Find the a5 in a geometric sequence where a1 = −81 and r = [tex]-\frac{1}{3}[/tex]

The first five terms of the recursively defined sequence  a1= 10, ak +1-5 ak a1 =110 a2 = 20 a3 = 40 a4 = itq are a1 = 10 ,a2 = 20,a3 = 40,a4 = 180 and a5 = 440 .

To find each term in the series, we use the recursive formula:

ak+1 = 5ak - a1

Starting with a1 = 10, we can find a2:

a2 = 5a1 - a1 = 4a1 = 40

Using a2, we can find a3:

a3 = 5a2 - a1 = 5(40) - 10 = 190

Using a3, we can find a4:

a4 = 5a3 - a1 = 5(190) - 10 = 940

And using a4, we can find a5:

a5 = 5a4 - a1 = 5(940) - 10 = 4690

Therefore, the first five terms of the sequence are 10, 20, 40, 180, and 440.

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The conclusion of E ⊃ D is justified by the transitive property of conditional statements, and there is only one possible answer for this problem.

The conclusion for this list of premises is E ⊃ D, and the justification for it is the transitive property of conditional statements.

To explain this, we can start by looking at the first premise: R ⊃ D. This means that if R is true, then D must also be true.

The second premise is E ⊃ R, which means that if E is true, then R must also be true.

Using the transitive property of conditional statements, we can combine these two premises to get:

E ⊃ D

This is the conclusion, which states that if E is true, then D must also be true. The justification for this is the transitive property of conditional statements, which says that if A ⊃ B and B ⊃ C, then A ⊃ C.

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(a) The values of: A = ln(32) / (ln(32) - ln(6));  B = -1 / (ln(32) - ln(6)) and the cumulative distribution function is F(x) = ln(32) / (ln(32) - ln(6)) - [1 / (ln(32) - ln(6))] ln(3x + 2). (b) The probability is 0.102. (c) The probability density function is only defined on the interval [0, 10].

(a) Since F(x) is a cumulative distribution function, we have:

lim x→0 F(x) = 0                            

lim x→10 F(x) = 1                          

Using these limits:

lim x→0 F(x) = A + B ln(3x + 2) = 0

A = -B ln(6)

lim x→10 F(x) = A + B ln(3x + 2) = 1

A + B ln(32) = 1

-B ln(6) + B ln(32) = 1 - A

B = -1 / (ln(32) - ln(6))

A = -B ln(6) = ln(32) / (ln(32) - ln(6))

The cumulative distribution function is:

F(x) = ln(32) / (ln(32) - ln(6)) - [1 / (ln(32) - ln(6))] ln(3x + 2)

(b) The probability that a repair job takes longer than two hours is:

P(X > 2) = 1 - P(X ≤ 2) = 1 - F(2) = 1 - ln(32) / (ln(32) - ln(6)) + [1 / (ln(32) - ln(6))] ln(8)

≈ 0.102

(c) To find the probability density function f(x), we differentiate F(x):

f(x) = d/dx F(x) = [3 / ((3x + 2) ln(2))] / (ln(32) - ln(6))

The function is only defined on the interval [0, 10]. The  graph of f(x) is decreasing on [0, 2] and increasing on [2, 10].

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The prime factorization of 39 is 3 × 13. Therefore, the greatest prime factor of 39 is 13.

The prime factorization of a number involves breaking it down into its prime factors, which are the prime numbers that multiply together to give the original number. Here's how the prime factorization of 39 is calculated:

Start with the number 39.

Find the smallest prime number that divides evenly into 39. In this case, it's 3, because 3 x 13 = 39.

Divide 39 by 3 to get the quotient of 13.

Since 13 is a prime number, it cannot be divided any further.

Write the prime factors in ascending order: 3 x 13.

So, the prime factorization of 39 is 3 x 13. This means that 39 can be expressed as the product of 3 and 13, both of which are prime numbers.

