Find the greatest common divisor of 252 and 60

Answer:

208.837209

Step-by-step explanation:

Data provided in the question:

The output of 35920 / 172 on calculator = 208.837209

Now,

All the digits or figures that are non-zero are considered as significant figures.

also, the number zero between any two consecutive number is considered as significant figure.

Thus,

for the given output the all the non-zero digits are significant, also the zeros are between the two consecutive number

Hence, the correct answer is  208.837209

Answer:

624 tourists only visited the Magic Kindgom.

Step-by-step explanation:

To solve this problem, we must build the Venn's Diagram of this set.

I am going to say that:

-The set A represents the tourists that visited LEGOLAND

-The set B represents the tourists that visited Universal Studios

-The set C represents the tourists that visited Magic Kingdown.

-The value d is the number of tourists that did not visit any of these parks, so: [tex]d = 74[/tex]

We have that:

[tex]A = a + (A \cap B) + (A \cap C) + (A \cap B \cap C)[/tex]

In which a is the number of tourists that only visited LEGOLAND, [tex]A \cap B[/tex] is the number of tourists that visited both LEGOLAND and Universal Studies, [tex]A \cap C[/tex] is the number of tourists that visited both LEGOLAND and the Magic Kingdom. and [tex]A \cap B \cap C[/tex] is the number of students that visited all these parks.

By the same logic, we have:

[tex]B = b + (B \cap C) + (A \cap B) + (A \cap B \cap C)[/tex]

[tex]C = c + (A \cap C) + (B \cap C) + (A \cap B \cap C)[/tex]

This diagram has the following subsets:

[tex]a,b,c,d,(A \cap B), (A \cap C), (B \cap C), (A \cap B \cap C)[/tex]

There were 1,168 tourists suveyed. This means that:

[tex]a + b + c + d + (A \cap B) + (A \cap C) + (B \cap C) + (A \cap B \cap C) = 1,168[/tex]

We start finding the values from the intersection of three sets.

The problem states that:

16 tourists had visited all three theme parks. So:

[tex]A \cap B \cap C = 16[/tex]

91 tourists had visited both LEGOLAND and Universal Studios. So:

[tex](A \cap B) + (A \cap B \cap C) = 91[/tex]

[tex](A \cap B) = 91-16[/tex]

[tex](A \cap B) = 75[/tex]

68 tourists had visited both the Magic Kingdom and Universal Studios. So

[tex](B \cap C) + (A \cap B \cap C) = 68[/tex]

[tex](B \cap C) = 68-16[/tex]

[tex](B \cap C) = 52[/tex]

87 tourists had visited both the Magic Kingdom and LEGOLAND

[tex](A \cap C) + (A \cap B \cap C) = 87[/tex]

[tex](A \cap C) = 87-16[/tex]

[tex](A \cap C) = 71[/tex]

295 tourists had visited Universal Studios

[tex]B = 295[/tex]

[tex]B = b + (B \cap C) + (A \cap B) + (A \cap B \cap C)[/tex]

[tex]295 = b + 52 + 75 + 16[/tex]

[tex]b + 143 = 295[/tex]

[tex]b = 152[/tex]

266 tourists had visited LEGOLAND

[tex]A = 266[/tex]

[tex]A = a + (A \cap B) + (A \cap C) + (A \cap B \cap C)[/tex]

[tex]266 = a + 75 + 71 + 16[/tex]

[tex]a + 162 = 266[/tex]

[tex]a = 104[/tex]

How many tourists only visited the Magic Kingdom (of these three)?

This is the value of c, the we can find in the following equation:

[tex]a + b + c + d + (A \cap B) + (A \cap C) + (B \cap C) + (A \cap B \cap C) = 1,168[/tex]

[tex]104 + 152 + c + 74 + 75 + 71 + 52 + 16 = 1,168[/tex]

[tex]c + 544 = 1,168[/tex]

[tex]c = 624[/tex]

624 tourists only visited the Magic Kindgom.

Answer:

The equation is an differential equation of second order.

The dependent variable is x, while t is the independent variable.

Step-by-step explanation:

The order of the equation depends on the greatest grade of the derivative, in this case it's the second derivative (x'')

Since x is a function of t, we would have that t is the independent variable while x is the dependent variable.

Final answer:

The given equation is mx'' + bx' + kx = 8t^6. The order of the equation is 2 and the independent and dependent variables are t and x, respectively. The parameters in the equation are m, b, and k.

Explanation:

The given equation is mx'' + bx' + kx = 8t^6, where x is a function of t. Let's break down the equation:

Therefore, the order of the equation is 2, as it contains the second derivative of x.

