find the laplace transformation of g(t) = 5te^-5t Us (t) use laplace transforms theorms g

Answer:

number of subsets of a set with 13 elements are: [tex]2^{13}[/tex]

Step-by-step explanation:

In order to solve this intuitively, we can start by a set with lesser elements. This will reveal a pattern that will be used to solve for the subsets of the 13 element set.

If we start with a set B. which contains only 3 elements.

[tex]B = \{1,2,3\}[/tex]

how many subsets of B are there? well we can count them. [the set containing {1,2} and {2,1} are the same, arrangement doesn't matter]

[tex]B_{0} = \{\}\\B_{1a}=\{1\}\\B_{1b}=\{2\}\\B_{1c}=\{3\}\\B_{2a}=\{1,2\}\\B_{2b}=\{2,3\}\\B_{2c}=\{3,1\}\\B_{3a}=\{1,2,3\}\\[/tex]

there are a total of 9 subsets here.

Similarly, if you try a with a subset with only two elements you'll find that it has a total of 4 subsets.

We can see that combinatorics is at play here.

for the set B. the number of subsets can be written as:

[tex]\text{\# of subsets of B} = ^3C_0+^3C_1+^3C_2+^3C_3\\\text{\# of subsets of B} = 1+3+3+1\\\\text{\# of subsets of B} = 8[/tex]

if we try with a 2-element set:

[tex]\text{\# of subsets} = ^2C_0+^2C_1+^2C_2\\\text{\# of subsets} = 1+2+1\\\ \text{\# of subsets} = 4[/tex]

We can use the same technique to find the number of subsets of the 13 element set.

But if you recognize a pattern here that this sets of combinations are actually part of the pascal triangle, the sum of each row of the triangle is 2^{the row's number}. hence.

[tex]\text{\# of subsets of B} = 2^3\\\ \text{\# of subsets of B} = 8[/tex]

So finally, the subsets of a 13-element set A will be

[tex]\text{\# of subsets of A} = ^{13}C_0+^{13}C_1+^{13}C_2+^{13}C_3\cdots+^{13}C_{12}+^{13}C_{13}\\OR\\\text{\# of subsets of A} = 2^{13}\\\text{\# of subsets of A} = 8192[/tex]

If the set A has 13 elements, the number of different subsets is [tex]2^{13}=8192[/tex]

All the possible subsets that can be formed from any given set is called the Power set of that set. Generally, if we had a set [tex]H[/tex] such that

[tex]|H|=k[/tex]

Where [tex]|H|[/tex] denotes the cardinality, or number of elements, in [tex]H[/tex], the power set of [tex]H[/tex], denoted by [tex]P(H)[/tex], has the following formula

[tex]P(H)=2^k\text{ elements}[/tex]

So, given the set [tex]A[/tex] such that

[tex]|A|=13[/tex]

the power set of [tex]A[/tex] will have [tex]2^{13} \text{ or } 8192 \text{ elements}[/tex]

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Answer:

its 5000

Step-by-step explanation:

for it to be 50000 like the other person said it would be asking what is 0.5 not 0.05 like it says in the question

Step-by-step explanation:

The largest whole number can be 9,999,999,999 if it is  decimal representation is simple. The device with me  supports up to 10 ^ 100 exponents, so that 9.9999999E99 could be a candidate too. If your exponents are not limited, then 9E9999999 is the largest (above what the calculator can demonstrate).

Answer:

At least two of the restaurant have different mean delivery time.

Step-by-step explanation:

The ANOVA known as Analysis of variance involves the hypothesis testing of means on the basis of variances. The null hypothesis in ANOVA is taken as the equality mean i.e. H0: All means are equal. Whereas alternative hypothesis consists of at least two means are not equal.

The problem states that a pizza lover is comparing average delivery time. The null and alternative hypothesis in this case would be

H0: All restaurant have equal mean delivery time.

H1: At least two restaurant have different mean delivery time.

In an ANOVA comparing four local pizza restaurants' delivery times, the alternative hypothesis is that at least two restaurants have different mean delivery times.

When performing an ANOVA with data comparing the average delivery times for four local pizza restaurants, the alternative hypothesis is that at least two of the restaurants have different mean delivery times. This is because ANOVA tests the null hypothesis that all group means are equal against the alternative that at least one group mean is different. Specifically, the null hypothesis for a one-way ANOVA test with four groups is μ1 = μ2 = μ3 = μ4, where each μ represents the mean delivery time of a different restaurant. The alternative hypothesis, which the student is inquiring about, is Ha: μi ≠ μj for some i ≠ j.

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Answer:

a) 38.3% probability that an 18-year-old man selected at random is between 68 and 70 inches tall.

b) 95.44% probability that the mean height x is between 68 and 70 inches.

Step-by-step explanation:

To solve this question, it is important to know the normal probability distribution and the Central Limit Theorem.

Normal probability distribution

Problems of normally distributed samples can be solved using the z-score formula.

In a set with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the zscore of a measure X is given by:

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

Central Limit Theorem

The Central Limit Theorem estabilishes that, for a random variable X, with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], a large sample size can be approximated to a normal distribution with mean [tex]\mu[/tex] and standard deviation [tex]\frac{\sigma}{\sqrt{n}}[/tex].