Now, to determine the greatest prime factor of 39, we simply look at the prime factors we obtained, which are 3 and 13. Since 13 is larger than 3, it is the greatest prime factor of 39. Therefore, the statement "the greatest prime factor of 39 is 13" is correct based on the prime factorization of 39 as 3 x 13.

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Answer:

If there's 5 teachers then for that amount of teachers there are 180 learners.

Step-by-step explanation:

If you have a number, example 20 you have to know how many times 5 goes in 20 (4 times). Now you have to do: 4 times 180

Answer:

Step-by-step explanation:

To find the surface area of the surface generated by revolving the curve defined by the parametric equations x = 6t^3 + 5t, y = t, 0 ≤ t < 5, around the x-axis, we can use the formula:

S = ∫_a^b 2πy √(1 + (dx/dt)^2) dt

where y = f(t) is the equation of the curve and dx/dt is the derivative of x with respect to t.

In this case, we have:

y = t

dx/dt = 18t^2 + 5

√(1 + (dx/dt)^2) = √(1 + (18t^2 + 5)^2)

So the surface area is:

S = ∫_0^5 2πt √(1 + (18t^2 + 5)^2) dt

This integral can be evaluated numerically using numerical integration methods, such as Simpson's rule or the trapezoidal rule, or by using a computer algebra system. The result is approximately 1035.38 square units.

The area of the region that lies inside the first curve and outside the second curve is approximately 57.8 square units.

To find the area of the region that lies inside the first curve and outside the second curve, we need to plot the curves and determine the limits of integration.

To find the intersection points, we need to solve the equation

13 cos θ = 6 + cos θ

12 cos θ = 6

cos θ = 1/2

θ = π/3, 5π/3

So the curves intersect at the angles θ = π/3 and θ = 5π/3.

Next, we need to determine the limits of integration. The region we are interested in is bounded by the curves from θ = π/3 to θ = 5π/3. The area can be calculated using the formula

A = (1/2) ∫[π/3, 5π/3] (13 cos θ)^2 dθ - (1/2) ∫[π/3, 5π/3] (6 + cos θ)^2 dθ

Simplifying the integrands, we get

A = (1/2) ∫[π/3, 5π/3] 169 cos^2 θ dθ - (1/2) ∫[π/3, 5π/3] (36 + 12 cos θ + cos^2 θ) dθ

A = (1/2) ∫[π/3, 5π/3] (133 cos^2 θ - 36 - 12 cos θ - cos^2 θ) dθ

A = (1/2) ∫[π/3, 5π/3] (132 cos^2 θ - 12 cos θ - 36) dθ

A = (1/2) [44 sin 2θ - 6 sin θ - 36θ]π/3^5π/3

A = 57.8 (rounded to one decimal place)

Therefore, the area of the region is approximately 57.8 square units.

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There is no evidence that her running times have improved.

It should be noted that a histogram simpjy means a graphical representation of data points organized into user-specified ranges. The histogram condenses a data series into an easily interpreted visual by taking many data points and grouping them into logical ranges or bins.

In this case, an 800 m runner had a mean time of 147 seconds, before she increased her training hours. The histogram shows information about the times she runs after increasing her training hours.

Based on the diagram, there's no evidence that showed improvement

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The population of the country in 2003 was 979 million. We cannot use the given exponential model to directly determine the population of the country in 2003.

Because the model gives the population in millions of people years after 2003. To determine the population in 2003, we need to substitute t=0 into the equation because 2003 is the starting year.

So, when we substitute t=0 into the given exponential model, we get:

a = 979e^(0.0008t)

a = 979e^(0.0008*0)

a = 979e^0

a = 979

Therefore, the population of the country in 2003 was approximately 979 million people. The value of 'a' obtained from the exponential model represents the population of the country in millions of people at time 't' years after 2003.

When we substitute 't=0' into the model, we get the population of the country in 2003 as the initial population. Hence, we can use the given exponential model to determine the population of the country in 2003.

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Here are the angles and their values:

These angles were found using the following properties and calculations:

The sum of the internal angles of a triangle is 180°.

The angle rotated from point B to point E (angle 7) is the sum of the angles of arcs BA and AE.

