Furthermore, each license plate string must contain exactly 8 distinct characters (including the space character). For example, CMSC250 is not a valid license plate string, but ’CM8Z 2S0’ and ’BIGCARSZ’ are. How many license plate strings are possible?

Answer: Our required probability is 0.947.

Step-by-step explanation:

Since we have given that

Number of light bulbs selected = 30

Probability that the light bulb produced in a facility are defective = 2.8% = 0.028

We need to find the probability that fewer than 3 defective bulbs are found.

We will use "Binomial distribution":

n = 30, p = 0.028

so, P(X>3)=P(X=0)+P(X=1)+P(X=2)

So, it becomes,

[tex]P(X=0)=(1-0.0.28)^{30}=0.426[/tex]

and

[tex]P(X=1)=^{30}C_1(0.028)(0.972)^{29}=0.368\\\\P(X=2)=^{30}C_2(0.028)^2(0.972)^28=0.153[/tex]

So, the probability that fewer than three defective bulbs are defective is given by

[tex]0.426+0.368+0.153\\\\=0.947[/tex]

Answer: Option(c) is correct.

Step-by-step explanation:

Data engineering is a set of guidelines and approaches that characterize the kind of information gathered and how it is put away and utilized.  

The Data engineering includes the means, for example, gathering of data, storage of information in the databases and access the information at whatever point required.  

So,data engineering refers to information storage, database plan and information structures.  

Consequently Data quality isn't clarified in Data engineering.

Answer:

[tex]x=-2,0,3[/tex]

Step-by-step explanation:

We have been given a function [tex]f(x)=15x^3-15x^2-90x[/tex]. We are asked to find the zeros of our given function.

To find the zeros of our given function, we will equate our given function by 0 as shown below:

[tex]15x^3-15x^2-90x=0[/tex]

Now, we will factor our equation. We can see that all terms of our equation a common factor that is [tex]15x[/tex].

Upon factoring out [tex]15x[/tex], we will get:

[tex]15x(x^2-x-6)=0[/tex]

Now, we will split the middle term of our equation into parts, whose sum is [tex]-1[/tex] and whose product is [tex]-6[/tex]. We know such two numbers are [tex]-3\text{ and }2[/tex].

[tex]15x(x^2-3x+2x-6)=0[/tex]

[tex]15x((x^2-3x)+(2x-6))=0[/tex]

[tex]15x(x(x-3)+2(x-3))=0[/tex]

[tex]15x(x-3)(x+2)=0[/tex]

Now, we will use zero product property to find the zeros of our given function.

[tex]15x=0\text{ (or) }(x-3)=0\text{ (or) }(x+2)=0[/tex]

[tex]15x=0\text{ (or) }x-3=0\text{ (or) }x+2=0[/tex]

[tex]\frac{15x}{15}=\frac{0}{15}\text{ (or) }x-3=0\text{ (or) }x+2=0[/tex]

[tex]x=0\text{ (or) }x=3\text{ (or) }x=-2[/tex]

Therefore, the zeros of our given function are [tex]x=-2,0,3[/tex].

Answer: 160

Step-by-step explanation:

Given : The options to fill science requirement =4

The options to fill diversity requirement =2

The options to fill English requirement =5

The options to fill math requirement = 4

The Fundamental Counting Principle say that the number of total outcomes is equal to the product of the number of ways of all the events occur in the problem.

Using Fundamental Counting Principle, we have the total number of possible outcomes for the given situation :-

[tex]4\times2\times5\times4=160[/tex]

Hence, the total number of possible outcomes = 160

Answer:

  see below

Step-by-step explanation:

Filling the given numbers into the given formulas, you have ...

Line AB:

  x = 1 +(4-1)t, y = 1 +(3-1)t

  x = 1+3t, y = 1+2t . . . . . . simplify

Line BC:

  x = 4 +(1-4)t, y = 3 +(7-3)t

  x = 4 -3t, y = 3 +4t . . . . . simplify

Line AC:

  x = 1 +(1-1)t, y = 1 +(7-1)t

  x = 1, y = 1+6t . . . . . . . . . .simplify

Answer: Mean = 4.8

Standard deviation = 1.96

Step-by-step explanation:

The mean and standard deviation of the binomial distribution is given by :-

[tex]\mu=np\\\sigma=\sqrt{np(1-p)}[/tex], where n is the total number of trials , p is the the probability of success.