In this problem, we have that:

[tex]\mu = 69, \sigma = 2[/tex]

(a) What is the probability that an 18-year-old man selected at random is between 68 and 70 inches tall?

This is the pvalue of Z when X = 70 subtracted by the pvalue of Z when X = 68. So

X = 70

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

[tex]Z = \frac{70 - 69}{2}[/tex]

[tex]Z = 0.5[/tex]

[tex]Z = 0.5[/tex] has a pvalue of 0.6915.

X = 68

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

[tex]Z = \frac{68 - 69}{2}[/tex]

[tex]Z = -0.5[/tex]

[tex]Z = -0.5[/tex] has a pvalue of 0.3085.

So there is a 0.6915 - 0.3085 = 0.383 = 38.3% probability that an 18-year-old man selected at random is between 68 and 70 inches tall.

(b) If a random sample of sixteen 18-year-old men is selected, what is the probability that the mean height x is between 68 and 70 inches?

Now we use the Central Limit Theorem, with [tex]n = 16, s = \frac{2}{\sqrt{16}} = 0.5[/tex]

The probability is also the pvalue of Z when X = 70 subtracted by the pvalue of Z when X = 68, but with s as the standard deviation. So

X = 70

[tex]Z = \frac{X - \mu}{s}[/tex]

[tex]Z = \frac{70 - 69}{0.5}[/tex]

[tex]Z = 2[/tex]

[tex]Z = 2[/tex] has a pvalue of 0.9772.

X = 68

[tex]Z = \frac{X - \mu}{0.5}[/tex]

[tex]Z = \frac{68 - 69}{0.5}[/tex]

[tex]Z = -2[/tex]

[tex]Z = -2[/tex] has a pvalue of 0.0228.

So there is a 0.9772 - 0.0228 = 0.9544 = 95.44% probability that the mean height x is between 68 and 70 inches.

In the case of normal distribution, the probability that a random 18-year-old man's height is between 68 and 70 inches is 0.383, while the probability that the mean height of a random sample of 16 men falls in the same range is approximately 0.999.

This question pertains to the statistical concept of normal distribution.

(a) To find the probability that a randomly selected 18-year-old man is between 68 and 70 inches tall, we first need to convert these heights to Z-scores. Given the mean height is 69 inches and the standard deviation is 2 inches, the Z-score for 68 inches is [tex](68-69)/2=-0.5[/tex] and for 70 inches it is [tex](70-69)/2=0.5[/tex]. The area between these Z-scores on a standard normal distribution chart represents the probability of a man being between 68 and 70 inches tall. This probability is approximately 0.383.

(b) In this part, we are dealing with a sample mean rather than individual values. Because of the Central Limit Theorem, a distribution of sample means will have a standard deviation equal to the population standard deviation divided by the square root of the sample size. So, our new standard deviation becomes 2/√16 = 0.5. Then we compute the Z-scores for 68 inches and 70 inches with this new standard deviation, and find the probability corresponding to the area between these Z-scores, which is considerably higher than individual case, almost 0.999.

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Answer:

The probability that we have a fair coin is 0.2717

Step-by-step explanation:

Let Ci be the event that the coin i is used for while we let i = 1,2,3 (representing the three coin)

We also let H be the event that the coin which is flipped lands (which is head)

Therefore, the problem arise:

That:  P(H/C1) = 1/2.............fair coin

          p(H/C2) = 2/3..........Chance of head

          p(H/C3) = 2/3............Chance of tail

Now, noting that the coin was picked at random,

We have, : p(Ci) = P(C1) = P(C2) = P(C3) = 1/3

We can then say that or calculate thus:

P( C2 | H ) = P (H ∩ C2) ÷ P(H)

Where P (H ∩ C2) means the probability of event intersection

= P(H ∩ C2) ÷ P(H ∩ C1) + P(H ∩ C2) + P(H ∩ C3)

= P(H | C2) P(C2) ÷ P(H | C1) P(C1) + P(H | C2) P(C2) + P(H | C3) P(C3)

= (1 / 2) (1 / 3) ÷  (1/2)(1/3) + (2/3)(1/3) + (2/3) (1/3)

0.5 × 0.3 ÷ (0.5 × 0.3) + (0.67 × 0.3) + (0.67 × 0.3)

0.15 ÷ 0.15 + 0.201 + 0.201

=0.15 / 0.552

= 0.2717

Answer:

A.positive and fairly strong

Step-by-step explanation:

Since an increase in food quantity usually means an increase in price, quantity and price are directly proportional, which configures a positive correlation.

Since it is stated that this relationship in observed at most restaurants, it can be concluded that there is a fairly strong correlation between  the amount of food served per person and the price paid for it.

Therefore, the answer is, A.positive and fairly strong

Answer:

The answer is A which is Positive and fairly strong.

Step-by-step explanation:

To properly do justice to the selected answer above, we describe a simple scenario.

Imagine yourself, content and perhaps a bit over-full after a lovely meal at a local restaurant. Then, the extravagant bill arrives.

Does  this high cost affirm your belief that this meal was valuable and thereby  influence your reordering of it?

