Given:
An = [6 n/(-4 n + 9)]
For both of the following answer blanks, decide whether the given sequence or series is convergent or divergent. If convergent, enter the limit (for a sequence) or the sum (for a series). If divergent, enter INF if it diverges to infinity, MINF if it diverges to minus infinity, or DIV otherwise.

(a) The sequence {An }._________________
(b) The series ∑n=1[infinity]( An )________________

Answers

Answer 1
Final answer:

The given sequence is convergent with a limit of -3/2, while the series is divergent since its terms do not approach zero

Explanation:

The sequence in question is An = [6*n/(-4*n + 9)]. To find out if this sequence is convergent or divergent, we need to take the limit as n approaches infinity. As n approaches infinity, the 'n' in the numerator and the 'n' in the denominator will dominate, making the sequence asymptotically equivalent to -6/4 = -3/2. Thus, the sequence is convergent, and its limit is -3/2.

On the other hand, the series ∑n=1[infinity]( An ) is the sum of the terms in the sequence. We can see that as n approaches infinity, the terms of this series do not approach zero, which is a necessary condition for a series to be convergent (using the nth term test). Therefore, the series is divergent.

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Related Questions


5. (6 marks) Use mathematical induction to prove that for each integer n ≥ 4, 5^n ≥ 2^2n+1 + 100.

(please take +100 into considersation since previous solution didnt )

Answers

Step-by-step explanation:

We will prove by mathematical induction that, for every natural [tex]n\geq 4[/tex],  

[tex]5^n\geq 2^{2n+1}+100[/tex]

We will prove our base case, when n=4, to be true.

Base case:

[tex]5^4=625\geq 612=2^{2*4+1}+100[/tex]

Inductive hypothesis:  

Given a natural [tex]n\geq 4[/tex],  

[tex]5^n\geq 2^{2n+1}+100[/tex]

Now, we will assume the induction hypothesis and then use this assumption, involving n, to prove the statement for n + 1.

Inductive step:

[tex]2^{2(n+1)+1}+100=2^{2n+1+2}+100=\\=4*2^{2n+1}+100\leq 4(2^{2n+1}+100)\leq 4*5^n<5^{n+1}[/tex]

With this we have proved our statement to be true for n+1.  

In conlusion, for every natural [tex]n\geq4[/tex].

[tex]5^n\geq 2^{2n+1}+100[/tex]

An item normally $15.99 is listed as being on sale for 30% off its original price, what must you pay?
A discount store promises that all the items it sells are 40% of their normal asking retail price. If one buys shoes that normally retail for $60.99 what is the price you would expect to pay?
Describe how you would answer each question
Then rewrite the percent off problem as a percent of problem.

Answers

Answer: a. You must pay $11.19 for the item.

b.The price you would expect to pay would be $36.59

Step-by-step explanation:

Hi, for first the question you need to calculate the 30 percent of the price of the item, and then subtract that result to the original price.

So: $15.99 × 0,30 = $4.797

$15.99 - $4.797 = $11.19

You must pay $11.19 for the item.

Question 2:

It´s a similar resolution, first you calculate the 40 percent of the retail price, and then subtract that result to the retail price.

So:

$60.99 × 0,40= $24.396

$60.99 - $24.396 = $36.59

The price you would expect to pay would be $36.59

Find all optimal solutions to the following LP using the Simplex Algorithm:
maxz = x1 + 2x2 + 3x3
s.t.
x1 + 2x2 + 3x3 ≤ 10

x1 + x2 ≤ 5

x1 ≤ 1

x1,x2,x3 ≥ 0

Answers

Answer:

z=10

x1=0

x2=0

x3=3.33

Step-by-step explanation:

First Step convert your constraints in standard equations

so we have

x1 + 2x2 + 3x3+x4 = 10

x1 + x2 +x5= 5

x1 +x6= 1

Now we pass it all to the simplex table

Remember that we choose the column with the most negative value

Pivot Element=3

Divide all elements on Pivot Line by Pivot Element

Line x5= 0*Pivot Line +Line x5

Line x6= 0*Pivot Line+ Line X6

Line Z= 3* Pivot Line + Line Z

We finish when all the elements from the line Z are positive

Hence we have that x3=3.33 and x1=0, x2=0 and the max of z is 10

Answer:

z=10

x1=0

x2=0

x3=3.33

Step-by-step explanation:

Reduce 1.256 g to micrograms, to milligrams, and to kilograms.

Answers

Final answer:

The conversion of 1.256 grams is as follows: 1,256,000 micrograms, 1256 milligrams, and 0.001256 kilograms.

Explanation:

To convert 1.256 grams (g) to different units of mass, you use the following conversion factors:

Micrograms (μg): 1 g = 1 × 10⁶ μg. So, 1.256 g = 1.256 × 10⁶ μg = 1,256,000 μg. Milligrams (mg): 1 g = 1000 mg. Thus, 1.256 g = 1.256 × 10³ mg = 1256 mg. Kilograms (kg): 1 g = 1 × 10⁻³ kg. Hence, 1.256 g = 1.256 × 10⁻³ kg = 0.001256 kg.

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1.256 grams is equal to 1,256,000 micrograms, 1,256 milligrams, and 0.001256 kilograms. These conversions use basic multiplication and division of the metric system.