Answer: The tan 130° is expressed as [tex]\dfrac{\sqrt{1-a^2}}{a}[/tex]

Step-by-step explanation:

Since we have given that

[tex]\cos 50^\circ=a[/tex]

As we know that

cos (π - θ ) = -cos θ

so, cos(180-50)=-cos 130° = -a

so, sin 130° would become

[tex]\sqrt{1-(-a)^2}\\\\=\sqrt{1-a^2}[/tex]

So, tan 130° is given by

[tex]\dfrac{\sin 130^\circ}{\cos 130^\circ}\\\\=\dfrac{\sqrt{1-a^2}}{a}[/tex]

Hence, the tan 130° is expressed as [tex]\dfrac{\sqrt{1-a^2}}{a}[/tex]

To express tan(130 degrees) in terms of cos(50 degrees), use the identities tan(180 - x) = -tan(x) and sin²(x) + cos²(x) = 1. tan(130 degrees) = -tan(50 degrees) simplifies to -√(1 - a²) / a. Therefore, tan(130 degrees) in terms of a is -√(1 - a²) / a.

Given that cos(50 degrees) = a, we need to express tan(130 degrees) in terms of a.

First, recall the relationship between the cosine and tangent of angles. We know that:

Therefore, tan(130 degrees) can be written as:

Next, using the identity for tangent in terms of cosine, we have:

Since sin²(x) + cos²(x) = 1, we can express sin(50 degrees) as:

Thus,

Finally, substituting back, we get:

Therefore, tan(130 degrees) in terms of a is -√(1 - a²) / a.

Answer: A) Representative

Explanation: A representative study is the study which is regarding the a cluster of people that has near about features that is similar to the most of the people of population. The characteristic in the cluster of people are considered as common which display the accurate sample features for the study.

The case mentioned in the question is a representative study case because it includes the residents of the country which tend to has closely matching features of the whole population of the country.

Answer:

a)There are [tex]7,2236*10^{26}[/tex] possible different passwords.

b)There are [tex]3,0784*10^{19}[/tex] possible different passwords.

Step-by-step explanation:

a) Each character of the string could be any of the 10 numerals or 21 letters, that means, each character may be any of these 31 different options.

If the password had only one character, there would be 31 different possible passwords, if it were a two characters string, there would be [tex]31^{2}=961[/tex], because in the first place we can place any of the 31 characters, and then in the second place, we can do the same, we are asked of combinations without taking into account order of the characters.

A generalized form, to know how many different passwords we can make of n characters is [tex]31^{n}[/tex].

If we have more than one possible lengths, we add those cases, in our case, we have 14 thru 18 characters, all that could be filled by any of the 31 possible characters.

[tex]\[\sum_{i=14}^{18}31^{i} =31^{14}+31^{15} +31^{16}+31^{17}+31^{18}=7,2236*10^{26}     \\\][/tex].

b) If each password must contain at least 15 numerals, that means the minimum length of a password is 15 now.

We proceed the same as the a) point, but taking into account that for 15 characters, there is not 31 possible characters anymore, but 10.

Doing each individual length:

[tex]Passwords_{length=15} =10^{15}[/tex]

That is because we would multiply 10 possible options by the next possible 10 options and so on 15 times.

[tex]Passwords_{length=16} =10^{15}*31[/tex]

Now, for the sixteenth character, we multiply by the now 31 possible options.

[tex]Passwords_{length=17} =10^{15}*31^{2} \\Passwords_{length=18} =10^{15}*31^{3}[/tex]

For lengths 17 and 18, we do similar stuff, multiplying by 31 for each new character.

To know all the possible combinations, we add the combinations for each length:

[tex]\sum_{i=15}^{18}Passwords_{lenght=i}=3,0784*10^{19}[/tex]

We can see there is approximately 7 orders of magnitude less with the restriction than there is without it.

Answer:

Robert can give writing supplies to at most 7 campers.

Step-by-step explanation:

The problem states that Robert must give the same number of supplies to each camper and use all the supplies. It means that the greatest number of campers that Robert can give writing supplies is the greatest common divisor(gcd) between the number of pencils and the number of note pads.

The gcd between two integers is the largest positive number that divides each of the integers. We can find this value by prime factorization.

The problem states that he has 14 pencils and 21 note pads. So we have to find gcd(14,21).

21 is not divisible by 2, so we try factoring by 3

14 is not divisible by 3, so we try factoring by 5

None of them are divisible by 5, so we move to 7

Both are divisible by 7, so

14 - 21 | 7

2  - 3

2<7, 3<7, so gcd(14,21) = 7.

Robert can give writing supplies to at most 7 campers.