Given : The probability that the produced by a particular manufacturer are less than 1.680 inches in​ diameter = 20%=0.2

Sample size : n=24                                                 [since 1 dozen = 12]

Now, the  mean and standard deviation of the binomial distribution is given by :-

[tex]\mu=24\times0.2=4.8\\\\\sigma=\sqrt{24(0.2)(1-0.2)}\\\\=1.95959179423\approx1.96[/tex]

The mean of the binomial distribution for the balls less than 1.680 inches in diameter is 4.8, and the standard deviation is approximately 1.96.

To find the mean and standard deviation for a binomial probability distribution where 20% of all balls are less than 1.680 inches in diameter from a sample of two dozen (24) balls, we use the formulas for a binomial distribution. The mean (μ) of a binomial distribution is calculated as μ = n * p, where n is the number of trials and p is the probability of success on a single trial. In this case, n = 24 and p = 0.20.

The mean is μ = 24 * 0.20 = 4.8.

To calculate the standard deviation (σ), we use the formula σ = √(n * p * (1 - p)), where (1 - p) is the probability of failure. The standard deviation is σ = √(24 * 0.20 * 0.80) = √(3.84) ≈ 1.96.

Answer:

100⁄25  60⁄15  4⁄1

Step-by-step explanation:

20/5 = 4

We need to see what equals 4

100⁄25 = 4

10⁄2=5

60⁄15=4

40⁄2=20

4⁄1=4

Answer:

It should be 314.9

Step-by-step explanation:

In every centimeter, there are about .3937 inches

So if you multiple .3937 by 800 and round to the nearest tenth, you get that answer

Let

[tex]S=23+24+25+\cdots+101+102+103[/tex]

This sum has ___ terms. Its terms form an arithmetic progression starting at 23 with common difference between terms of 1, so that the [tex]n[/tex]-th term is given by the sequence [tex]23+(n-1)\cdot1=22+n[/tex]. The last term is 103, so there are

[tex]103=22+n\implies n=81[/tex]

terms in the sequence.

Now, we also have

[tex]S=103+102+101+\cdots+25+24+23[/tex]

so that adding these two ordered sums together gives

[tex]2S=(23+103)+(24+102)+\cdots+(102+24)+(103+23)[/tex]

[tex]\implies2S=\underbrace{126+126+\cdots+126+126}_{81\text{ times}}=81\cdot126[/tex]

[tex]\implies S=\dfrac{81\cdot126}2\implies\boxed{S=5103}[/tex]

Step-by-step Answer:

Calculating Pi using Archimedes method of polygons.

We know that the definition of pi is the ratio of circumference of a circle divided by the diameter.  Starting with Pythagorean Theorem, and proposition 3 of Euclid’s Elements, Archimedes was able to approximate pi to any precision arithmetically, without further resort to geometry!

He figured that the perimeter of any regular polygon (all sides and vertex angles equal) is an approximation to a circle.  More sides will make closer approximations.

Starting with a hexagon, he bisects the central angles to make polygons 12-, 24-, 48- and 96-sides, whose perimeters approaches that of a circle, and hence the approximation to pi since the diameter remains known and constant.

Proposition 3 is also commonly referred to as the angle bisector theorem, which states that in a triangle, an angle bisector subdivides the opposite sides in the ratio of the two remaining sides.

 Please refer to the attached image for the nomenclature of the geometry.

The accompanying diagram shows that the perimeter of a hexagon is 12 times the length of AB, or 12*(1.0/2) = 6.  With the diameter equal to 2*1.0 = 2, the approximation to pi is 6/2=3.0.