Or does the cost of the meal overshadow your  enjoyment of it and leave you wishing you had chosen a simple meal at a better price point?

What features must an expensive restaurant  provide you over a bargain one to justify the extra cost?

Based of the question asked in respect to the scenario above, researchers went on a an experiment collecting different data from a restaurant and used the Latent Dirichlet Allocation (LDA) model to classify each review of the data and key words like ( e.g. food, service, price, ambiance, anecdotes, miscellaneous) were paid attention to.

After analysis, it was perceived that the value is such a tricky parameter to measure, marketers,  restaurateurs and economists often overlook it, instead focusing on objective  restaurant price and quality’s effect on customer satisfaction.

Answer:

a) There is a 63.02% probability that no one has done an one time fling.

b) There is a 36.98% probability that at least one person has done a one time fling

c) There is a 99.15% pprobability that no more than two people have done a one time fling.

Step-by-step explanation:

For each adult, there are only two possible outcomes. Either they have done an one time fling, or they have not. This means that we use the binomial probability distribution to solve this problem.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]

In which [tex]C_{n,x}[/tex] is the number of different combinatios of x objects from a set of n elements, given by the following formula.

[tex]C_{n,x} = \frac{n!}{x!(n-x)!}[/tex]

And p is the probability of X happening.

In this problem we have that:

[tex]n = 9, p = 0.05[/tex]

A. no one has done a one time fling

This is P(X = 0).

[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]

[tex]P(X = 0) = C_{9,0}.(0.05)^{0}.(0.95)^{9} = 0.6302[/tex]

There is a 63.02% probability that no one has done an one time fling.

B. at least one person has done a one time fling

Either no one has done a one time fling, or at least one person has. The sum of the probabilities of these events is decimal 1.

So

[tex]P(X = 0) + P(X \geq 1) = 1[/tex]

[tex]P(X \geq 1) = 1 - `(X = 0) = 1 - 0.6302 = 0.3698[/tex]

There is a 36.98% probability that at least one person has done a one time fling

c. no more than two people have done a one time fling

This is

[tex]P(X \leq 2) = P(X = 0) + P(X = 1) + P(X = 2)[/tex]

[tex]P(X = 1) = C_{9,1}.(0.05)^{1}.(0.95)^{8} = 0.2985[/tex]

[tex]P(X = 2) = C_{9,2}.(0.05)^{2}.(0.95)^{7} = 0.0628[/tex]

[tex]P(X \leq 2) = P(X = 0) + P(X = 1) + P(X = 2) = 0.6302 + 0.2985 + 0.0628 = 0.9915[/tex]

There is a 99.15% pprobability that no more than two people have done a one time fling.

In a group of 9 adults, the probability that no one has done a 'one-time fling' is 0.630, the probability that at least one person has done it is 0.370, and the probability that no more than two people have done it is approximately 0.853.

The question here involves probability given certain conditions. The probability of an event happening can be found by dividing the number of times the event occurs by the total number of occurrences. Here we have the probability of the 'one-time fling' happening as 5% or 0.05, the complement of which (it not happening) is 1 - 0.05 = 0.95.

A. The probability of none out of nine friends having done a one-fling time is (0.95)^9 = 0.630. So, there’s a 0.630 chance that none of them have done a one-time fling.

B. The probability of at least one person having done a one-time fling would be the complement of no one having done it, which equals 1 - (the probability of no one having done it), 1 - 0.630 = 0.370.

C. The probability of no more than two people having done a one-time fling is found by adding the probabilities of exactly zero, one, and two persons having done it. Using the binomial probability formula, this would amount to 0.630 (for 0 person) + 9(0.05)*(0.95)^8 (for one person) + 36*(0.05)^2*(0.95)^7 (for two people) = 0.853, approximately.

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Answer:

A researcher studies length of time in college, first through fourth year, and its relation to academic motivation. To get the most detail out of her measures, she assesses each student in both the fall and spring semesters of each of their four years in school. She finds that students have increasingly higher motivation from their first semester to their seventh semester (the start of their fourth year), with a trailing off in the last semester. The independent variable are:

Step-by-step explanation:

Answer:

Step-by-step explanation:

i) Confidence level = 90%

ii) therefore significance level = [tex]\alpha[/tex]  =  [tex]\frac{100 - 90}{100}\hspace{0.1cm} = \hspace{0.1cm} 0.1[/tex]

iii) this problem is for a two tailed test.

iv) therefore significance level = [tex]\frac{\alpha }{2}[/tex] = [tex]\frac{0.1}{2}[/tex] = 0.05

iii) sample size = 15

iv) therefore degrees of freedom  = sample size  - 1   =   15 - 1    =  14

v) the upper ( or right) chi squared value, [tex]\chi^2_{R}[/tex]  from Chi squared critical value tables = 23.685

The correct critical value χR is 23.685.

To find the critical value χR corresponding to a sample size of 15 and a confidence level of 90%, we first determine the degrees of freedom.

The degrees of freedom (df) for a chi-square distribution is calculated as n - 1, where n is the sample size. Here, n = 15, so df = 14.

Next, look up the critical value for the chi-square distribution with 14 degrees of freedom and a 90% confidence level.