To convert 1.256 grams to different units, you should be familiar with the respective conversion factors:

1 gram (g) = 1,000 milligrams (mg)1 gram (g) = 1,000,000 micrograms (µg)1 gram (g) = 0.001 kilograms (kg)

Steps to Convert:

Micrograms: Multiply the number of grams by 1,000,000:
1.256 g × 1,000,000 µg/g = 1,256,000 µgMilligrams: Calculate the milligrams by 1,000:
1,000 mg/g × 1.256 g = 1,256 mgIn kilograms: The grams should be divided by 1,000:
1.256 g ÷ 1,000 g/kg is 0.00012556 kg.

Therefore, the conversions are:

1.256 grams = 1,256,000 micrograms1.256 grams = 1,256 milligrams1.256 grams = 0.001256 kilograms

3. Write a recursive algorithm of the sequence t(1)=1 and t(n)=n2 t(n-1) as a function.

Answers

Answer:

int t(int n){

     if(n == 1)

          return 1;

     else

          return n*n*t(n-1);

}

Step-by-step explanation:

A recursive function is a function that calls itself.

I am going to give you an example of this algorithm in the C language of programming.

int t(int n){

     if(n == 1)

          return 1;

     else

          return n*n*t(n-1);

}

The function is named t. In the else clause, the function calls itself, so it is recursive.

Prove that the trajectory of a projectile is parabolic, having the form y = ax + bx2. To obtain this expression, solve the equation x = v0xt for t and substitute it into the expression for y = v0yt − 1 2 gt2. (These equations describe the x and y positions of a projectile that starts at the origin.) You should obtain an equation of the form y = ax + bx2 where a and b are constants.

Answers

Answer: y = v₀tgθx - gx²/2v₀²cos²θ

a = v₀tgθ

b = -g/2v₀²cos²θ

Step-by-step explanation:

x = v₀ₓt

y = v₀y.t - g.t²/2

x = v₀.cosθt → t = x/v₀.cosθ

y = v₀y.t - g.t²/2

v₀y = v₀.senθ

y = v₀senθ.x/v₀cosθ - g/2.(x/v₀cosθ)²

y = v₀.tgθ.x - gx²/2v₀²cos²θ

a = v₀tgθ → constant because v₀ and θ do not change

b = - g/2v₀²cos²θ → constant because v₀, g and θ do not change

Final answer:

The trajectory of a projectile is shown to be parabolic by substituting time from the x-direction motion equation into the y-direction motion equation and rearranging, yielding a parabolic form y = ax + bx², with constants determined by initial velocity and gravity.

Explanation:

To prove that the trajectory of a projectile is parabolic, we start with the equations of motion in the x and y directions for a projectile that starts at the origin. For the x-direction, we have x = v_0x t, where v_0x is the initial velocity component in the x-direction and t is the time.

In the y-direction, the equation is y = v_0y t - (1/2) g t², where v_0y is the initial velocity component in the y-direction and g is the acceleration due to gravity.

To find t from the x equation: t = x / v_0x. Substituting this into the y equation yields: y = v_0y (x / v_0x) - (1/2) g (x / v_0x)^2.

Now, simplifying we get: y = (v_0y / v_0x) x - (g/2 v_0x^2) x^2, which is of the form y = ax + bx². Here, a = v_0y/v_0x and b = -g/2 v_0x²are constants depending on the initial velocity components and acceleration due to gravity. The equation y = ax + bx² represents a parabolic path, confirming that the projectile's trajectory is indeed a parabola.

Find the cube root of 10 upto 5 signaficant figures by newton raphson method

Answers

Answer: The cube root of 10 is 2.1544 using an Xo value of -0.003723

Step-by-step explanation: The Newton-Raphson is a root finding method and its formula is NR: X=Xo-(f(x)/f'(x). Once you have the equation you also need to find the derivative of that equation before applying the formula. Since the problem stated that X =10, the method was applied to find the best root in order to find the cube root of 10 up to 5 significant figures. The best method is to use a software like Excel that helps you calculate those iterations faster. The root finding for this example was -0.003723.


You deposit $10,000 into a bank account at 2% annual interest. How long will it take for the $10,000 to compound to $30,000?

N= I/Y= PV= PMT= FV= P/Y=

Answers

Answer: time = 55.48 years

Explanation:

Given:

Principal amount = $10000

Interest rate = 2% p.a

Amount = $30000

We can evaluate the time taken using the following formula:

[tex]Amount=Principal(1+\frac{r}{100})^{t}[/tex]

[tex]30000 = 10000\times(1+\frac{2}{100})^{t}[/tex]

Solving the above equation, we get

[tex]1.02^{t} = 3[/tex]

Now taking log on both sides, we get;

[tex]t\ log(1.02) = log(3)[/tex]

time = 55.48 years

For what value(s) of k will the relation not be a function?
A = {(1.5k−4, 7), (−0.5k+8, 15)}

Answers

Answer:

K=6

Step-by-step explanation:

we know that

A function is: a relation from a set of inputs to a set of possible outputs where each input is related to exactly one output

so

In this problem

If the x-coordinate of the first point is equal to the x-coordinate of the second point, then the relation will not be a function

Remember that the x-coordinate is the input value and the y-coordinate is the output value

Equate the x-coordinates and solve for k

[tex]1.5k-4=-0.5k+8[/tex]

[tex]1.5k+0.5k=8+4[/tex]

[tex]2k=12[/tex]

[tex]k=6[/tex]

therefore

If the value of k is 6 the, the relation will not be a function

Final answer:

To determine when the relation A is not a function, set the first elements of the ordered pairs equal to each other. Solving for k, we find that k must equal 6 to make the relation not a function, as this value of k would result in the same input for both pairs, violating the definition of a function.