Pi(6) = 3.0

If we divide the central angle by two, we end up with a 12-sided polygon (dodecagon), with the half central angle of 15 degrees (triangle A’BC).  To calculate the new perimeter, we need to calculate the length A’B, which is given by the angle-bisector theorem as

A’B / A’A  = BC / AC

All other sides are known in terms of A’B

A’B / (0.5-A’B) = sqrt(3)/2 / 1

Solve for A’B by cross-multiplication and solving for A’B, we get

A’B = sqrt(3)/(2sqrt(3)+4) = 0.2320508 (to 7 decimals)

At the same time, the radius has been reduced to  

A’C = sqrt(A’B^2+BC^2) = 0.896575

That brings the approximation of pi as 12*A’B/A’C

P(12) = 3.1058285 (7 decimals)

Continuing bisecting, now using a polygon of 24 sides, we only have to replace

AB by A’B, AC by A’C, and 12 by 24 to get

Pi(24) = 3.132629 (7 decimals)

Repeating again for a polygon of 48 sides,  

Pi(48) = 3.1393502

Pi(96) = 3.1410320

Pi(192) = 3.1414525

Pi(384) = 3.1415576

Etc.

The accurate value of pi to 10 digits is 3.1415926536

And we conclude that Pi(48) is the first approximation the provides 2 decimal places of accuracy.

Note: What was calculated was actually the lower bound value of pi.

We can obtain the upper bound value of pi using the length of BC as the radius, which gives the upper bound.  The average of the two bounds for a 384-sided polygon gives P_mean(384) = 3.1416102, which is accurate to 2 units in the 5th decimal place.

The Archimedes exhaustion method is a geometric approach to estimate the value of pi. By inscribing and circumscribing regular polygons within and around a circle, Archimedes determined lower and upper bounds for the value of pi.

As the number of sides for the polygons increased, the approximation of pi became more accurate. With a 96-sided polygon, Archimedes found that pi was greater than 3.1408 and less than 3.1429. By taking the average of these bounds, a more precise estimation of pi accurate to two decimal places is achieved.

Thus, using the Archimedes exhaustion method, we can estimate pi to be approximately 3.14, making it precise to two decimal places.

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The position of the particle when t equals 6 is equal to -27 feet.

The total distance this particle travels during the 6-s time interval is equal to 69 feet.

Based on the information provided above, we can logically deduce the following polynomial function that models the position of a particle along a straight line;

[tex]s = 1.5t^3 - 13.5t^2 + 22.5t[/tex]

In order to determine the position of the particle when t is equal to 6, we would substitute 6 for t in the polynomial function as follows;

[tex]s(6) = 1.5(6)^3 - 13.5(6)^2 + 22.5(6)[/tex]

s(6) = 324 - 486 + 135

s(6) = -27 feet.

In order to determine the total distance this particle travels during the 6-s time interval, we would have to plot a graph of the velocity of the particle. Also, the velocity of the particle can be determined by taking the first derivative of the position with respect to time;

[tex]s = 1.5t^3 - 13.5t^2 + 22.5t\\\\s' = 4.5t^2 - 27t + 22.5[/tex]

Based on the graph, the particle changes directions at t equal 1 seconds and again t equal 5 seconds. Hence, the velocity of the particle drops to zero at these positions.

In this context, we would find the distance between these intervals;

Distance (0 ≤ t ≤ 1) = 10.5 - 0 = 10.5 feet.

Distance (1 ≤ t ≤ 5) = 10.5 - (-37.5) = 48 feet.

Distance (5 ≤ t ≤ 6) = -27 - (-37.5) = 10.5 feet.

For the total distance, we have;

Total distance = 10.5 + 48 + 10.5

Total distance = 69 feet.

Complete Question:

The position of a particle along a straight line is given by [tex]s = 1.5t^3 - 13.5t^2 + 22.5t[/tex] ft, where t is in seconds. Determine the position of the particle when t = 6 s and the total distance it travels during the 6-s time interval. Hint: Plot the path to determine the total distance traveled.

Answer:

42 gtt/min

Step-by-step explanation:

Amount of fluid to be infused = 500 mL

Time = 3 hours = 3×60 = 180 minutes

Tubing drop factor/mL = 15 gtt/mL

Fussion rate = (Amount of fluid to be infused / time in minutes)

Fussion rate = 500/180 = 2.78 mL/min

gtt/min = Tubing drop factor/mL× Fusion rate

⇒gtt/min = 15×(500/180)

⇒gtt/min = 15×(25/9)

⇒gtt/min = 125/3

⇒gtt/min = 41.67

⇒gtt/min = 42

∴42 drops/min (gtt/min) should be given.