In statistical tables, the value corresponding to a 90% confidence level (or equivalently, 10% significance level, which leaves 5% in each tail of the distribution) for 14 degrees of freedom is 23.685.

Therefore, the correct critical value χR is 23.685.

Answer:

[tex]f(t) = -0.025t + 0.61[/tex]

In which t is measured in decades.

Step-by-step explanation:

A linear function f(t) has the following format:

[tex]f(t) = at + b[/tex]

We are given two points of the function, so we can solve a system of equations to find the values for a and b.

61% of all adults in 1971 (t = 0)

This means that [tex]f(0) = 0.61[/tex]

So

[tex]f(t) = at + b[/tex]

[tex]0.61 = a*(0) + b[/tex]

[tex]b = 0.61[/tex]

51% in 2011 (t = 4)

This means that [tex]f(4) = 0.51[/tex]

So

[tex]f(t) = at + 0.61[/tex]

[tex]0.51 = 4a + 0.61[/tex]

[tex]4a = -0.10[/tex]

[tex]a = -0.025[/tex]

So the linear function is:

[tex]f(t) = -0.025t + 0.61[/tex]

In which t is measured in decades.

Hi, you haven't provided the system of linear equations that you need to solve. Therefore, I'll just explain how to use Gauss-Jordan in a system of equations and you can apply the same method to the system of equations you have.

Answer with explanation and step by step solution:

1. For the system of equations:

[tex]4X_{1} + 8X_{2} + 12X_{3} = 36\\8X_{1} + 10X_{2} + 12X_{3} = 48\\4X_{1} + 14X_{2} + 24X_{3} = 60\\[/tex]

2. We can represent it as a matrix by placing every number of the equation as follow:  

[tex]\left[\begin{array}{ccccc}4&8&12&|&36\\4&5&6&|&24\\2&7&12&|&30\end{array}\right][/tex]

3. As you can see all the coefficients in the equation are divisible by two, so we can express the system of equations as follow:  

 [tex]\left[\begin{array}{ccccc}2&4&6&|&18\\4&5&6&|&24\\2&7&12&|&30\end{array}\right][/tex]

4. Gauss-Jordan method solves the system of equations by applying simple operations to the Matrix: Multiplication by non-zero numbers, adding a multiple of one row to another and swapping rows.

Step by step solution:  

Divide both sides of equation one by two:

[tex]\left[\begin{array}{ccccc}1&2&3&|&9\\4&5&6&|&24\\2&7&12&|&30\end{array}\right][/tex]

Subtract two times the equation two to the equation three:  

[tex]\left[\begin{array}{ccccc}1&2&3&|&9\\4&5&6&|&24\\-6&-3&0&|&-18\end{array}\right][/tex]

Divide equation number three by minus three and subtract two times the equation one to equation two:

[tex]\left[\begin{array}{ccccc}1&2&3&|&9\\2&1&0&|&6\\2&1&0&|&6\end{array}\right][/tex]

Subtract the equation two to the equation three:  

[tex]\left[\begin{array}{ccccc}1&2&3&|&9\\2&1&0&|&6\\0&0&0&|&0\end{array}\right][/tex]

Because now we have two equations for three unknown values X1, X2 and X3 the system has an infinite number of solutions.  

Equivalente system (From matrix to equation notation):  

[tex]1X_{1} + 2X_{2} + 3X_{3} = 9\\2X_{1} + 1X_{2} = 6\\[/tex]

Conclusion:  

For whatever system you have you need to convert the system into a matrix notation and using the basic operations, described here, reduce the complexity of the system until:  

You have a solution, you discover that the system has an infinite number of solutions or the system of equation is inconsistent.  

Example of inconsistency

If after making the basic operations to your system you get a result like this

[tex]\left[\begin{array}{ccccc}7&0&4&|&9\\2&1&0&|&6\\0&0&0&|&-1\end{array}\right][/tex]  

You can say that the system is inconsistent because zero is not equal to minus one.  

Example of solution  

If after making the basic operations to your system you get a result like this

[tex]\left[\begin{array}{ccccc}1&0&0&|&9\\0&1&0&|&-6\\0&0&1&|&-1\end{array}\right][/tex]

You can say that the system have a solution in which X1 = 9, X2 = -6 and X3 = -1

To solve a system of linear equations using Gaussian elimination with back-substitution or Gauss-Jordan elimination, follow the steps of writing the equations in matrix form, performing row operations, creating a diagonal of 1's, and using back-substitution to find the variable values.

To solve a system of linear equations using Gaussian elimination with back-substitution or Gauss-Jordan elimination, follow these steps:

Example: Solve the system of equations 2x + 3y = 8 and 4x - 2y = 10 using Gaussian elimination with back-substitution.

The solution to the system of equations is x = 23/8 and y = 3/4.

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Answer:

If a 1% level of significance is used to test a null hypothesis, there is a probability of ____ less than 1%______ of rejecting the null hypothesis when it is true

Step-by-step explanation:

Given that a hypothesis testing is done.

Level of significance used is 1%

i.e. alpha = 1%

When we do hypothesis test, we find out test statistic Z or t suitable for the test and find p value

If p value is < 1% we reject null hypothesis otherwise we accept null hypothesis.