Answer:

a) 56x = 16800 - 14p  

b) $1200

c) 300 pounds

Step-by-step explanation:

Given:

At p₁ = $900 ; x₁ = 75 pounds

at p₂ = $956 ; x₂ = 75 - 14 = 61 pounds

Now,

from the standard equation of line, we have

[tex](x - x_1)=\frac{(x_2-x_1)}{(p_2-p_1)}\times(p-p_1)[/tex]

on substituting the respective values, we get

[tex](x - 75)=\frac{(61-75)}{(956-900)}\times(p-900)[/tex]

or

( x - 75 ) × 56 = -14p + 12600

or

56x - 4200 = -14p + 12600

or

56x = 16800 - 14p        (relation between the unit price p and demand x)

b) For no consumers x = 0

thus, substituting in the relation we get

56 × 0 = 16800 - 14p

or

14p = 16800

or

p = $1200

c) For free , p = $0

on substituting in the above relation derived, we get

56x = 16800 - ( 14 × 0 )

or

x = 300 pounds

Answer:

134

Step-by-step explanation:

to find how many gifts it can make, you must find how many times 34 can go into 4580. do this by deciding 4580 by 34.

4580/34=134.7

Since you can't make less than a whole gift you must round it down to 134

Jordan can make 134 number of gifts with 34 M & M s in each with 4580 personalized M and M s.

Jordan wants to gift the volunteer so he is making the gifts for volunteers.

For gifting purpose the number of personalized M and M s she ordered is given by = 4580.

The number of personalized M and M s she puts in each volunteer 's gift is given by = 34.

So the number of total gifts she can pack with this number of personalized M and M s given by = 4580/34 = 134.7 = 134 approximately.

So Jordan can make 134 gifts with 34 M and M s in each with 4580 personalized M and M s.

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3x+3x-5= 1

6x-5=1

Whenever moving a number, the sign always changes.

6x-5+5= 1+5

6x= 1+5

6x= 6

divide both sides by 6

6x/6= 6/6

x= 1

Check solution by using the substitution method

3(1)+3(1)-5=1

3+3-5=1

6-5= 1

1=1

Answer: x=1

Answer is provided in the image attached.

Answer:

p1. If I pay attention to class, therefore I'll take good grades on test next week.

p2. I like cheese.

Step-by-step explanation:

Propositions are statements that declare something from/for someone. It always states something. It may be classified as simple (p1) or combined (p2) depending on the presence or absence of logical connectors p1 (If...therefore). A combined proposition is made up of two simple propositions.

We can and ought work with symbolic operators.

Let p1 be translated into symbolic language,

I pay attention to class = q

I'll take good grades on test next week r

q→r

p2= I like cheese. We can just simply call it p2.

Answer:

Probability of rain on vacation = 0.7875

Step-by-step explanation:

Given,

So,

Probability of rain on Friday,P(F) = 0.1

Probability of rain on Saturday, P(S) = 0.25

Probability of rain on Sunday, P(T)= 0.3

Probability of rain on both Friday and Saturday, P(F∩S)= 0.1×0.25

                                                                                           = 0.025

Probability of rain on both Friday and Saturday, P(S∩T)=0.25×0.3

                                                                                           = 0.075

Probability of rain on both Friday and Saturday, P(T∩F)=0.3×0.1

                                                                                           =0.03

Probability of rain on whole vacation, P(F∩S∩T)=0.1×0.25×0.3

                                                                               = 0.0075

Probability that there will be rain on vacation,

P(A)= P(F)+P(S)+P(T)+P(F∩S)+P(S∩T)+P(T∩F)+P(F∩S∩T)

     = 0.1+0.25+0.3+0.025+0.075+0.03+0.0075

     = 0.7875

Hence, the probability that there will be rain on vacation is 0.7875.

Final answer:

The probability that it will rain at least once on a weekend with varying rain chances each day cannot be found by adding probabilities. To estimate this, calculate the combined chance of no rain throughout the weekend and subtract it from 100%. The result for the given percentages is a 52.75% chance of rain during the weekend.

Explanation:

Understanding Probability in Weather Forecasts

When looking at the chance of rain during a long weekend with different percentages each day, we cannot simply add the probabilities to find the overall chance of rain. Instead, the best method to estimate the probability of it raining at least once during the weekend is to calculate the probability that it does not rain on any of the days and subtract this from 100%.

For the individual chances of no rain: Friday (90%), Saturday (75%), and Sunday (70%), we multiply these probabilities together to find the cumulative chance of no rain all weekend, which gives us: 0.9 * 0.75 * 0.7 = 0.4725, or 47.25%. Thus, the probability of it raining at least once during the weekend is 1 - 0.4725 = 0.5275, or 52.75%.