Answer:

16.59 inches

Step-by-step explanation:

Mean value = u = 14.5 inches

Standard deviation = [tex]\sigma[/tex] = 0.9 in

We need to find the 99th percentile of the given distribution. This can be done by first finding the z value associated with 99th percentile and then using that value to calculate the exact value of hip breadth that lies at 99th percentile

From the z-table, the 99th percentile value is at a z-value of:

z = 2.326

This means 99% of the z-scores are below this value. Now we need to find the equivalent hip breadth for this z-score

The formula to calculate the z score is:

[tex]z=\frac{x-u}{\sigma}[/tex]

where, x is the hip breadth which is equivalent to this z-score.

Substituting the values we have:

[tex]2.326=\frac{x-14.5}{0.9}\\\\ 2.0934=x-14.5\\\\ x=16.5934[/tex]

Rounded to 2 decimal places, engineers should design the seats which can fit the hip breadth of upto 16.59 inches to accommodate the 99% of all males.

To find the hip breadth for men that separates the smallest 99% from the largest 1%, we can use the z-score formula and the standard normal distribution table. The hip breadth that separates the smallest 99% is approximately 16.197 inches.

To find the hip breadth for men that separates the smallest 99% from the largest 1%, we need to determine the z-score corresponding to a 99% percentile. Firstly, we will calculate the z-score using the formula: z = (x - μ) / σ, where x is the hip breadth, μ is the mean (14.5 in.), and σ is the standard deviation (0.9 in.). Secondly, we use the standard normal distribution table or a z-score calculator to find the z-score that corresponds to a 99% percentile. Finally, we can solve for x using the formula: x = z * σ + μ.

Substituting the values, we have z = (x - 14.5) / 0.9. From the standard normal distribution table, the z-score that corresponds to a 99% percentile is approximately 2.33.

Plugging the values into the equation, we get 2.33 = (x - 14.5) / 0.9. Solving for x gives us x = 2.33 * 0.9 + 14.5 = 16.197 in.

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Answer:

  525%

Step-by-step explanation:

The relationship between the variables is ...

  cost + markup = selling price

  cost + 84%(selling price) = selling price

  cost = selling price(100% -84%) = 16%(selling price)

Then the markup based on cost is ...

  markup/cost = (84%(selling price))/(16%(selling price)) = 84/16

  markup/cost = 5.25 = 525%

Answer:

So y(3)=1

Step-by-step explanation:

Given that

[tex]\dfrac{dy}{dx}=x-y[/tex]

y(0)=1,step size h=1

From Euler's method

[tex]\dfrac{dy}{dx}=f(x,y)=x-y[/tex]

[tex]y_{n+1}=y_n+hf(x_n,y_n),x_n=x_0+nh[/tex]  

[tex]y_1=y_0+hf(x_0,y_0)[/tex]

[tex]y_1=1+1f(0,1)[/tex]

f(0,1)=0-1= -1

[tex]y_1=1-1[/tex]=0

[tex]y_{2}=y_1+hf(x_1,y_1)[/tex]

[tex]y_{2}=0+1f(1,0)[/tex]

f(1,0)=1

[tex]y_{2}=1[/tex]

[tex]y_{3}=y_2+hf(x_2,y_2)[/tex]

[tex]y_{3}=1+1f(2,1)[/tex]

f(2,1)=1

[tex]y_{3}=1+1[/tex]=2

[tex]y_{4}=y_3+hf(x_3,y_3)[/tex]

[tex]y_{4}=2+1f(3,2)[/tex]

f(3,2)= -1

[tex]y_{4}=2-1[/tex]=1

So y(3)=1

Answer:

Data warehouse

Step-by-step explanation:

Data warehouse  is a technology that combines data from one or more sources so it can be compared for making business decisions.

Moreover, Data warehouse can be explained as  a technology that combines from one or more sources so it can be prepared for making business decisions. A data warehouse is constructed by Integrating data from multiple heterogeneous sources that support analytical reporting, and decision making.