Explanation:

For a relation to be considered a function, each input value must be associated with exactly one output value. Looking at the provided relation A = {(1.5k-4, 7), (-0.5k+8, 15)}, we can see that there are two ordered pairs. For A to not be a function, the first element of both ordered pairs must be the same because this would mean that a single input is associated with two different outputs, which violates the definition of a function.

We set the first elements equal to each other to find the value of k that would make the relation not a function: 1.5k - 4 = -0.5k + 8. Solving this equation, we add 0.5k to both sides and add 4 to both sides to obtain 2k = 12, and then divide both sides by 2 to find that k = 6.

Therefore, when k equals 6, the relation A will not be a function because both ordered pairs will have the first element equal to 5, which corresponds to two different output values (7 and 15).

A closed rectangular container with a square base is to have a volume of 686 in3. The material for the top and bottom of the container will cost $4 per in2, and the material for the sides will cost $2 per in2. Find the dimensions of the container of least cost.

Answers

Answer:

Length = Width = 7 inches

Height = 14 inches

Step-by-step explanation:

Let's call  

x = length and width of the bottom and top

y = heigth

The bottom and top are squares, so their total area is

[tex]A_{bt}=x^2+x^2=2x^2[/tex]

the area of one side is xy. As we have 4 sides, the total area of the sides is

[tex]A_s=4xy[/tex]

The cost for the bottom and top would be

[tex]C_{bt}=\$4.2x^2=\$8x^2[/tex]

The cost for the sides would be

[tex]C_s=\$2.4xy=\$8xy[/tex]

The total cost to made the box is

[tex]C_t=\$(8x^2+8xy)[/tex]

But the volume of the box is  

[tex]V=x^2y=686in^3[/tex]

isolating

[tex]y=\frac{686}{x^2}[/tex]

Replacing this expression in the formula of the total cost

[tex]C_t(x)=8x^2+8x(\frac{686}{x^2})=8x^2+\frac{5488}{x}[/tex]

The minimum of the cost should be attained at the point x were the derivative of the cost is zero.

Taking the derivative

[tex]C'_t(x)=16x-\frac{5488}{x^2}[/tex]

If C' = 0 then

[tex]16x=\frac{5488}{x^2}\Rightarrow x^3=\frac{5488}{16}\Rightarrow x^3=343\Rightarrow x=\sqrt[3]{343}=7[/tex]

We have to check that this is actually a minimum.

To check this out, we take the second derivative

[tex]C''_t(x)=16+\frac{10976}{x^3}[/tex]

Evaluating this expression in x=7, we get C''>0, so x it is a minimum.

Now that we have x=7, we replace it on the equation of the volume to get

[tex]y=\frac{686}{49}=14in[/tex]

The dimensions of the most economical box are

Length = Width = 7 inches

Height = 14 inches

Final answer:

To find the dimensions of the container of least cost, assume the length of the base is x and the width of the base is also x. The height of the container is 686/(x^2). The dimensions of the container of least cost are approximately 10.33 inches by 10.33 inches by 6.68 inches.

Explanation:

To find the dimensions of the container of least cost, we need to minimize the cost function. Let's assume the length of the base is x, then the width of the base will also be x since it's a square. The height of the container will be 686/(x^2) since the volume is given as 686 in³. The total cost, C, is given by C = 4(2x^2) + 2(4x(686/(x^2))). Simplifying this expression gives C = 8x^2 + 5488/x. To find the minimum cost, we can take the derivative of C with respect to x and set it equal to zero:

dC/dx = 16x - 5488/x^2 = 0

Solving this equation gives x = √(343/2) ≈ 10.33. Therefore, the dimensions of the container of least cost are approximately 10.33 inches by 10.33 inches by 6.68 inches.

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Taxes reduce your paycheck by 22% each month. In an Excel spreadsheet, the salary earned for a month stored in cell A35. Write an Excel formula that would calculate the dollar amount of taxes.

a.
=A35*1.22

b.
=A35/0.22

c.
=A35*0.22

d.
=0.22/A35

e.
=A35/22

f.
None of the above.

Answers

Answer:  Option 'c' is correct.

Step-by-step explanation:

Since we have given that

The salary earned for a month stored in = A35

Rate of tax reduces by = 22%

We need to remove the % sign by dividing 22 by 100 and it becomes 0.22.

So, the dollar amount of taxes is given by

[tex]Taxes=A35\times \dfrac{22}{100}\\\\Taxes=A35\times 0.22[/tex]

Hence, Option 'c' is correct.

25% of all who enters a race do not complete. 30 haveentered.
what is the probability that exactly 5 are unable tocomplete the
race?

Answers

Answer:

The probability that exactly 5 are unable to complete the  race is 0.1047

Step-by-step explanation:

We are given that 25% of all who enters a race do not complete.

30 have entered.

what is the probability that exactly 5 are unable to complete the  race?