Addressing the incorrect statements:

a. A 60% chance of rain on Saturday and a 70% chance on Sunday does not result in a 130% chance over the weekend. Probabilities cannot exceed 100%, indicating that this statement is erroneous.

b. The probability that a baseball player hits a home run cannot be directly compared to the probability of getting a hit without knowing specific statistics. Home runs are a subset of hits, so naturally, the chance of any hit is higher than a home run specifically.

Answer:

3 Numbers: 15, 21, 35

Step-by-step explanation:

We want number that divide 109 with a remainder of 4.

Thus, it can divide of 109 - 4 = 105

Factors of 105 = 3, 5, 7

Thus for getting two digit number it must be Multiple of any two factors of 105.

i.e. 15, 21 and 35

Hence there are only 3 numbers that divide 109 with a remainder of 4.

Answer:

15, 21, &35

Step-by-step explanation:

109-4=105

Factorize 105:

1x105

3x35

5x21

7x15

1, 3, 5, & 7 are all one digits numbers and 105 is a three digit number.

The 3 two digit numbers are 15, 21, & 35.

Answer:

The probability that for the second sample of 500 college students, the mean number of hours worked will be less than 14.6 is 0.6554

Step-by-step explanation:

The sampling distribution of the sample mean is given by a normal distribution with mean [tex]\mu[/tex] and variance [tex]\frac{\sigma^2}{n}[/tex], where [tex]\mu[/tex] is the mean and [tex]\sigma^2[/tex] is the variance of the population that generates the data. In this way the random variable;

[tex]Z=\frac{\bar x - \mu_{\bar x}}{\sigma_{\bar x}}[/tex] is a standard normal variable. As [tex]\bar {x}-\mu_{\bar x} = 0.4\sigma_{\bar x}[/tex], then [tex]Z = 0.4[/tex].

[tex]P (X <14.6) = P (Z <0.4) = 0.6554[/tex]

The probability that the mean number of hours worked per week in the second sample of 500 students will be less than 14.6 hours is approximately [tex]\( {0.345} \).[/tex]

To solve this problem, we need to understand the relationship between the sample mean, the population mean, and the standard deviation of the sampling distribution. Here’s the breakdown of the steps needed:

1. Determine the z-score for the sample mean in the first sample:

The mean number of hours worked per week in the first sample is 14.6, which lies 0.4 standard deviations below the mean of the sampling distribution. This means the z-score is -0.4.

2. Find the corresponding probability:

The z-score tells us how many standard deviations away from the mean our sample mean is. We need to find the probability that a second sample will have a mean number of hours worked that is less than 14.6.

3. Use the standard normal distribution:

The z-score formula for the sampling distribution is given by:

[tex]\[ z = \frac{\bar{x} - \mu}{\sigma_{\bar{x}}} \][/tex]

where [tex]\(\bar{x}\)[/tex] is the sample mean, [tex]\(\mu\)[/tex] is the population mean, and [tex]\(\sigma_{\bar{x}}\)[/tex] is the standard error of the mean.

In this case, we know that the mean of the sampling distribution [tex](\(\mu_{\bar{x}}\))[/tex] is such that:

[tex]\[ z = \frac{14.6 - \mu_{\bar{x}}}{\sigma_{\bar{x}}} = -0.4 \][/tex]

4. Determine the probability:

To find the probability that the mean number of hours worked in the second sample is less than 14.6, we look up the z-score of -0.4 in the standard normal distribution table or use a cumulative distribution function (CDF) for the normal distribution.

The z-score of -0.4 corresponds to a cumulative probability (or area to the left of z) of approximately 0.3446.

Answer:

The period of the [tex]\sin[/tex] function is [tex]2\pi[/tex]. Then three times this period is [tex]6\pi[/tex].

Step-by-step explanation:

Using Mathematica you can use the command Plot as follows:

```

Plot[Sin[x], {x,0,6Pi}]

```

The output is the graph shown below.

Using Mathematica, you can plot a Sin function that covers a range three times its period by using the Plot[] function and specifying the range of x-values from 0 to 6*Pi. This results in a plot from x=0 to x=6π, a range that covers three full periods of the Sin function.

To begin, once you've installed and opened Mathematica, you'll want to make use of the Plot[] function to indeed plot the function. The Sin function is periodic and has a period of 2π. If you'd like to plot a function that covers a range three times this period, you'd want a range of 6π. The code to achieve this is as follows:

Plot[Sin[x], {x, 0, 6*Pi}]

In this code, 'Sin[x]' is the function you want to plot and '{x, 0, 6*Pi}' defines the range of x-values over which to plot the function. The result is a plot of the Sin function from x=0 to x=6π, which covers three full periods of the Sin function.