Taking the transform of both sides gives

[tex]\mathcal L_s\{x''+8x'+15x\}=0[/tex]

[tex](s^2X(s)-sx(0)-x'(0))+8(sX(s)-x(0))+15X(s)=0[/tex]

where [tex]X(s)[/tex] denotes the Laplace transform of [tex]x(t)[/tex], [tex]\mathcal L_s\{x(t)\}[/tex]. Solve for [tex]X(s)[/tex] to get

[tex](s^2+8s+15)X(s)=2s+13[/tex]

[tex]X(s)=\dfrac{2s+13}{s^2+8s+15}=\dfrac{2s+13}{(s+3)(s+5)}[/tex]

Split the right side into partial fractions:

[tex]\dfrac{2s+13}{(s+3)(s+5)}=\dfrac a{s+3}+\dfrac b{s+5}[/tex]

[tex]2s+13=a(s+5)+b(s+3)[/tex]

If [tex]s=-3[/tex], then [tex]7=2a\implies a=\dfrac72[/tex]; if [tex]s=-5[/tex], then [tex]3=-2b\implies b=-\dfrac32[/tex]. So

[tex]X(s)=\dfrac72\dfrac1{s+3}-\dfrac32\dfrac1{s+5}[/tex]

Finally, take the inverse transform of both sides to solve for [tex]x(t)[/tex]:

[tex]x(t)=\dfrac72e^{-5t}-\dfrac32e^{-3t}[/tex]

The initial value problem is a second-order homogeneous differential equation that can be solved using the Laplace Transform. After substituting the initial conditions and simplifying the equation, one can decompose the equation using partial fraction decomposition and finally find the solution in the time domain.

Your given initial value problem is a second-order homogeneous differential equation. You should use the Laplace Transform to solve it. The Laplace transform of this equation is: L{x''(t) + 8x'(t) + 15x(t)} = 0 which simplifies to s²X(s) - sx(0) - x'(0) + 8[sX(s) - x(0)] + 15X(s) = 0. Substituting the initial conditions x(0) = 2 and x'(0) = -3, we get s²X(s) - 2s - (-3) + 8[sX(s) - 2] + 15X(s) = 0, then simplify to (s² + 8s + 15)X(s) = 2s + 3.

The roots of the quadratic equation s² + 8s + 15 = 0 are -5 and -3. So, the solution of the equation X(s) = (2s + 3) / (s² + 8s + 15) can be solved by using partial fraction decomposition. Therefore, the solution in the time domain would be x(t) = 2e⁻³ᵗ - e⁻⁵ᵗ.

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Answer:

$1343.25

Step-by-step explanation:

995*1.35=1343.25

Answer:

Step-by-step explanation:

For these we have to assume the price and cost functions are linear.

Let p, c, r, P, q represent price, cost, revenue, Profit, and quantity (of items), respectively. The 2-point form of the equation for a line is ...

  y = (y2 -y1)/(x2 -x1)(x -x1) +y1

Price Function

Using the two-point form for price, we get ...

  p = (60 -65)/(20 -10)(q -10) +65 = -5/10(q -10) +65

  p = (-1/2)q +70 . . . . price per item

Cost Function

Using the two-point form for cost, we get ...

  c = (1050 -650)/(20 -10)(q -10) +650 = 40(q -10) +650

  c = 40q +250 . . . . cost for q items

Revenue Function

Revenue is the product of price and quantity:

  r(q) = qp

  r(q) = (1/2)q(140 -q) . . . . revenue from sale of q items

Profit Function

Profit is the difference between revenue and cost.

  P(q) = r(q) -c = 1/2q(140 -q) -(40q +250)

  P(q) = -1/2q^2 +30q -250

  P(q) = (-1/2)(q -10)(q -50) . . . . factored form

Break-Even Points

The profit function will be zero when its factors are zero, at q=10 and q=50. The price function tells us the corresponding prices are $65 and $45 per item, respectively.

Maximum Profit

The profit function is a maximum at the quantity halfway between the break-even points. There, q = (10+50)/2 = 30, and P(30) is ...

  P(30) = -1/2(30-10)(30-50) = 1/2(20^2) = 200 . . . . dollars

Quantity for Maximum Profit

This was found to be 30 in the previous section.

Answer: Matrix B is non- invertible.

Step-by-step explanation:

A matrix is said to be be singular is its determinant is zero,

We know that if a matrix is singular then it is not invertible.    (1)

Or if a matrix is invertible then it should be non-singular matrix.      (2)

Given :  A and B are n x n matrices from which A is invertible.