So p value can be atmost 1% only for accepting null hypothesis.

So the answer is 1%

If a 1% level of significance is used to test a null hypothesis, there is a probability of ____less than 1%______ of rejecting the null hypothesis when it is true.

Answer:

The question is not so clear and complete

Step-by-step explanation:

But for questions like this, since the equation has been given, what is expected is for us to make comparison, compare the RHS with the LHS or by method of comparing coefficients.

We follow the stated conditions since we are told that b and c are both integers which are greater than 1 and b is less than the product of cb. from these conditions, we can compare and get the values of b , c and d.

Another approach is to assume values, make assumptions with the stated conditions, however, our assumptions must be valid and correct if we substitute the assumed values of b, c and d in the equation, it must arrive at the same answer for the RHS. i.e LHS = RHS

Answer:

[tex]y = \frac{\sqrt{2}x}{3} - \frac{\sqrt{2}\pi}{3} + 1.5[/tex]

Step-by-step explanation:

The equation to the line normal to the curve has the following format:

[tex]y - y(x_{0}) = m(x - x_{0})[/tex]

In whicm m is the derivative of y at the point [tex]x_{0}[/tex]

In this problem, we have that:

[tex]x_{0} = \pi[/tex]

[tex]y(x) = 3\cos{\frac{x}{3}}[/tex]

[tex]y(\pi) = 3\cos{\frac{\pi}{3}} = \frac{3}{2}[/tex]

The derivative of [tex]\cos{ax}[/tex] is [tex]a\sin{ax}[/tex]

So

[tex]y(x) = 3\cos{\frac{x}{3}}[/tex]

[tex]y'(x) = 3*\frac{1}{3}\sin{\frac{x}{3}} = \sin{\frac{x}{3}}[/tex]

[tex]m = \sin{\frac{\pi}{3}} = \frac{\sqrt{2}}{3}[/tex]

The equation of the line normal to the curve of y=3cos1/3x is:

[tex]y - y(x_{0}) = m(x - x_{0})[/tex]

[tex]y - \frac{3}{2} = \frac{\sqrt{2}}{3}(x - \pi)[/tex]

[tex]y = \frac{\sqrt{2}}{3}(x - \pi) +  \frac{3}{2}[/tex]

[tex]y = \frac{\sqrt{2}x}{3} - \frac{\sqrt{2}\pi}{3} + 1.5[/tex]

Answer:the total discounted cost for your stay is $1948.2

Step-by-step explanation:

If you book your room six months in advance, you will save 15% off the price of the room. You decide to take advantage of the offer and book a room for 6 nights. If the original price per night for the room is $382, this means that the value of the discount would be

15/100 × 382 = 0.15 × 382 = $57.3

The cost of the room for a night would be

382 - 57.3 = $324.7

Since you are staying for six nights,

the total discounted cost for your stay would be

324.7 × 6 = $1948.2

Answer:

Step-by-step explanation:

The relative frequencies are 0.15, 0.08, 0.39, and 0.38 for the respective classes while 38% of the observations are at least 300 but less than 350.

The cumulative relative frequency distribution given can be used to construct a relative frequency distribution. To find the relative frequency for each class, subtract the cumulative frequency of the previous class from the cumulative frequency of the current class. Therefore:

The Total relative frequency is the sum of all the relative frequencies and should equal 1 (or almost equal to 1 due to rounding differences)

For the second part of your question: 'what percent of the observations are at least 300 but less than 350?', since this is the relative frequency distribution, it means that class 300 to 350 has already been expressed as a percentage (since relative frequency is a proportion and can be expressed as a percentage). So, 38% of the observations are at least 300 but less than 350.

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Answer:

[tex]r=\sqrt{6/sinucosu}[/tex]

Step-by-step explanation:

To convert to polar form:

[tex]x=rcosu[/tex]

[tex]y=rsinu[/tex]

We substitute into our function:

[tex]y=6/x[/tex]

[tex]rsinu=6/rcosu[/tex]

multiply both sides by r:

[tex]r^2sinu=6/cosu[/tex]

solve for r by dividing by sinu:

[tex]r=\sqrt{6/sinucosu}[/tex]

Answer: Normal approximation can be used for discrete sampling distributions, such as Binomial distribution and Poisson distribution if certain conditions are met.

Step-by-step explanation: We will give conditions under which the Binomial and Poisson distribitions, which are discrete, can be approximated by the Normal distribution. This procedure is called normal approximation.

1. Binomial distribution: Let the sampling distribution be the binomial distribution [tex]B(n,p)[/tex], where [tex]n[/tex] is the number of trials and [tex]p[/tex] is the probability of success. It can be approximated by the Normal distribution with the mean of [tex]np[/tex] and the variance of [tex]np(1-p)[/tex], denoted by [tex]N(np,np(1-p))[/tex] if the following condition is met:

[tex]n>9\left(\frac{1-p}{p}\right)\text{ and } n>9\left(\frac{p}{1-p}\right)[/tex]

2. Poisson distribution: Let the sampling distribution be the Poisson distribution [tex]P(\lambda)[/tex] where [tex]\lambda[/tex] is its mean. It can be approximated by the Normal distribution with the mean [tex]\lambda[/tex] and the variance [tex]\lambda[/tex], denoted by [tex]N(\lambda,\lambda)[/tex] when [tex]\lambda[/tex] is large enough, say [tex]\lambda>1000[/tex] (however, different sources may give different lower value for [tex]\lambda[/tex] but the greater it is, the better the approximation).