As a side note, if you wanted to create a scatter plot or calculate a best-fit line as you might do in a statistics application, you would use different commands within Mathematica or possibly use a different software or calculator that specializes in statistical analysis, such as a TI-83/84 calculator. However, these operations are not needed to simply plot a sin function.

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Answer:  The required inverse of the given matrix is

[tex]P^{-1}=\left[\begin{array}{ccc}a&b&c\\d&e&f\\g&h&i\end{array}\right].[/tex]

Step-by-step explanation:  We are given to find the inverse of the following orthogonal matrix :

[tex]P=\left[\begin{array}{ccc}a&d&g\\b&e&h\\c&f&i\end{array}\right] .[/tex]

We know that

if M is an orthogonal matrix, then the inverse matrix of M is the transpose of M.

That is, [tex]M^{-1}=M^T.[/tex]

The transpose of the given matrix P is given by

[tex]P^T=\left[\begin{array}{ccc}a&b&c\\d&e&f\\g&h&i\end{array}\right].[/tex]

Therefore, according to the definition of an orthogonal matrix, the inverse of matrix P is given by

[tex]P^{-1}=P^T=\left[\begin{array}{ccc}a&b&c\\d&e&f\\g&h&i\end{array}\right].[/tex]

Thus, the required inverse of the given matrix is

[tex]P^{-1}=\left[\begin{array}{ccc}a&b&c\\d&e&f\\g&h&i\end{array}\right].[/tex]

Answer: There are 1820 sets of four marbles other than lavender.

Step-by-step explanation:

Since we have given that

Number of red marbles = 3

Number of green marbles = 5

Number of lavender marbles = 1

Number of yellow marbles = 2

Number of orange marbles = 6

So, Total number of marbles = 3 + 5+ 1 + 2 + 6 = 17

We need to find the sets of 4 marbles other than lavender.

so, Number of total marbles other than lavender  becomes = 17 -1 =16

Number of marbles in a set = 4

So, Number of ways becomes

[tex]^{16}C_4\\\\=1820[/tex]

Hence, there are 1820 sets of four marbles other than lavender.

Answer:

[tex]\$21.1[/tex]

Step-by-step explanation:

We know that for principal amount P , time period T and rate of interest [tex]R\%[/tex] , simple interest is given by [tex]S.I. = \frac{P\times R\times T}{100}[/tex] .

Here ,

[tex]P=\$1600\\T=2\,\,months=\frac{2}{12}\,\,years=\frac{1}{6}\,\,years\\S.I=\$56.24[/tex]

To find :  simple interest rate i.e., [tex]R\%[/tex]

On putting values of [tex]P\,,\,T\,,\,S.I[/tex] in formula , we get [tex]S.I. = \frac{P\times R\times T}{100}[/tex]

[tex]56.24 = \frac{1600\times R\times 1}{600}\\R=\frac{56.24\times 600 }{1600}=\frac{703\times 3}{100}=\$21.09[/tex]

Now we need to round off the answer to the nearest tenth .

So, simple interest rate is % = [tex]\$21.09[/tex] = [tex]\$21.1[/tex]

Answer:  The required age of Nate is 4 years.

Step-by-step explanation:  Given that Alice is 3 years old and Nate is half of the sum of Alice's age and 5.

We are to find the age of Nate.

Let x and y represents the ages of Alice and Nate respectively in years.

Then, according to the given information, we have

[tex]x=3.[/tex]

and also,

[tex]y=\dfrac{1}{2}(x+5)\\\\\\\Rightarrow y=\dfrac{1}{2}(3+5)\\\\\\\Rightarrow y=\dfrac{1}{2}\times8\\\\\Rightarrow y=4.[/tex]

Thus, the required age of Nate is 4 years.

Statistics are divided into two main branches: the descriptive one that, as the name implies, seeks to describe the data through mathematical processes that allow them to be analyzed, find numerical patterns, representative data, group values ​​and summarize the information; the other branch is the inferential, which seeks to make estimates or predictions, inferences, on a set of data obtained, usually samples, making use of probabilities and distributions.

Answer

The study described is a descriptive study since it performs a series of mathematical processes on a set of data to be able to analyze them, find representative patterns and data, describe their behavior and make a comparison.

Historian Jane Flaherty's research on the U.S. Treasury's state before the Civil War is a descriptive statistical study, summarizing the surplus/deficit data from 1854 to 1861 to depict the financial condition at the time.