Then A must be non-singular matrix.                            ( from 2 )

If AB is singular.

Then either A is singular or B is singular but A is a non-singular matrix.

Then , matrix B should be a singular matrix.                   ( from 2 )

So Matrix B is non- invertible.                                     ( from 1 )

Final answer:

The probability that the sample mean carapace length is more than 4.25 inches is 0.0764.

Explanation:

To find the probability that the sample mean carapace length is more than 4.25 inches, we need to use the properties of the normal distribution. First, we need to calculate the z-score for the sample mean using the formula:
z = (x - μ) / (σ / sqrt(n))
Where x is the sample mean, μ is the population mean, σ is the population standard deviation, and n is the sample size.

Plugging in the values:
z = (4.25 - 4.125) / (1.05 / sqrt(175))

Simplifying:
z = 1.428571

Next, we need to find the cumulative probability from the z-table. The table will give us the probability of getting a z-score less than or equal to a given value. Since we want the probability that the sample mean is more than 4.25 inches, we need to subtract the cumulative probability from 1:
Probability = 1 - cumulative probability

Looking up the cumulative probability in the z-table, we find that it is approximately 0.9236. Therefore, the probability that the sample mean carapace length is more than 4.25 inches is:
Probability = 1 - 0.9236 = 0.0764

Answer:

Step-by-step explanation:

Given function is

(a)F(x)=[tex]\left ( x^{3}+5\right )^{0.25}-15e^{x^{3}}[/tex]

[tex]F^{'}\left ( x\right )[/tex]=[tex]0.25\left ( x^{3}+5\right )^{-0.75}\frac{\mathrm{d} x^{3}}{\mathrm{d} x}-15e^{x^{3}}\frac{\mathrm{d} x^{3}}{\mathrm{d} x}[/tex]

[tex]F^{'}\left ( x\right )=0.25\left ( x^{3}+5\right )^{-0.75}\times 3x^{2}-15e^{x^{3}}\times \left ( 3x^{2}\right )[/tex]

(b)F(x)=[tex]\frac{\left ( x-3\right )^2\left ( x-5\right )}{\left ( x-4\right )^2\left ( x^{2}+3\right )^5}[/tex]

[tex]F^{'}\left ( x\right )[/tex]=[tex]\frac{\left [ 2\left ( x-3\right )\right \left ( x-5\right )+\left ( x-3\right )^2]\left [ \left ( x-4\right )^2\left ( x^2+3\right )^5\right ]-\left [ 2\left ( x-4\right )^{3}\left ( x^2+3\right )^5+5\left ( x^2+3\right )^4\left ( 2x\right )\left ( x-4\right )^2\right ]\left [ \left ( x-3\right )^2\left ( x-5\right )\right ]}{\left [\left ( x-4\right )^2\left ( x^2+3\right )^5\right ]^2}[/tex]

Answer:

The answer is zero or option D.

Step-by-step explanation:

Proponents of rational expectations theory argued that, in the most extreme case, if policymakers are credibly committed to reducing inflation and rational people understand that commitment, and quickly lower their inflation expectations, the sacrifice ratio could be as small as 0.

The answer is:

The only value of "x" that ARE NOT in the domain of the function g, are -2 and 3.

Restriction: {-2,3}

Since we are working with a quotient (or division), we must remember that the only restriction for this kind of functions are the values that make the denominator equal to 0, so, the domain of the function will include all the values of "x" that are different than the zeroes or roots of the denominator.

We have the function:

[tex]h(x)=\frac{x-3}{x^2-x-6}[/tex]

Where its denominator is :

[tex]x^2-x-6[/tex]

Now, finding the roots or zeroes of the expression, by factoring, we have:

We need to find two numbers which product is equal to -6 and its addition is equal to -1, these numbers are -3 and 2, we have:

[tex]-3*2=-6\\-3+2=-1[/tex]

So, the factorized form of the expression will be:

[tex](x-3)*(x+2)[/tex]

We have that the expression will be equal to 0 if "x" is equal to "-2" and "3", so, the values that are not in the domain of g are: -2,3.

Hence, we have:

Restriction: {-2,3}

Domain: (-∞,-2)U(-2,3)U(3,∞)

Have a nice day!