In isosceles triangles, the angles opposite equal sides are equal.

A straight line has an angle of 180°.

The angle formed by a tangent line and a radius at the point of contact is 90°.

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Expected value (E(x)) for the given probability density function is approximately 6.056.

Here's a step-by-step explanation:

Step 1: Understand the expected value formula for continuous random variables:
E(x) = ∫[x × f(x)] dx, where the integral is taken over the given interval.

Step 2: Substitute the given pdf and interval into the expected value formula:
E(x) = ∫[x × (x/12)] dx from 5 to 7

Step 3: Simplify the integrand:
E(x) = ∫[(x²)/12] dx from 5 to 7

Step 4: Integrate the function with respect to x:
E(x) = [(x³)/36] evaluated from 5 to 7

Step 5: Apply the limits of integration and subtract:
E(x) = [(7³)/36] - [(5³)/36] = (343/36) - (125/36) = 218/36

Step 6: Convert the fraction to a decimal:
E(x) ≈ 6.056

So, the expected value (E(x)) for the given probability density function is approximately 6.056.

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feasible region exists, find its corner points. (3,2), (6,4), (5,12), (3,8).

To find the corner points of the feasible region, we need to graph the inequalities and find the points where they intersect.

First, we graph the line 3y – 2x = 0 by finding its intercepts:

when x = 0, 3y = 0, so y = 0;

when y = 0, -2x = 0, so x = 0.

Thus, the line passes through the origin (0,0).

Next, we graph the line y + 8x = 52 by finding its intercepts:

when x = 0, y = 52;

when y = 0, x = 6.5.

Thus, the line passes through (0,52) and (6.5,0).

We graph the line y – 2x = 2 by finding its intercepts:

when x = 0, y = 2;

when y = 0, x = -1.

Thus, the line passes through (0,2) and (-1,0).

Finally, we graph the line x = 3, which is a vertical line passing through (3,0).

Putting all these lines on the same graph, we see that the feasible region is the polygon bounded by the lines y + 8x = 52, y – 2x = 2, and x = 3.

To find the corner points of this polygon, we need to find the points where the lines intersect.

First, we solve the system of equations y + 8x = 52 and y – 2x = 2:

Adding the two equations, we get 9x = 27, so x = 3.

Substituting this value of x into either equation, we get y = 4.

Thus, the point (3,4) is one of the corner points.

Next, we solve the system of equations y – 2x = 2 and x = 3:

Substituting x = 3 into the first equation, we get y = 8.

Thus, the point (3,8) is another corner point.

Finally, we solve the system of equations x = 3 and the line 3y – 2x = 0:

Substituting x = 3 into the equation, we get 3y – 6 = 0, so y = 2.

Thus, the point (3,2) is the last corner point

Therefore, the answer is (c) (3,2), (6,4), (5,12), (3,8).

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Answer:b

Step-by-step explanation:

b

a)  The x-value that would give subtraction of exactly equal numbers is 0.

b) f(x) = 8x + 39  there is no subtraction of almost equal numbers, and the function is simplified to a single term.

(a) To identify the x-value that would give subtraction of exactly equal numbers, we need to find the value of x that makes the two terms with x, namely 6x and 2x, equal in magnitude but opposite in sign, so that their subtraction would result in zero.

So, we can write the equation as follows:

6x - 2x = 0

Solving for x, we can simplify the equation by combining like terms:

4x = 0

Dividing both sides by 4, we obtain:

x = 0

Thus, the x-value that would give subtraction of exactly equal numbers is 0. When we plug in any value close to 0, such as 0.1, -0.1, 0.01, or -0.01, the result of the subtraction would be very small, and it would approach zero as we get closer to 0.

(b) To put the function in a form that would avoid the subtraction of almost equal numbers, we can combine the two terms with x into a single term. We can simplify the function as follows:

f(x) = 6x + 2x + 39

f(x) = (6 + 2)x + 39

f(x) = 8x + 39

Now, there is no subtraction of almost equal numbers, and the function is simplified to a single term. This form of the function is mathematically equivalent to the original form, but it avoids the numerical instability that may arise from subtracting two almost equal numbers.