The study by historian Jane Flaherty that researched the condition of the U.S. Treasury on the eve of the Civil War represents a descriptive statistical study. This type of study involves summarizing and describing aspects of a specific set of information. Flaherty used information about the annual surplus/deficit figures from the periods 1854-1857 under President Franklin Pierce and 1858-1861 under President James Buchanan to illustrate the financial state of the Union prior to Abraham Lincoln's presidency. The data clearly indicate a shift from consistent surpluses during Pierce's term to substantial deficits under Buchanan, which contributed to the exhausted condition of the Treasury as the nation approached the Civil War. Descriptive statistics, like those presented, help to paint a historical picture without making predictive claims or inferences about other data sets or future outcomes.

Answer:

1) False 2) True 3) True 4) True

Step-by-step explanation:

1)FALSE

We can prove this by giving a counterexample,

Take the arithmetic sequence  

[tex]\left \{ a_1,a_2,a_3,... \right \}[/tex]

where  

[tex]a_n=(n-1)-15[/tex]

in this case d=1

Then

[tex]a_1+a_2+...+a_12=-15-14-...-4<0[/tex]

2)TRUE

Given that for an arithmetic sequence

[tex]a_n=a_1+(n-1)d[/tex]

Where d is a constant other than 0, then

[tex]\lim_{n \to \infty}a_n\neq 0 [/tex]

and so, the series  

[tex]\sum_{n=1}^{\infty}a_n[/tex]

diverges.

3)TRUE

This is the definition of infinite sum.

If [tex]S_n=a_1+a_2+...+a_n[/tex]

then [tex]\sum_{n=1}^{\infty}a_n=\lim_{n \to \infty}S_n[/tex]  

4)TRUE

If  

[tex]\left \{ a_1,a_2,a_3,... \right \}[/tex]

is a geometric sequence, then the n-th partial sum is given by

[tex]S_n=\frac{a_1r^n-a_1}{r-1}[/tex]

Since r<1

[tex]\lim_{n \to\infty}r^n=0[/tex]

and so, the geometric series

[tex]\sum_{n=1}^{\infty}a_n=\lim_{n \to\infty}S_n=\frac{a_1}{1-r}[/tex]

The statements about arithmetic and geometric sequences and series are evaluated for truthfulness, illustrating key concepts of convergence and divergence in series within mathematics.

Let's examine each statement regarding sequences and series in mathematics:

Answer:

(a) 450 minutes

(b) 275 minutes

(c) 725 minutes

Step-by-step explanation:

Given,

Number of units in one batch = 25

move time for raw materials to Cutting Department per unit =6 minutes

Time taken by Cutting Department per unit = 4 minutes

move time to Sewing Department per unit = 3 minutes

Time taken by sewing Department per unit = 10 minutes

move time to Packaging Department per unit = 2 minutes

time taken by packaging Department per unit = 4 minutes

(a) value added time for one unit = Time taken by Cutting Department per unit

                                                    +Time taken by sewing Department per unit

                                                +time taken by packaging Department per unit

                                                = (4 + 10 + 4) minutes

                                                 = 18 minutes

Value added time for 25 units = 18 x 25

                                                  = 450 minutes

(b) within batch wait time per unit =move time for raw materials to Cutting                                                                                department per unit +move time to Sewing Department per unit +move time to Packaging Department per unit

                                                       = (6+3+2) minutes

                                                        = 11 minutes

Total within batch wait time = 11 x 25

                                               = 275 minutes

(c) total lead time for batch = Value added time for 25 units

                                                  +Total within batch wait time

                                              = 450+ 275

                                               = 725 minutes

Answer:

[tex]S=\frac{cos(x-\frac{\alpha}{2})-cos(x+n\alpha-\frac{\alpha}{2})}{2sin\frac{\alpha}{2}}[/tex]

Step-by-step explanation:

We are given that [tex]S=sin(x) +sin(x+\alpha)+sin(x+2\alpha)+....+sin(x+n\alpha),n\in N[/tex]

We have to find the value of S

We know that

[tex]\sum_{k=0}^{n-1}sin(x+k.d)=\frac{sinn\times \frac{d}{2}}{sin\frac{d}{2}}\times sin(\frac{2x+(n-1)d}{2})[/tex]

We have d=[tex]\alpha[/tex]

Substitute the values then we get

[tex]\sum_{k=0}^{n-1}sin(x+k.\alpha)=\frac{sin\frac{n\alpha}{2}}{sin\frac{\alpha}{2}}\times sin(\frac{2x+(n-1)\alpha}{2})[/tex]

[tex]\sum_{k=0}^{n-1}sin(x+k.\alpha)=\frac{sin\frac{n\alpha}{2}\cdot sin(\frac{2x+(n-1)\alpha}{2})}{sin\frac{\alpha}{2}}[/tex]