So, We will use binomial

Formula : [tex]P(X=r) =^nC_r p^r q^{n-r}[/tex]

p is the probability of success i.e. 25% = 0.25

q is the probability of failure =  1- p  = 1-0.25 = 0.75

We are supposed to find the probability that exactly 5 are unable to complete the  race

n = 30

r = 5

[tex]P(X=5) =^{30}C_5 (0.25)^5 (0.75)^{30-5}[/tex]

[tex]P(X=5) =\frac{30!}{5!(30-5)!} \times(0.25)^5 (0.75)^{30-5}[/tex]

[tex]P(X=5) =0.1047[/tex]

Hence the probability that exactly 5 are unable to complete the  race is 0.1047

Some conference-goers saunter over to the Healthy Snack Box Machine, where they each choose one of five kinds of fruit, one of three herbal teas, and one of six flavors of wrap sandwich to get packed in a box. How many possible snack boxes are there?

Answers

Answer: There are 90 snack boxes.

Step-by-step explanation:

Given : The number of kinds of fruits = 5

The number of kinds of herbal teas = 3

The number of kinds of flavors of wrap sandwich = 6

Then by using the fundamental principal of counting, the number of possible snack are there will be :_

[tex]5\times3\times6= 90[/tex]

Therefore, the number of possible snacks = 90

The lodhl diner offers a meal combination consisting of an appetizer, a soup, a main course, and a dessert. There are five appetizers, five soups, four main courses, and five desserts. Your diet restricts you to choosing between a dessert and an appetizer. (You cannot have both.) Given this restriction, how many three-course meals are possible?

Answers

Answer:   100

Step-by-step explanation:

Given : The lodhl diner offers a meal combination consisting of an appetizer, a soup, a main course, and a dessert.

There are 5 appetizers, 5 soups, 4 main courses, and 5 desserts.

Also, a dessert and a appetizer are not allowed to take together.

By Fundamental counting principal ,

Number of three-course meals with dessert and without appetizer :

[tex]5\times4\times5=100[/tex]      (1)

Number of three-course meals with appetizer and without dessert :

[tex]5\times5\times4=100[/tex]      (2)

Now, the number of meals with either dessert or appetizer :-

[tex]100+100=200[/tex]           [Add (1) and (2)]

Answer:

500

Step-by-step explanation:

5x5x4x5

we do

5x5=25

25x4=100

100x5= 500

A worker performs a repetitive assembly task at a workbench to assemble products. Each product consists of 25 components. Various hand tools are used in the task. The standard time for the work cycle is 7.45 min, based on using a PFD allowance factor of 15%. If the worker completes 75 product units during an 8-hour shift, determine the number of standard hours accomplished.

Answers

Answer:

9.3125 hours

Step-by-step explanation:

Given:

Number of components consisting in a product = 25

Standard time for work cycle = 7.45 minutes

or

standard time for work cycle = [tex]\frac{\textup{7.45}}{\textup{60}}[/tex] hours

Number of units completed = 75

Now,

The number of standard hours

= Number of units completed × standard time for the work cycle

=  [tex]75\times\frac{\textup{7.45}}{\textup{60}}[/tex] hours

or

The number of standard hours = 9.3125 hours

Answer:

The answer is 9.3125 hours :)

Step-by-step explanation:

How does remote work relate to taking an online class or being an online student (fully online or hybrid)?

Answers

Answer:

Answered

Step-by-step explanation:

Online courses are those classes which are delivered entirely online. Students study via web cam,  chat rooms and smart boards. Whereas hybrid learning is a combination of both online learning and  traditional in class learning.  Remote work is working away from the work place at your own comfort and choice of location.

If venom A is four times as potent as venom B and venom C is 2.5 times as potent as venom B. How many times as potent is venom C compared to venom A?

Answers

Answer:

Venom C is  0.625 times as potent as venom A.

Step-by-step explanation:

We are given that venom  A is four times as potent as venom B.

Venom C is 2.5 times as potent as Venom B.

We have to find that how many times as potent is venom C is compared to venom A.

Let Venom B =x

Then Venom A=[tex]4\times x=4x[/tex]

Venom C=[tex]2.5\times x=2.5 x[/tex]

[tex]\frac{Venom\;C}{venom\;A}=\frac{2.5x}{4x}=\frac{25}{40}=\frac{5}{8}=0.625[/tex]

Hence, venom C is  0.625 times as potent as venom A.

Final answer:

In relative terms, venom C is 0.625 times as potent as venom A, using venom B as a common comparison base.

Explanation:

To solve this, we first need to establish a common comparison base. If venom A is four times as potent as venom B, let's assign a relative potency value to venom B, let's say 1. Therefore, the potency of venom A is 4. Similarly, if venom C is 2.5 times as potent as venom B, then the potency of venom C is 2.5.

To find out how much more potent venom C is compared to venom A, we divide the potency of venom C by the potency of venom A. Mathematically, it looks like this: Potency of C ÷ Potency of A = 2.5 ÷ 4 = 0.625. Therefore, venom C is 0.625 times as potent as venom A.

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2. Perform the calculations 11540+5972 in BCD arithmetic.