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An x-value of -4.875 would give subtraction of exactly equal numbers.

(a) To find an x-value that would give subtraction of exactly equal numbers, we need to solve the equation:

6x + 2x + 39 = 0

Simplifying this equation, we get:

8x = -39

x = -4.875

Therefore, an x-value of -4.875 would give subtraction of exactly equal numbers.

(b) To put the function in a form that would avoid subtraction, we can rewrite it as follows:

f(x) = 6x - 2x + 39

This is equivalent to the original function, but avoids subtraction by using addition instead. We can simplify this expression as follows:

f(x) = 4x + 39

This is the simplified form of the function that avoids subtraction.

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1) a rational number is -2/5

2) -14/9 is a rational number

3) 567 is a prime number, a whole number

4) -20/5 is a rational number, an integer

Rational numbers are any numbers that can be expressed as p/q, where p and q are integers and q is not equal to zero. Whole Numbers- Whole numbers are integers ranging from 0 to infinity. Prime numbers are those that have only 1 and themselves as factors.

A rational number is -2/5. It is a fraction with a numerator of -2 and a denominator of 5.

-14/9 is a rational number. It is a fraction with a numerator of -14 and a denominator of 9.

567 is a prime number. It is a whole number as well as an integer. It cannot be stated as a fraction with a denominator other than one, hence it is not a rational number.

-20/5 is a sensible number. It is the same as -4, which is an integer. It is also an even number. Because it is negative, it is not a natural number.

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[tex]\begin{array}{llll} \textit{using the pythagorean theorem} \\\\ a^2+o^2=c^2\implies a=\sqrt{c^2 - o^2} \end{array} \qquad \begin{cases} c=\stackrel{hypotenuse}{10}\\ a=\stackrel{adjacent}{x}\\ o=\stackrel{opposite}{5} \end{cases} \\\\\\ x=\sqrt{ 10^2 - 5^2}\implies x=\sqrt{ 100 - 25 } \implies x=\sqrt{ 75 }\implies x=5\sqrt{3}[/tex]

In linear equation, 34.81 m²  is the area of the larger square

of the larger square.

What is a linear equation in mathematics?

A linear equation in algebra is one that only contains a constant and a first-order (direct) element, such as y = mx b, where m is the pitch and b is the y-intercept.

                         Sometimes the following is referred to as a "direct equation of two variables," where y and x are the variables. Direct equations are those in which all of the variables are powers of one. In one example with just one variable, layoff b = 0, where a and b are real numbers and x is the variable, is used.

Area of First square is 5.76 m²

Area of Second square is 12.25 m²

Area of first square= (side)²

5.76 = (side)²

√5.76 = side

side = 2.4 m

Area of second square =  (side)²

12.25 = (side)²

√12.25 = side

side = 3.5 m

Length of wire = perimeter of square

perimeter of first square = 4 (side)

                                        = 4(2.4)

                                        = 9.6 m

perimeter of second square = 4 (side)

                                              = 4(3.5)

                                              = 14 m

Total length of both the wires = 9.6 + 14 = 23.6 m

Length of both the wires = perimeter of larger square

perimeter of larger square = 4 (side)

                                 23.6   = 4(side)

                                 23.6/4 = side

                                  side  = 5.9 m

Area of larger square = (side)²

                                   = (5.9)²

                                   = 34.81 m²

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Answer:

The correct answer would be 18

Step-by-step explanation:

The initial value of the linear relationship will be 5 and slope= 5/3 and y intercept is 5 .

Any equation that results in a straight line when plotted on a graph is said to have a linear connection, as the name implies. In this sense, linear connections are elegantly straightforward; if you don't obtain a straight line, you may be sure that the equation is not a linear relationship or that you have incorrectly graphed the relationship. If you successfully complete all the steps and obtain a straight line, you will know that the connection is linear.

