, the independent variable, is changed while the effect, the dependent variable, is measured and any unrelated factors are controlled.

It is possible to anticipate every outcome.

Chance determines how the experiment turns out.

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