To find the values of x that are not in the domain of the function g(x), we need to identify any values for x that would make the function undefined. The function g(x) = (x - 3) / (x^2 - x - 6) becomes undefined when the denominator is equal to zero, since division by zero is not allowed.
Thus, we need to find the values of x that make the denominator x^2 - x - 6 equal to zero. To do this, we'll solve the quadratic equation:
x^2 - x - 6 = 0
To solve this quadratic equation, we can factor the quadratic expression, or use the quadratic formula. We'll try factoring first:
x^2 - x - 6 = (x - 3)(x + 2)
Set each factor equal to zero and solve for x:
x - 3 = 0  -->  x = 3
x + 2 = 0  -->  x = -2
So, the values of x that are not in the domain of g(x) are -2 and 3, because these are the values that make the denominator equal to zero. Hence, g(x) is undefined at x = -2 and x = 3.
Therefore, the values that are NOT in the domain of g are:
-2, 3

Answer:

angular moment is 0.25 kg.m²/s

Step-by-step explanation:

given data in question

mass (m) = 0.25 kg

length of string i.e. radius (r) = 0.5 m

velocity = 2 m/s

to find out

angular momentum before mass release

solution

we know angular moment formula i.e.

angular moment = mass × velocity × radius   ................1

put the value mass velocity and radius in equation 1 we get angular moment i.e.

angular moment = mass × velocity × radius

angular moment = 0.25 × 2 × 0.5

angular moment = 0.25

so the angular moment is 0.25 kg.m²/s before release and 0.25 kg.m²/s after release because angular momentum is always conserved

The question deals with angular momentum and its conservation in rotational motion. The angular momentum before the mass is released remains constant and is carried by the mass upon release. Supplementary problems discuss changes in angular momentum and the effects of pulling in a spinning mass on its rotational dynamics.

The question regards the concept of angular momentum in classical mechanics, specifically within the realm of rotational motion. Angular momentum, denoted by L, is a physical quantity that represents the rotational inertia of a spinning object multiplied by its angular velocity, and it's given by the formula L = Iω, where I is the moment of inertia and ω is the angular velocity.

For the case where a mass is spinning over your head and then released, the angular momentum just before the release is conserved. This means that if we calculate the angular momentum while the mass is attached to the string and spinning, the same amount of angular momentum will be present in the mass's linear motion after it is released along the tangent. If we assume that the mass travels in a circular path while attached to the string, the angular momentum can be related to the linear momentum by L = mvr, where m is the mass, v is the linear velocity just before release, and r is the radius of the circular path.

Upon release, because angular momentum is conserved, the mass carries this angular momentum into its linear motion, causing it to move off on a tangent at a velocity that reflects this conservation. If there are no external torques acting on the system, the angular momentum will not change; therefore, it 'moves' with the mass as linear momentum.

Regarding the supplementary problems provided, when angular velocity is increased, the tendency for a spinning object is to move outward due to centrifugal force. Consequently, the string's angle with respect to the vertical will increase. To calculate the initial and final angular momenta, one would use the same conservation principle, taking into account the changes in angular velocity and the moment of inertia. A scenario where the rod spins fast enough to make the ball horizontal suggests an infinitely large angular velocity, which is not practically achievable. Therefore, the ball cannot be truly horizontal as it would require an infinite amount of energy.

Concerning the rock on a string example, as you pull the string in and reduce the radius, the angular momentum remains constant assuming no external torques are acting on the system. This leads to an increase in the angular velocity since L = Iω and I decreases with a smaller radius (I for a point mass is mr²). The increased speed will result in a shorter time required for one revolution (higher frequency of rotation) and a greater centripetal acceleration. The string is under more tension as a result of the increased centripetal force, which might lead to it breaking.

For the collision problem with the spinner and the rod spinning at different rates, conservation of energy or momentum principles would be employed to find the corresponding change in angular velocity. The initial and final angular momenta or energies are equated, considering that all the energy transferred is mechanical and that the rotational inertia of the spinner is required to calculate the angular velocity post-collision.

Answer:

Oil supply will run out in 44.283 years.

Step-by-step explanation:

There are 1.338 trillion barrels of oil in proven reserves.

If oil consumption is 82.78 million barrels per day then we have to calculate the number of years in which supply of oil runs out.