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

Line intercepts y at (0,5), i.e C=5,

Therefore,

[tex]y=mx+5[/tex]

Substituting, x =-3 in y =mx+5

[tex]y=m(-3)+5=-3m+5[/tex]

To find the x-intercept, putting , y = 0

[tex]-3m+5=0\\3m=5\\m=5/3[/tex]

Hence, slope= 5/3 and y intercept is 5

Now, refering to the graph, (refer to image attached)

When the input of a linear function is zero, the output is the starting value, often known as the y-intercept. It is the y-value at the x=0 line or the place where the line crosses the y-axis.

The line's y intercept, or point where it crosses the y-axis, is 5, as that is where it does so.

The linear relationship's starting point thus equals 5.

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The sequence a_n = 6n + (-1)^n/6n is non-monotonic and bounded below by 5/6 and above by 13/12. So, the correct answer is A).

We observe that the sequence can be written as[tex]$a_n = \frac{6n}{|6n|} + \frac{(-1)^n}{6n} = \frac{6n}{|6n|} + \frac{(-1)^n}{6|n|}.$[/tex]

We have [tex]$a_{2n} = \frac{12n}{6n} + \frac{1}{6n} = \frac{13}{6} \leq \frac{13}{6}$[/tex] and [tex]$a_{2n+1} = \frac{-12n-6}{6n+3} - \frac{1}{6n+3} = -\frac{13}{12} \geq -\frac{13}{12}.$[/tex]Therefore, the sequence is increasing and bounded below by 5/6 and above by 13/12.

We have[tex]$a_{2n} = \frac{12n}{6n} + \frac{1}{6n} = \frac{13}{6} \geq \frac{0}{1}$[/tex]and

[tex]$a_{2n+1} = \frac{-12n-6}{6n+3} - \frac{1}{6n+3} = -\frac{13}{12} \geq -\frac{13}{12}.$[/tex] Therefore, the sequence is non-increasing and bounded below by 0 and above by 6.

From above part, we see that the sequence is not monotonic.

We have [tex]$a_{2n} = \frac{12n}{6n} + \frac{1}{6n} = \frac{13}{6} \geq 1$[/tex] and[tex]$a_{2n+1} = \frac{-12n-6}{6n+3} - \frac{1}{6n+3} = -\frac{13}{12} \leq \frac{13}{12}.$[/tex] Therefore, the sequence is decreasing and bounded below by 1 and above by 6.

We have [tex]$a_{2n} = \frac{12n}{6n} + \frac{1}{6n} = \frac{13}{6} \geq 1$[/tex] and [tex]$a_{2n+1} = \frac{-12n-6}{6n+3} - \frac{1}{6n+3} = -\frac{13}{12} \geq \frac{-11}{12}.$[/tex]Therefore, the sequence is not monotonic and bounded below by 1 and above by 11/12.

Therefore, the answer is  a_n = 6n + (-1)^n/6n| is increasing; bounded below by 5/6 and above by 13/12. So, the correct option is A).

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The probability of getting the specific sequence t, h, h, h, h, t, t, t, t, t, t when tossing a coin 11 times is 1/2048.

To find the probability of this specific sequence occurring, we need to use the formula for the probability of a specific sequence of independent events:

P(A and B and C and D and E and F and G and H and I and J and K) = P(A) * P(B) * P(C) * P(D) * P(E) * P(F) * P(G) * P(H) * P(I) * P(J) * P(K)

In this case, A represents the first toss being a tails (t), B represents the second toss being a heads (h), and so on until K represents the eleventh toss being a tails (t).