Answer:72

Step-by-step explanation:

Answer: m1 = 4

m2 = 5

m3 = 2

Step-by-step explanation:

given (21/11, 6/11) = m1 (-1/3) + m2 (3, -2) + m3 (5, 2)

= (-m1 + 3m2 + 5m3) / 11 = 21/11

= (3m1 + (-2)m2 + 2m3) / 11 = 6/11

so that m1 + m2 +m3 = 11

-m1 + 3m2 + 5m3 = 21

3m1 - 2m2 + 2m3 = 6

from this, we get the augmented matrix as

\left[\begin{array}{cccc}-1&1&1&11\\-1&3&5&21\\3&-2&2&6\end{array}\right]

= \left[\begin{array{cccc}-1&1&1&11\\0&4&6&32\\0&-5&-1&-27\end{array}\right]  \left \{ {{R2=R2 + R1} \atop {R3=R3 -3R1 }]} \right.

= \left[\begin{array}{cccc}-1&1&1&11\\0&1&3/2&8\\0&-5&-1&-27\end{array}\right]

= \left[\begin{array}{cccc}-1&1&1&11\\0&1&3/2&8\\0&0&13/2&13\end{array}\right]

(R3 = R3 + 5R2)

this gives m1 + m2 + m3 = 11

m2 + 3/2 m3 = 8

13/2 m3 = 8

13/2 m3 = 13

m3 = 2

m2 = 8 -3/2 (2) = 5

= m1 = 11- 5 - 2 = 4

this gives

m1 = 4

m2 = 5

m3 = 2

To achieve the given center of mass, 4 kg of mass should be allocated to the vector u1, 6 kg to the vector u2, and 1 kg to the vector u3.

The provided equation for the center of mass expresses a weighted average of the position vectors, with each mass acting as the weight on its respective vector. We are given three vectors and a total mass, and must find the weights to apply to each vector in order to achieve a particular center of mass. Setting up equations for each component of the center of mass, we get two equations:

21/11 = (-1m1 + 3m2 + 5m3) / 11 and 6/11 = (3m1 - 2m2 + 2m3) / 11.

These equations can then be solved simultaneously to determine the values of m1, m2, and m3. Solving these equations gives m1 = 4 kg, m2 = 6 kg, and m3 = 1 kg. So, four kilograms of mass should be allocated to the vector u1, six kilograms to the vector u2, and one kilogram to the vector u3.

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Answer:

[tex]0.23 - 1.96\sqrt{\frac{0.23(1-0.23)}{200}}=0.172[/tex]

[tex]0.23 + 1.96\sqrt{\frac{0.23(1-0.23)}{200}}=0.288[/tex]

The 95% confidence interval would be given by (0.172;0.288)

Step-by-step explanation:

Previous concepts

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".  

The margin of error is the range of values below and above the sample statistic in a confidence interval.  

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".  

The population proportion have the following distribution

[tex]p \sim N(p,\sqrt{\frac{p(1-p)}{n}})[/tex]

[tex]X=46[/tex] number of vehicles that were not covered by insurance

[tex]n=200[/tex] random sample taken

[tex]\hat p=\frac{46}{200}=0.23[/tex] estimated proportion of vehicles that were not covered by insurance

[tex]p[/tex] true population proportion of vehicles that were not covered by insurance

Solution to the problem

In order to find the critical value we need to take in count that we are finding the interval for a proportion, so on this case we need to use the z distribution. Since our interval is at 95% of confidence, our significance level would be given by [tex]\alpha=1-0.95=0.05[/tex] and [tex]\alpha/2 =0.025[/tex]. And the critical value would be given by:

[tex]z_{\alpha/2}=-1.96, z_{1-\alpha/2}=1.96[/tex]

The confidence interval for the mean is given by the following formula:  

[tex]\hat p \pm z_{\alpha/2}\sqrt{\frac{\hat p (1-\hat p)}{n}}[/tex]

If we replace the values obtained we got:

[tex]0.23 - 1.96\sqrt{\frac{0.23(1-0.23)}{200}}=0.172[/tex]

[tex]0.23 + 1.96\sqrt{\frac{0.23(1-0.23)}{200}}=0.288[/tex]

The 95% confidence interval would be given by (0.172;0.288)

1 a) Pairs of Independent Events are A and B, A and D

b) Triples of Independent Events are A, B, C

2 a) Both HHHHHHHHH and HHHHHHHHT are equally likely.

b) Probability of each event:

HHHHHHHHH: 1/512

HHHHHHHHT: 1/512

c) The person is suffering from gambler's fallacy

d) The person is suffering from misconception of randomness.

1) Events

A: Heads on the first toss.

B: Tails on the second toss.

C: Heads on the third toss.