Answers

Answer:

0001 0110 1110 1011 0010

Step-by-step explanation:

First decimal number = 11540

second decimal number = 5972

BCD of 0 to 9 decimal number

Decimal number                               BCD

       0                                                  0000

       1                                                    0001

       2                                                   0010

       3                                                   0011

       4                                                   0100

       5                                                   0101

       6                                                   0110

       7                                                    0111

       8                                                    1000

       9                                                    1001

Hence, we can write above two numbers in BCD form as

11540 => 0001 0001 0101 0100 0000

5972 => 0101 1001 0111 0010

So, sum of 11540 and 5972 in BCD form can be given by

       0001 0001 0101 0100 0000

        +       01 01 1001 01 11  001 0

      0001  01 10 11 10 10 11  001 0

So, the sum of 11540 and 5972 in BCD form is

                  0001 0110 1110 1011 0010

2-23 Ace Machine Works estimates that the probability its lathe tool is properly adjusted is 0.8. When the lathe is properly adjusted, there is a 0.9 probability that the parts produced pass inspection. If the lathe is out of adjustment, however, the probability of a good part being produced is only 0.2. A part randomly chosen is inspected and found to be acceptable. At this point, what is the posterior probability that the lathe tool is properly adjusted?

Answers

Answer:

The posterior probability that the lathe tool is properly adjusted is 94.7%

Step-by-step explanation:

This can be formulated as the following problem:

What is the probability of B happening, knowing that A has happened.

It can be calculated by the following formula

[tex]P = \frac{P(B).P(A/B)}{P(A)}[/tex]

Where P(B) is the probability of B happening, P(A/B) is the probability of A happening knowing that B happened and P(A) is the probability of A happening.

In your problem we have that:

-A is the probability that the part chosen is found to be acceptable.

The problem states that the probability its lathe tool is properly adjusted is 0.8. When it happens, there is a 0.9 probability that the parts produced pass inspection. There is also a 0.2 probability of  the lathe is out of adjustment, when it happens  the probability of a good part being produced is only 0.2.

So, P(A) = P1 + P2 = 0.8*0.9 + 0.2*0.2 = 0.72 + 0.04 = 0.76

Where P1 is the probability of a good part being produced when lathe tool is properly adjusted and P2 is the probability of a good part being produced when lathe tool is not properly adjusted.

- P(B) is the the probability its lathe tool is properly adjusted. The problem states that P(B) = 0.8

P(A/B) is the probability of A happening given that B has happened. We have that A is the probability that the part chosen is found to be acceptable and B is the probability its lathe tool is properly adjusted. The problem states that when the lathe is properly adjusted, there is a 0.9 probability that the parts produced pass inspection. So P(A/B) = 0.9

So, probability of B happening, knowing that A has happened, where B is the lathe tool is properly adjusted and A is that the part randomly chosen is inspected and found to be acceptable is:

[tex]P = \frac{P(B).P(A/B)}{P(A)} = \frac{0.8*0.9}{0.76} = \frac{0.72}{0.76} = 0.947 = 94.7%[/tex]

The posterior probability that the lathe tool is properly adjusted is 94.7%

Using Bayes' Theorem, the posterior probability that the lathe tool is properly adjusted given a part passes inspection is approximately 0.947 (94.7%). We determined this by combining the prior probabilities and the conditional probabilities to find the total probability of a part passing inspection.

To find the posterior probability, we can use Bayes' Theorem. Let's define the events:

A : Lathe tool is properly adjustedA' : Lathe tool is out of adjustmentB : Part passes inspection

Prior Probabilities

The probability that the lathe tool is properly adjusted is 0.8. Therefore, P(A) = 0.8 and P(A') = 0.2.

Conditional Probabilities

When the lathe is properly adjusted, the probability that a part passes inspection is P(B|A) = 0.9.When the lathe is out of adjustment, the probability that a part passes inspection is P(B|A') = 0.2.

Applying Bayes' Theorem:

Bayes' Theorem states:

[tex]\[P(A|B) = \frac{P(B|A) \cdot P(A)}{P(B)}\][/tex]

First, we need to find P(B), the total probability that a part passes inspection:

[tex]\[P(B) = P(B|A) \cdot P(A) + P(B|A') \cdot P(A')\][/tex]

Substituting values:

[tex]\[P(B) = 0.9 \cdot 0.8 + 0.2 \cdot 0.2 = 0.72 + 0.04 = 0.76\][/tex]

Now, using Bayes' Theorem, we can calculate P(A|B):

[tex]\[P(A|B) = \frac{0.9 \cdot 0.8}{0.76} = \frac{0.72}{0.76} \approx 0.947\][/tex]

Thus, the posterior probability that the lathe tool is properly adjusted given that a part is found to be acceptable is approximately 0.947 (94.7%).

Are the rational numbers closed under addition?

Answers

Answer:

Yes, rational numbers are closed under addition.

Step-by-step explanation:

Rational numbers are number that can be expressed in the form of fraction [tex]\frac{x}{y}[/tex], where x and y are integers and y ≠ 0.

Now, the closure property of addition of rational number states that if we add two rational number, then the sum of these two rational number will also be a rational number.

Let a and b be two rational number, then,

a+b = c, where c is the sum of and b

c is also a rational number.

Thus, rational numbers are closed under addition.

This can be explained with the help of a example.