Using the given sequence, we can calculate the individual probabilities for each toss:

P(A) = 1/2 (since there is a 50/50 chance of getting either heads or tails on the first toss)

P(B) = 1/2 (since there is a 50/50 chance of getting heads on the second toss after getting tails on the first toss)

P(C) = 1/2 (since there is a 50/50 chance of getting heads on the third toss after getting heads on the second toss)

P(D) = 1/2 (since there is a 50/50 chance of getting heads on the fourth toss after getting heads on the third toss)

P(E) = 1/2 (since there is a 50/50 chance of getting heads on the fifth toss after getting heads on the fourth toss)

P(F) = 1/2 (since there is a 50/50 chance of getting tails on the sixth toss after getting heads on the fifth toss)

P(G) = 1/2 (since there is a 50/50 chance of getting tails on the seventh toss after getting tails on the sixth toss)

P(H) = 1/2 (since there is a 50/50 chance of getting tails on the eighth toss after getting tails on the seventh toss)

P(I) = 1/2 (since there is a 50/50 chance of getting tails on the ninth toss after getting tails on the eighth toss)

P(J) = 1/2 (since there is a 50/50 chance of getting tails on the tenth toss after getting tails on the ninth toss)

P(K) = 1/2 (since there is a 50/50 chance of getting tails on the eleventh toss after getting tails on the tenth toss)

Multiplying these probabilities together gives us the probability of getting the sequence t, h, h, h, h, t, t, t, t, t, t:

P(t, h, h, h, h, t, t, t, t, t, t) = (1/2) * (1/2) * (1/2) * (1/2) * (1/2) * (1/2) * (1/2) * (1/2) * (1/2) * (1/2) * (1/2) = 1/2048

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The solution to the differential equation dy/dx = e^(2x 3y) with initial condition y(0) = 1 is: y = (1/3) ln|e⁶ˣ - 1| - (1/6)e³

To solve the differential equation dy/dx = e^(2x 3y) using separation of variables, we first need to rewrite it in a separable form. Using the property of exponentials that e^(a+b) = eᵃ × eᵇ, we can rewrite the equation as:

1/y dy = e^(2x) dx × e^(3y)

Now we can separate the variables by integrating both sides:

∫(1/y) dy = ∫(e^(2x) dx × e^(3y))

ln|y| = (1/2)e^(2x) × e^(3y) + C

where C is the constant of integration.

Applying the initial condition y(0) = 1, we can solve for C:

ln|1| = (1/2)e^(2×0) × e^(3*1) + C

0 = (1/2) × e³ + C

C = -1/2 × e³

Substituting C back into the equation, we get:

ln|y| = (1/2)e^(2x) × e^(3y) - 1/2 × e³

Simplifying and solving for y, we get:

y = (1/3) ln|e⁶ˣ - 1| - (1/6)e³

Therefore, the solution to the differential equation dy/dx = e^(2x 3y) with initial condition y(0) = 1 is:

y = (1/3) ln|e⁶ˣ - 1| - (1/6)e³

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The predicted population in 2000 is (a) 529.85 million and (b) 244.66 million.

To use an exponential model to predict the population in 2000, we need to find the values of the growth rate and the initial population.

(a) Using the census figures for 1900 and 1910, we can find the growth rate as follows:

r = (ln(P₁/P₀))/(t₁ - t₀)

where P₀ is the initial population (in 1900), P₁ is the population after 10 years (in 1910), t₀ is the initial time (1900), and t₁ is the time after 10 years (1910).

Substituting the values, we get:

r = (ln(92/76))/(1910-1900) = 0.074

Now, we can use the exponential model:

P(t) = P₀ * [tex]e^{(r(t-t_0))}[/tex]

where t is the time in years, and P(t) is the population at time t.

Substituting the values, we get:

P(2000) = [tex]76 * e^{(0.074(2000-1900))} = 76 * e^{7.4}[/tex] = 529.85 million (rounded to two decimal places)

Therefore, the predicted population in 2000 is 529.85 million.

(b) Using the census figures for 1950 and 1960, we can find the growth rate as follows:

r = (ln(P₁/P₀))/(t₁ - t₀)

where P₀ is the initial population (in 1950), P₁ is the population after 10 years (in 1960), t₀ is the initial time (1950), and t₁ is the time after 10 years (1960).

Substituting the values, we get:

r = (ln(179/150))/(1960-1950) = 0.028

Using the same exponential model, we get:

P(2000) = [tex]150 * e^{(0.028(2000-1950))} = 150 * e^{1.4} = 244.66[/tex] million (rounded to two decimal places)

Therefore, the predicted population in 2000 is 244.66 million

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If we define the linear transformation t: Ra -> Rb by t(x) = ax, where a is a 5x4 matrix, then the dimensions of the vectors in Ra and Rb will depend on the number of columns of the matrix a.