D: All three outcomes the same (HHH or TTT).

E: Exactly one head turns up.

a) To check if two events are independent, we see if [tex]P(A \cap B) = P(A) \times P(B)[/tex]

Calculating Individual Probabilities

P(A) = 1/2 (Heads on the first toss)

P(B) = 1/2 (Tails on the second toss)

P(C) = 1/2 (Heads on the third toss)

P(D) = P(HHH) + P(TTT) = 1/8 + 1/8 = 1/4

P(E) = 3/8 (Exactly one head: H, T, T; T, H, T; T, T, H)

Checking Independence

i. A and B

P(A ∩ B): Probability of A and B happening together (H on first and T on second):

There are 4 outcomes: HTH, HTT, HHH, HHT. Only HTT has T on second.

P(A ∩ B) = 1/4.

P(A) × P(B) = 1/2 × 1/2 = 1/4.

They are independent

ii. A and D

P(A ∩ D): Probability of A and D (first is H and all are the same):

Only HHH satisfies this.

P(A ∩ D) = 1/8.

P(A) × P(D) = 1/2 × 1/4 = 1/8.

They are independent

iii. A and E

P(A ∩ E): Probability of A and E (first is H and only one head):

Possible outcomes: H, T, T (1 possibility).

P(A ∩ E) = 1/8.

P(A) × P(E) = 1/2 × 3/8 = 3/16.

They are not independent.

iv. D and E

P(D ∩ E): Probability of D and E (all the same and exactly one head):

Impossible since D can only be HHH or TTT, and E requires one head.

P(D ∩ E) = 0.

P(D) × P(E) = 1/4 × 3/8 = 3/32.

They are not independent

b) To check if three events are independent, we check if P(A ∩ B ∩ C) = P(A) × P(B) × P(C)

i. A, B, C

P(A ∩ B ∩ C): Probability of H on first, T on second, H on third:

Outcome: HTH.

P(A ∩ B ∩ C) = 1/8.

P(A) × P(B) × P(C) = 1/2 × 1/2 × 1/2 = 1/8.

They are independent

ii. A, B, D

P(A ∩ B ∩ D): Cannot have T on second and all the same

P(A ∩ B ∩ D) = 0.

P(A) × P(B) × P(D) = 1/2 × 1/2 × 1/4 = 1/16.

They are not independent

iii. C, D, E

P(C ∩ D ∩ E): Cannot have all the same (D) and exactly one head (E).

P(C ∩ D ∩ E) = 0.

P(C) × P(D) × P(E) = 1/2 × 1/4 × 3/8 = 3/64.

They are not independent.

2a) Both events are equally likely since they are specific sequences of tosses.

b) Probability of each event:

Probability of HHHHHHHHH

9 heads in a row:

Probability = (1/2)^9 = 1/512.

Probability of HHHHHHHHT

8 heads followed by 1 tail:

Probability = (1/2)^9 = 1/512.

c) They are suffering from the gambler's fallacy, which is the incorrect belief that past events influence independent events.

d) They may be suffering from the misconception of randomness, believing that certain sequences are more likely due to patterns they perceive.

To solve for X, you first add the integers together and the X variables together. Then you subtract 3 from each side, followed by dividing by 3 on each side. Here is the work to show our math:

X + X + 1 + X + 2 = 154

3X + 3 = 154

3X + 3 - 3 = 154 - 3

3X = 151

3X/3 = 151/3

X = 50 1/3

Since 50 1/3 is not an integer, there is no true answer to this problem.

To solve for three consecutive even integers where the sum of the first two equals 154, we assign x, x + 2, and x + 4 to those integers. Solving the equation results in x = 76, thus the three integers are 76, 78, and 80.

In Mathematics, specifically in algebra, this problem involves solving for unknowns. Let's call our three consecutive even numbers x, x + 2, and x + 4. This is because two integers are even if they differ by 2. The equation given by the problem is x + (x + 2) = 154, as we are told that the sum of the first and second of the three numbers is 154.

To solve this equation, firstly, combine like terms, you get 2x + 2 = 154. Secondly, isolate the variable x by subtracting 2 from both sides of this equation, you get 2x = 152. Thirdly, to solve for x, divide both sides of the equation by 2. The result is x = 76. This gives you the first integer. Since we know that the integers are consecutive even numbers, the other two integers are 76 + 2 = 78 and 76 + 4 = 80. Therefore, the three consecutive even integers are 76, 78, and 80.