[tex]\frac{1}{7} + \frac{2}{7}[/tex] = [tex]\frac{2}{7}[/tex]

It is clear that [tex]\frac{1}{7}, \frac{2}{7}, \frac{3}{7}[/tex] are rational numers.

Thus, rational numbers are closed under addition.

A lot of 50m by 36 m. A house is 31m by 9m built on the lot how much space is left over?

Answers

Answer:

Step-by-step explanation:

First you would find the area of the lot:

50 x 36 = 1800 m^2

Then you would find the area of the house:

31 x 9 = 279 m^2

Then you subtract the are of the lot minus the area of the house:

1800 - 279 = 1521 m^2

And the final answer is:

1521 square meters is left

The area of space which is left over is [tex]1521m^{2} [/tex]

Required steps shown below,

Area of lot [tex]=length*width[/tex]

                  [tex]=50*36=1800m^{2} [/tex]

Area of house[tex]=length*width[/tex]

                       [tex]=31*9=279m^{2} [/tex]

To calculate area of space is left over, subtract area of house from area of lot.

The space left over is,

                                [tex]1800-279=1521m^{2} [/tex]

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Determine the truth values of these statements

a) The product of x 2 and x 3 is x 6 .

b) 2π + 5π = 7π

c) x 2 > 0 for any real number.

d) The integer 315 − 8 is even.

e) The sum of two odd integers is even.

f) √ 2 ∈ Z

g) −1 ∈/ Z +

h) π ∈ Q

Answers

Answer:

a) False b) True c) False d ) False e) True f) False g) False h) False

Step-by-step explanation:

a) False

x² × x³ = [tex]x^5[/tex]

When we multiply two exponential number with same base, their powers add up.

b) True

2π + 5π = π(2+5) = 7π

The coefficients of π are added together.

c) False

x² ≥ 0. For any value of x, negative or positive x² is always positive. But for x = 0, x² = 0×0 = 0

d) False

315 - 8 = 307, which is clearly an odd number.

e)True

The sum of all integers is always even.

Let m and n be two odd integers.

Thus, they can be expressed as m = 2r + 1 and n = 2s +1, where r and s are even integers.

m + n = 2r +2s + 2, which is clearly even.

f) False

Since √2 is an irrational number. It cannot belong to z, which is collection of all integer number.

√2 ∉ z

g) False

Since -1 is a negative integer, it cannot belong to [tex]z^+[/tex], as it is collection of all positive integers.

h) False

π cannot belong to Q because Q is a collection of all rational numbers and π is not a rational number. The decimal expansion of π is non- terminating that is it does not end.

Use the binomial theorem to compute (2x-1)^5

Answers

Answer:

The expended form of the provided expression is: [tex]32x^5-80x^4+80x^3-40x^2+10x-1[/tex]

Step-by-step explanation:

Consider the provided expression.

[tex](2x-1)^5[/tex]

The binomial theorem:

[tex](a+b)^{n}=\sum _{r=0}^{n}{n \choose r}a^{n-r}b^r[/tex]

Where,

[tex]{n \choose r}= ^nC_r =\frac{n!}{(n-r)!r!}[/tex]

Now by using the above formula.

[tex]\frac{5!}{0!\left(5-0\right)!}\left(2x\right)^5\left(-1\right)^0+\frac{5!}{1!\left(5-1\right)!}\left(2x\right)^4\left(-1\right)^1+\frac{5!}{2!\left(5-2\right)!}\left(2x\right)^3\left(-1\right)^2+\frac{5!}{3!\left(5-3\right)!}\left(2x\right)^2\left(-1\right)^3+\frac{5!}{4!\left(5-4\right)!}\left(2x\right)^1\left(-1\right)^4+\frac{5!}{5!\left(5-5\right)!}\left(2x\right)^0\left(-1\right)^5[/tex]

[tex]2^5\cdot \:1\cdot \:1\cdot \:x^5-1\cdot \frac{2^4\cdot \:5x^4}{1!}+1\cdot \frac{2^3\cdot \:20x^3}{2!}-1\cdot \frac{2^2\cdot \:20x^2}{2!}+1\cdot \frac{5\cdot \:2x}{1!}+1\cdot \frac{\left(-1\right)^5}{\left(5-5\right)!}[/tex]

[tex]32x^5-80x^4+80x^3-40x^2+10x-1[/tex]

Hence, the expended form of the provided expression is: [tex]32x^5-80x^4+80x^3-40x^2+10x-1[/tex]


Given any integer m≥2, is it possible to find a sequence of m−1 consecutive positive integers none of which is prime? Explain your answer.

7

4

Answers

Final Answer:

No, it is not possible to find a sequence of m−1 consecutive positive integers none of which is prime for any integer m≥2.

Explanation:

In order to determine whether it is possible to find a sequence of m−1 consecutive positive integers none of which is prime, we need to consider the properties of prime numbers. A prime number is a natural number greater than 1 that has no positive divisors other than 1 and itself. Therefore, if we take any m−1 consecutive positive integers, at least one of them will be divisible by a prime number. This means that it is not possible to find a sequence of m−1 consecutive positive integers none of which is prime.

For example, let’s consider the sequence of 4 consecutive positive integers: 10, 11, 12, and 13. Among these numbers, 11 and 13 are prime. Even if we consider larger sequences, we will always encounter at least one prime number within the consecutive positive integers.