[tex]S=\frac{sin\frac{n\alpha}{2}\cdot sin(\frac{2x+(n-1)\alpha}{2})}{sin\frac{\alpha}{2}}[/tex]

[tex]S=\frac{2sin\frac{n\alpha}{2}\cdot sin(\frac{2x+(n-1)\alpha}{2})}{2sin\frac{\alpha}{2}}[/tex]

[tex]S=\frac{cos(x+\frac{n\alpha}{2}-\frac{\alpha}{2}-\frac{n\alpha}{2})-cos(x+\frac{n\alpha}{2}-\frac{\alpha}{2}+\frac{n\alpha}{2})}{2sin\frac{\alpha}{2}}[/tex]

Because [tex]cos(x-y)-cos(x+y)=2 sinxsiny[/tex]

[tex]S=\frac{cos(x-\frac{\alpha}{2})-cos(x+n\alpha-\frac{\alpha}{2})}{2sin\frac{\alpha}{2}}[/tex]

Answer:

The line under the symbols will have the same effect for the subset and proper subset symbols and the less than and less than or equal symbols.

By concept, a proper subset is the set that will have some but not all of the values of a given set, for example:

Imagine we had the sets:

A={a,e,i,o,u}

B={a,o,u}

C={a,e,i,ou}

we can say that

B⊂A   (B is a proper subset of A)

but we cannot say that:

C⊂A

because B has some elements of A, while C has all the elements in A, so C is not a proper subset of A.

Now, a subset can contain some or all of the elements contained in another set.

we can for sure say that:

B⊆A   and also that C⊆A

Because C has all the elements of A, so it fits into the subset definition.

Comparing this to the < and ≤ symbols, the < symbol means that a value will be less than another value. This doesn't include the greater value, for

example, we can say that:

2<5

but we cannot say that 5<5 because they are both the same. That statement is false.

On the other hand, the ≤ stands for, less than or equal to. This symbol can be used when a number is less than another one or equal to it, for example, we can say that:

2≤5 and we can also say that 5≤5 because they are the same and the symbol does include the actual value of 5.

So as you may see, the relation is that the line under the symbol includes the values or sets while if the symbols don't have a line under them, this means that the greater value or the original set is not to be included.

The symbols for subset (⊆) and proper subset (⊂) in sets are similar to ≤ and < in numbers, respectively. A subset includes the possibility of equality, while a proper subset does not, analogous to how ≤ and < function with numbers.

The symbols for subset (⊆) and proper subset (⊂) in set theory are analogous to the symbols for ≤ (less than or equal to) and < (less than) for numbers. A subset can be thought of as ≤ because it includes the possibility of being equal to the set it is compared with (similar to how 5 ≤ 5 is true). A proper subset is like < because it does not include the set itself; it must be a strict part of the set, excluding equality (as 5 < 6 is true but 5 is not < 5).

Example: Let's consider two sets A = {1, 2, 3} and B = {1, 2, 3, 4}. Here, A is a proper subset of B, which we denote as A ⊂ B, akin to stating A < B if they were numbers since A does not contain all elements of B. However, if A was {1, 2, 3, 4}, then A ⊆ B, similar to A ≤ B in number terms because A contains all elements of B.

Answer with Step-by-step explanation:

Let A and B be non- empty bounded subset of R

a.We have to find why [tex]sup(A\cup B)[/tex]exist

If A and B are bounded set

Then there exist  constant  such that

[tex]a\leq A\leq b[/tex] and [tex]c\leq B\leq d[/tex]

Then , sup of A =b and sup of B=d

When  a set is bounded then all elements lie in the set are lie between the constants s and t.

All elements are less than or equal to t then t is supremum of set.

Because both set are bounded and sup of both set A and B are exist.All elements A union B are less than or equal to sup A or sup B.

[tex]sup(A\cup B)=max(sup A, sup B)[/tex]

Then, [tex]sup (A\cup B)[/tex] exist.

b.We have to prove that

[tex] sup (A\cup B)=max(sup A,sup B)[/tex]

Suppose ,A =(1,2) and B=(2,3)

Sup A=2 , sup B=3

[tex](A\cup B)=(1,2)\cup (2,3)[/tex]

Upper bound of [tex]A\cup B)=3[/tex]

Hence, [tex]Sup (A\cup B)=3[/tex]

If A=(4,5),B=(2,3)

Sup A=5,Sup B=3

[tex]A\cup B=(4,5)\cup (2,3)[/tex]

[tex]Sup(A\cup B)=5[/tex]

Hence, [tex]Sup(A\cup B)=5[/tex]

Hence, we can say that [tex]sup(A\cup B)=max(sup A,sup B)[/tex].