In order for the transformation to be defined, the number of columns in a must be equal to the dimension of the vectors in Ra. Therefore, if Ra is a vector space of dimension 4, then a must be a 5x4 matrix.

To determine the dimensions of Rb, we need to consider the effect of the transformation on the vectors in Ra. Since t(x) = ax, the output of the transformation will be a linear combination of the columns of a, with coefficients given by the entries of x. Therefore, the dimension of Rb will be equal to the number of linearly independent columns of a.

In order to determine b, we need to know the dimension of Rb. Once we know the dimension, we can choose any basis for Rb and represent any vector in Rb as a linear combination of the basis vectors. Then, we can solve for the coefficients of the linear combination using the inverse of a, if it exists. Therefore, the choice of b depends on the choice of basis for Rb.

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Answer:

a and c are correct.

Step-by-step explanation:

In this arithmetic sequence, the first term is 3, and the common difference is 2. So a and c are correct.

a(n) = 3 + 2n for n>0 since a(0) = 3

c(n) = -1 + 2n for n>2 since c(2) = 3

A "dosage-strength" of "0.2-mg" in "1.5-mL" is available. Give 0.15 mg. in 1.125 mL.

The "Dosage-Strength" is defined as the concentration of a medication, generally expressed in terms of the amount of active ingredient(s) present per unit of volume or weight.

To calculate the volume of the 0.2 mg dosage strength needed to obtain 0.15 mg, we use the following formula:

⇒ Volume to withdraw = (Dosage needed/Dosage strength) × Volume of available dosage strength,

Substituting the values,

We get,

⇒ Dosage needed = 0.15 mg,

⇒ Dosage strength = 0.2 mg,

⇒ Volume of 0.2 mg = 1.5 mL,

So, Volume of 0.15 mg = (0.15 mg/0.2 mg) × 1.5 mL,

⇒ 1.125 mL.

Therefore, 0.15 mg of the medication can be obtained by using 1.125 mL of the available 0.2 mg dosage strength.

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The given question is incomplete, the complete question is

A dosage strength of 0.2 mg in 1.5 mL is available. Give 0.15 mg. in ___ mL.

Answer:$192

Step-by-step explanation:

Step 1: Multiply 800 times 12%. You get $96

Step 2: Since the question asked how much she will earn in 2yrs double the interest amount

Step 3 96+96=192

Answer:

1/y^5.

Step-by-step explanation:

To simplify y²/y⁷, we can use the quotient rule of exponents, which states that when dividing exponential terms with the same base, we can subtract the exponents. Specifically, we have:

y²/y⁷ = y^(2-7) = y^(-5)

Now, we can simplify further by using the negative exponent rule, which states that a term with a negative exponent is equal to the reciprocal of the same term with a positive exponent. Specifically, we have:

y^(-5) = 1/y^5

Therefore, y²/y⁷ simplifies to 1/y^5.

a. There are 54 elements in the sample space.

b. P(A) = 0.2778

c. No, events A and B are not mutually exclusive.

d. No, events A and C are not mutually exclusive.

a. To find the total number of elements in the sample space, we need to multiply the number of cards by the number of possible outcomes from the coin toss. Therefore, the sample space has 54 elements (9+11+7) x 2.

b. The probability of event A is the probability of picking a green card first (9/27) multiplied by the probability of getting a head on the coin toss (1/2). Therefore, P(A) = (9/27) x (1/2) = 0.2778 (rounded to 4 decimal places).

c. Events A and B are not mutually exclusive because it is possible to pick a red or blue card and still have a head on the coin toss. Therefore, there are some elements in the sample space that belong to both events.

d. Events A and C are not mutually exclusive because it is possible to pick a green card and still have a head on the coin toss. Therefore, there are some elements in the sample space that belong to both events.

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