In conclusion, due to the nature of prime numbers and their distribution among positive integers, it is not possible to find a sequence of m−1 consecutive positive integers none of which is prime for any integer m≥2.

YP=+8+10+12+...+ 106 and Q=2+4+6+8+ ... 104 are sums of arithmetic sequences, determine which is greater, Por Q, and by how much

Answers

Final answer:

The sum of sequence P (YP) is greater than the sum of sequence Q (Q) by 94, with P summing to 2850 and Q summing to 2756.

Explanation:

To determine which sum is greater between P and Q, we need to first understand that P and Q represent the sum of arithmetic sequences. The first sequence, P, begins with 8 and increments by 2, ending at 106. The second sequence, Q, starts at 2 and increments by 2, ending at 104.

To find out the sums, we can use the formula for the sum of an arithmetic sequence: S = n/2 (a1 + an), where S is the sum, n is the number of terms, a1 is the first term, and an is the last term. For the sequence P, the first term a1 is 8, and the last term an is 106. To find n, the number of terms, we can use the formula n = (an - a1) / d + 1, where d is the common difference, which is 2 in this case.

Applying the formula to the sequence P, we get:

The number of terms in P: n(P) = (106 - 8) / 2 + 1 = 50

Sum of P: S(P) = 50/2 x (8 + 106) = 25 x 114 = 2850

For the sequence Q:

The number of terms in Q: n(Q) = (104 - 2) / 2 + 1 = 52

Sum of Q: S(Q) = 52/2 x (2 + 104) = 26 x 106 = 2756

Comparing these sums, we can see that the sum of sequence P, 2850, is greater than the sum of sequence Q, 2756. The difference between the sums is 2850 - 2756 = 94.

Therefore, P is greater than Q by 94.


What of the following basic rules is NOT true about geometry?

A. A straight line intersecting two parallel lines produces corresponding equal angles

B. Similar triangles have the same sides

C. Similar triangles have the same angles

D. The sum of the interior angles of any triangle equals 180°

E. None of the above

Answers

Answer:

A. A straight line intersecting two parallel lines produces corresponding equal angles.

Step-by-step explanation:

If two straight lines are parallel, and a straight line intersects them, then we get alternate angles equal, exterior angle equal to the opposite interior angle on the same side and the interior angles on the same sides equal to two right angles.

Hence, here the rule that is not true is : A. A straight line intersecting two parallel lines produces corresponding equal angles.

Compare the dotplot to a histogram of the data. (Select all that apply.)

(A) Histograms show the frequency of individual data values.
(B) The raw data can be retrieved from the dotplot, but not the histogram.
(C) Dotplots show the frequency of individual data values.
(D) The raw data can be retrieved from the histogram, but not the dotplot.

Answers

Final answer:

In summary, the raw data can only be retrieved from a dot plot, not a histogram, as the dot plot represents individual data points, whereas a histogram represents data in grouped intervals.

Explanation:

When comparing a dot plot to a histogram, there are several key points to consider:

(A) Histograms do not show the frequency of individual data values; instead, they show the frequency of data within a range of values.(B) The raw data can be retrieved from the dot plot, as each dot represents an individual data point. This is not the case with a histogram, where individual data points are not shown, and only grouped data within intervals are represented by bars.(C) Dotplots are used to show the frequency of individual data values, with each dot representing one occurrence of a data point.(D) You cannot retrieve raw data from a histogram as you can from a dot plot, because a histogram represents data in grouped intervals, not specific data points.

A frequency table or a frequency polygon can also be useful tools for data representation in complement to histograms and dot plots, especially when comparing distributions.

Sketch these four lines y = 2x+1, y =-x, and x = 0 and x = 2. Then use integrals to find the area of the region bounded by these lines. Finally, check your answer by computing this area using simple geometry.

Answers

Answer:

Area = 8

Step-by-step explanation:

A skecth is given in the attached file, there are two extra lines used to calculate the area with simple geometry:

We must use a double integral to obtain the area:

[tex]\int\limits^2_0 {\int\limits^b_a  \, dy } \, dx[/tex]

Where

b stands for y=2x+1

a stands for y=-x

Carring out the integrals we find the area:

[tex]\int\limits^2_0 {(2x+1 - (-x))} \, dx =  \int\limits^2_0 {3x+1} \, dx = (3x^{2}/2+x) \left \{ {{2} \atop {0}} \right.\\ A =( 3*2^{2}/2) + 2 =8[/tex]

Geometrically we can divide the area bounded by this lines as two triangles and a rectangle from the figure and the intersection of these lines we kno that the three figures have a base of 2. The heigth of the rectangle is 1 and for the triangles we have 4 for the upper triangle and 2 for the lower.

Therefore:

[tex]A = A_{upperT}+ A_{rect}+ A_{lowerT}[/tex]

and

[tex]A_{upperT}=2*4/2=4\\A_{rect}=2*1=2\\ A_{lowerT}=2*2/2=2[/tex]

Summing the four areas we have:

A=8

Greeting!

A consumer products company is formulating a new shampoo and is interested in foam height (in mm). Foam height is approximately normally distributed and has a standard deviation of 20 mm. The company wishes to test H0: μ = 175 mm versus H1: μ > 175 mm, using a random sample of n = 10 samples.(a) Find P-value if the sample average is = 185? Round your final answer to 3 decimal places.(b) What is the probability of type II error if the true mean foam height is 200 mm and we assume that α = 0.05? Round your intermediate answer to 1 decimal place. Round the final answer to 4 decimal places.(c) What is the power of the test from part (b)? Round your final answer to 4 decimal places.