Answer:

90% of people marry there 7th grade love. since u have read this, u will be told good news tonight. if u don't pass this on nine comments your worst week starts now this isn't fake. apparently if u copy and paste this on ten comments in the next ten minutes you will have the best day of your life tomorrow. you will either get kissed or asked out in the next 53 minutes someone will say i love you

Step-by-step explanation:

Final answer:

To express 160 pounds (lbs) in kilograms (kg), multiply 160 pounds by the conversion factor of 1 kilogram to 2.205 pounds. The result is approximately 72.6 kilograms.

Explanation:

To express 160 pounds (lbs) in kilograms (kg), we can use the conversion factor that 1 kg is equal to 2.205 pounds (lb). We can set up a proportion to solve for the weight in kilograms:

160 lb = x kg

1 kg = 2.205 lb

By multiplying both sides of the equation by 1 kg, we get:

160 lb * 1 kg / 2.205 lb = x kg

Simplifying the expression, we find that x is approximately 72.6 kg when rounded to the nearest hundredths.

The result is approximately 72.56 kg when rounded to the nearest hundredths place. To convert 160 pounds to kilograms, you divide by 2.205.

To convert pounds (lbs) to kilograms (kg), you can use the conversion factor where 1 kg is approximately equal to 2.205 lbs.

Given:

Substituting the given value:

Rounding to the nearest hundredths place, we get:

Thus, 160 pounds is approximately equal to 72.56 kilograms.

Answer:

h =  v/lw .

Step-by-step explanation:

v=lwh

Divide both sides by lw:

v  / lw = h.

Answer:

H.''n is not divisible by 6 and n is not divisible by both 2 and 3.

Step-by-step explanation:

We are given that  a statement ''n is divisible by 6 or n is divisible by both 2 and 3.''

We have to write the negation of the  given statement.

Negation: If  a statement p is true then its negations is  p is false.

n is divisible by 6 then negation is n is not divisible by 6.

n is divided by both 2 and 3 then negation is n is not divisible  by both 2 and 3.

Therefore, negation of given statement

''n is not divisible by 6 and n is not divisible by both 2 and 3.

Hence, option H is true.

Answer:

H.''n is not divisible by 6 and n is not divisible by both 2 and 3.

Step-by-step explanation:

We are given that  a statement ''n is divisible by 6 or n is divisible by both 2 and 3.''

We have to write the negation of the  given statement.

Negation: If  a statement p is true then its negations is  p is false.

n is divisible by 6 then negation is n is not divisible by 6.

n is divided by both 2 and 3 then negation is n is not divisible  by both 2 and 3.

Therefore, negation of given statement

''n is not divisible by 6 and n is not divisible by both 2 and 3.

Hence, option H is true.

Step-by-step explanation:

Answer:

a) 6561

b) 3024

c) 1296

Step-by-step explanation:

Given : Using the digits 1 through 9.

To find : The number of different 4-digit numbers such that :

(a) Digits can be used more than once.

(b) Digits cannot be repeated. 2 .

(c) Digits cannot be repeated and must be written in increasing order.

Solution :

Digits are 1,2,3,4,5,6,7,8,9

We have to form different 4-digit number let it be _ _ _ _

(a) Digits can be used more than once.

For first place there are 9 possibilities.

For second place there are 9 possibility as number repeats.

Same for third and fourth we have 9 possibility.

The number of ways are [tex]9\times 9\times 9\times 9=6561[/tex]

(b)  Digits cannot be repeated.

For first place there are 9 possibilities.

For second place there are 8 possibility as number do not repeats.

For third place there are 7 possibility as number do not repeats.

For fourth place there are 6 possibility as number do not repeats.

The number of ways are [tex]9\times 8\times 7\times 6=3024[/tex]

c) Digits cannot be repeated and must be written in increasing order.

The number which we can use on first position are 1,2,3,4,5,6 i.e. 6

The number which we can use on second position are 2,3,4,5,6,7 i.e. 6

The number which we can use on third position are 3,4,5,6,7,8 i.e. 6

The number which we can use on fourth position are 4,5,6,7,8,9 i.e. 6

Total number of ways are [tex]6\times 6\times 6\times 6=1296[/tex]

There are 6561 different 4-digit numbers when digits can be used more than once, 3024 different 4-digit numbers when digits cannot be repeated, and 3024 different 4-digit numbers when digits cannot be repeated and must be written in increasing order.

Answer:

The function tends to negative infinity

Step-by-step explanation:

The Y and X axes in this case are asymptotes, it means that the function will never touch them. When x is negative and is so small, the function tends to negative infinity, because the function try to cut it but it will never happen.

:)

Answer:

F(x) is a negative number with a large absolute value

Step-by-step explanation:

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