In this sum we will convert 1.338 million barrels of oil into million barrels first then apply unitary method to calculate the time in which oil supply runs out.

Since 1 trillion = [tex]10^{6}[/tex] million

Therefore, 1.338 trillion = 1.338×[tex]10^{6}[/tex] million

∵ 82.78 million barrels oil is the consumption of = 1 day

∴ 1 million barrels oil is the consumption of = [tex]\frac{1}{82.78}[/tex] day

∴ 1.338×[tex]10^{6}[/tex] barrels will be consumed in = [tex]\frac{1.338(10^{6})}{82.78}[/tex] days

= 16163.3245 days

≈ [tex]\frac{16163.3245}{365}[/tex] years

≈ 44.283 years

Therefore, oil supply will run out in 44.283 years

Answer:

σ = 13 miles

Step-by-step explanation:

Let us consider X continuous random variable and λ be the parameter of exponential density function.

where E(x) = [tex]\frac{1}{\lambda}[/tex]

where  E(x) = is expected value=13

we have to find λ=[tex]\frac{1}{E(x)}[/tex]

                          λ=[tex]\frac{1}{13}[/tex]

                          λ=0.076

standard deviation = V(X) = σ =  [tex]\frac{1}{\lambda}[/tex]

now , σ =  [tex]\frac{1}{0.076}[/tex]

          σ = 13 miles is the distance between the two major crack.

Final answer:

The standard deviation of the distance between two major cracks in a highway, which follows an exponential distribution with a mean of 13 miles, is 13 miles.

Explanation:

The question asks for the standard deviation of the distance between two major cracks in a highway, given that this distance follows an exponential distribution with a mean of 13 miles.

In an exponential distribution, the mean (μ) and standard deviation (σ) are equal.

Therefore, the standard deviation of the distance between two major cracks is also 13 miles.

Answer: 2.14 %

Step-by-step explanation:

Given : pH measurements of a chemical solutions have

Mean :  [tex]\mu=6.8[/tex]

Standard deviation : [tex]\sigma=0.02[/tex]

Let X be the pH reading of a randomly selected customer chemical solution.

We assume  pH measurements of this solution have a nearly symmetric/bell-curve distribution (i.e. normal distribution).

The z-score for the normal distribution is given by :-

[tex]z=\dfrac{x-\mu}{\sigma}[/tex]

For x = 6.74

[tex]z=\dfrac{6.74-6.8}{0.02}=-3[/tex]

For x = 6.76

[tex]z=\dfrac{6.76-6.8}{0.02}=-2[/tex]

The p-value =[tex]P(6.74<x<6,76)=P(-3<z<-2)[/tex]

[tex]P(z<-2)-P(z<-3)=0.0227501- 0.0013499=0.0214002\approx0.0214[/tex]

In percent, [tex]0.0214\times=2.14\%[/tex]

Hence, the percent of pH measurements reading below 6.74 OR above 6.76 = 2.14%

To solve the equation [tex]25^3^k[/tex] = 625, rewrite the base as a power of 5 and use the exponentiation rule. Simplify the equation and equate the exponents to solve for k.

To solve the exponential equation 253k = 625, we need to rewrite the base of 25 as a power of 5. Since 25 = 52, we can rewrite the equation as (52)3k = 625.

Using the rule (ab)c = ab × c, we can simplify the equation to 52 × 3k = 625.

Now, we can rewrite 625 as a power of 5 by realizing that 625 = 54. Therefore, we have 52 × 3k = 54.

Since the bases are the same, we can equate the exponents and solve for k:

2 × 3k = 4

6k = 4

k = 4/6

k = 2/3

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To solve the exponential equation 25^3k = 625, rewrite the base 25 as a power of 5. Simplify the equation and set the exponents equal to each other to solve for k.

To solve the exponential equation 253k = 625, we can rewrite the base 25 as a power of 5, since 52 = 25. So, the equation becomes (52)3k = 625. Using the rule of exponents (am)n = amn , we can simplify it to 56k = 625.

Next, we can rewrite 625 as a power of 5: 625 = 54. So, the equation becomes 56k = 54.

Since the bases are the same, we can set the exponents equal to each other: 6k = 4. Solving for k, we divide both sides by 6 to get k = 4/6 = 2/3.

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