Answers

Answer:

a) 0.057

b) 0.5234

c) 0.4766

Step-by-step explanation:

a)

To find the p-value if the sample average is 185, we first compute the z-score associated to this value, we use the formula

[tex]z=\frac{\bar x-\mu}{\sigma/\sqrt N}[/tex]

where

[tex]\bar x=mean\; of\;the \;sample[/tex]

[tex]\mu=mean\; established\; in\; H_0[/tex]

[tex]\sigma=standard \; deviation[/tex]

N = size of the sample.

So,

[tex]z=\frac{185-175}{20/\sqrt {10}}=1.5811[/tex]

[tex]\boxed {z=1.5811}[/tex]

As the sample suggests that the real mean could be greater than the established in the null hypothesis, then we are interested in the area under the normal curve to the right of  1.5811 and this would be your p-value.

We compute the area of the normal curve for values to the right of  1.5811 either with a table or with a computer and find that this area is equal to 0.0569 = 0.057 rounded to 3 decimals.

So the p-value is  

[tex]\boxed {p=0.057}[/tex]

b)

Since the z-score associated to an α value of 0.05 is 1.64 and the z-score of the alternative hypothesis is 1.5811 which is less than 1.64 (z critical), we cannot reject the null, so we are making a Type II error since 175 is not the true mean.

We can compute the probability of such an error following the next steps:

Step 1

Compute [tex]\bar x_{critical}[/tex]

[tex]1.64=z_{critical}=\frac{\bar x_{critical}-\mu_0}{\sigma/\sqrt{n}}[/tex]

[tex]\frac{\bar x_{critical}-\mu_0}{\sigma/\sqrt{n}}=\frac{\bar x_{critical}-175}{6.3245}=1.64\Rightarrow \bar x_{critical}=185.3721[/tex]

So we would make a Type II error if our sample mean is less than 185.3721.  

Step 2

Compute the probability that your sample mean is less than 185.3711  

[tex]P(\bar x < 185.3711)=P(z< \frac{185.3711-185}{6.3245})=P(z<0.0586)=0.5234[/tex]

So, the probability of making a Type II error is 0.5234 = 52.34%

c)

The power of a hypothesis test is 1 minus the probability of a Type II error. So, the power of the test is

1 - 0.5234 = 0.4766

Final answer:

The p-value is 0.014, indicating strong evidence against the null hypothesis. The probability of a type II error is 0.0907, and the power of the test is 0.9093.

Explanation:

To find the p-value, we need to determine the probability of observing a sample average of 185 or higher, given that the true mean foam height is 175. Since the sample size is small, we'll use the t-distribution instead of the normal distribution. With a sample size of 10, the degrees of freedom is 9. Using the t-distribution table or a calculator, we find the p-value to be 0.014 (rounded to 3 decimal places).

The probability of a type II error is the probability of failing to reject the null hypothesis when it is actually false. In this case, the null hypothesis is μ = 175, but the true mean foam height is 200. We assume α = 0.05, so the critical value is 1.645 (from the t-distribution table for a one-tailed test). Using the formula for the standard error of the sample mean, σ/√n, we can calculate the standard deviation of the sample mean to be 20/√10 = 6.32 (rounded to 2 decimal places). The difference between the critical value and the true mean is (200 - 175)/6.32 = 3.96 (rounded to 2 decimal places). Using a t-distribution table or calculator to find the area to the right of 3.96 with 9 degrees of freedom, we find the probability of a type II error to be 0.0907 (rounded to 4 decimal places).

The power of the test is 1 minus the probability of a type II error. So the power of the test is 1 - 0.0907 = 0.9093 (rounded to 4 decimal places).

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In △ABC, CD is an altitude, such that AD = BC. Find AC, if AB = 3 cm, and CD = 3 cm.

PLEASE ANSWER !!!!!

Answers

Final answer:

To find the length of AC in the right-angled isosceles triangle △ADC with altitude CD and AD equals to BC, we use the Pythagorean theorem, substituting known lengths to solve for AC, which is √11.25 cm.

Explanation:

The student is asking to find the length of side AC in a triangle △ABC where CD is an altitude and AD is equal to BC, with given lengths AB = 3 cm and CD = 3 cm.

To solve for AC, we can use the Pythagorean theorem which states that for a right triangle, the square of the length of the hypotenuse (the side opposite the right angle) is equal to the sum of the squares of the lengths of the other two sides. Since CD is an altitude and AD = BC, △ADC is a right-angled isosceles triangle.

Using the Pythagorean theorem in △ADC, we have:

AD2 + CD2 = AC2

Since AD = BC and CD = 3 cm, let's assume AD = x cm. Then,

x2 + 32 = AC2

x2 + 9 = AC2

But, AB = 3 cm, and AB = AD + DB = x + x = 2x,

Therefore, x = AB/2 = 1.5 cm. Substituting this value back into the previous equation:

(1.5)2 + 9 = AC2 = 2.25 + 9 = 11.25

AC = √11.25 cm

Therefore, the length of AC is √11.25 cm.

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