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Answer:

Step-by-step explanation:

Given that price, p, and quantity, x, are inearly related

We are given two points on this line as (p,x) = (96.90, 121) and (40.20,310)

Using two point formula we find linear equation as

[tex]\frac{y-121}{310-121} =\frac{0-96.90}{40.20-96.90} \\(x-121)(-56.70)=189(p-96,.90)\\189p+56.70x =25174.80[/tex]

a)[tex]189p+56.70x =25174.80[/tex] is the linear equation

b) A(x)

Substitute to get

189(50.70)+56.70x = 25174.80

x=275 units

c) This is linear function hence no local maxima or minima

d) No maximia or minima

Answer:

Substitution of x+1

Step-by-step explanation:

We are given that

[tex]\int \frac{1}{x^2+2x+2}dx[/tex]

[tex]\int\frac{1}{(x^2+2x+1)+1}dx[/tex]

[tex]\int\frac{1}{(x+1)^2+1^2}dx[/tex]

By using identity

[tex](a+b)^2=a^2+b^2+2ab[/tex]

Substitute x+1=t

Differentiate w.r.t x

dx=dt

Substitute the values

[tex]\int\frac{1}{t^2+1^2}dx[/tex]

[tex]\frac{1}{1}tan^{-1}\frac{t}{1}+C[/tex]

By using formula :[tex]\int\frac{1}{x^2+a^2}dx=\frac{1}{a}tan^{-1}\frac{x}{a}+C[/tex]

[tex]tan^{-1}(x+1)+C[/tex]

To complete the square for the quadratic polynomial x^2 + 2x + 2, we obtain (x + 1)^2 + 1. Substituting u = x + 1 simplifies the integral, which is then recognized as the derivative of arctan(u), resulting in the solution arctan(x + 1) + C.

The student asked to complete the square for the integrand x2 + 2x + 2 and to propose a substitution to compute the integral ∫ 1 / (x2 + 2x + 2) dx.

First, we complete the square for the quadratic polynomial x2 + 2x + 2. That gives us:

(x + 1)2 + 1 = x2 + 2x + 1 + 1 = x2 + 2x + 2.

We have now written our quadratic in the form of a perfect square plus a constant, which simplifies our integral:

∫ 1 / ((x + 1)2 + 1) dx.

We can substitute u = x + 1, which implies du = dx.

This transforms our integral into:

∫ 1 / (u2 + 1) du,

which can be recognized as the derivative of arctan(u), hence the integral is:

arctan(u) + C, where C is the constant of integration.

Substituting back our original variable, the solution is:

arctan(x + 1) + C.

Answer: 5/36

Step-by-step explanation:

We assume it's a fair die and the probability of any number coming up is 1/6.

Let's denote first die with it's number as A2, that is first die roled number 2

Let's denote second die with it's number as B4, that is Second die rolled 4.

For us to have the sum of those numbers to be 8, then we have possibilities of rolling the numbers

A2 and B6, A3 and B5, A4 and B4, A5 and B3, A6 and B2

This becomes :

[pr(A2) * pr(B6)] + [pr(A3) * pr(AB5)] + [pr(A4) * pr(B4)] + [pr(A5) * pr(3)] + [pr(A6) * pr(B2)]

Which becomes:

[1/6 * 1/6] + [1/6 * 1/6] + [1/6 * 1/6] + [1/6 * 1/6 + [1/6 * 1/6]

Which becomes :

1/36 + 1/36 + 1/36 + 1/36 + 1/36 = 5/36

Hence, the probability of the sum of the two numbers on the dice being 8 is 5/36

Answer:

Step-by-step explanation:

x2 = 36?

solution

x^2-(36)=0

Factoring:  x2-36

Check : 36 is the square of 6

Check :  x2  is the square of  x1  

Factorization is :       (x + 6)  •  (x - 6)

(x + 6) • (x - 6)  = 0

x+6 = 0

x = -6

x-6 = 0

add 6 to both side

x = 6

x = -6

The solutions of an equation are the true values of the equation.

The expression that represents the solution(s) of [tex]\mathbf{x^2 = 36}[/tex] is [tex]\mathbf{x = \pm6}[/tex]

The equation is given as:

[tex]\mathbf{x^2 = 36}[/tex]

Take square roots of both sides

[tex]\mathbf{x = \pm\sqrt{36}}[/tex]

Express the square root of 36 as 6.

So, we have

[tex]\mathbf{x = \pm6}[/tex]

So, the expression that represents the solution(s) of [tex]\mathbf{x^2 = 36}[/tex] is [tex]\mathbf{x = \pm6}[/tex]

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Answer: N >/= 7 bits

Minimum of 7 bits

Step-by-step explanation:

The minimum binary bits needed to represent 65 can be derived by converting 65 to binary numbers and counting the number of binary digits.

See conversation in the attachment.

65 = 1000001₂

65 = 7 bits :( 0 to 2^7 -1)

The number of binary digits is 7

N >/= 7 bits

To represent the unsigned decimal integer 65 in binary format, we need a minimum of 7 binary bits.

This question is related to the conversion of decimal numbers to binary format. To present an unsigned decimal 65 in binary, we first need to find the highest power of 2 that is less than or equal to 65, which is 2^6=64.

Then we continue to find the next highest power of 2 for the remainder of the value until we reach zero. This process would require a total of 7 bits. Therefore, to represent 65 as an unsigned binary integer, we need a minimum of 7 bits.

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Answer: x =6 1/3

Step-by-step explanation:

Whichever value of x from the given option that make f(x) = 0 is the zero of f(x). x = 6 1/3 satisfies this, you can check by replacing 6 1/3 by x in the equation. 6 1/3 can be written as 19/3

f(19/3) = 9(19/3)² - 54(19/3) - 19

= 9(361/9) - 54(19/3) - 19

= 361 - 342 - 19

= 0.

The remaining values don't give 0.