Consider the following statement: There exist no integers a and b such that 18a + 6b = 1. (a) Rewrite this statement using quantifiers, propositional variables, and predicates. (b) Rewrite the negation of this statement using quantifiers, propositional variables, and predicates. (c) Using contradiction, prove the statement is false.

Answer:

  yes

Step-by-step explanation:

You can always separate an equation into two parts and see where those graphs intersect.

Joel's method works well.

_____

Additional comments

Preston should know that the invention of logarithms makes it easy to solve equations like this. x = log₂(14) = log(14)/log(2) ≈ 3.8073549.

As for Joel's method, I prefer to subtract the right side to get the equation ...

  2^x -14 = 0

Then graphing y = 2^x -14, I look for the x-intercept. Most graphing calculators make it easy to find x- and y-intercepts. Not all make it easy to find points of intersection between different curves.

Answer:

Yes, the graph intersects around (3.807,14), so 3.807 is a good estimate of the solution to 2^x=14.

Step-by-step explanation:

Answer: The answers in order are: Equal, 120°, Equal, 120°, Equal, no

Step-by-step explanation:

The positive sequence components have equal magnitudes and 120° phase displacements in positive sequence.  Their phase sequence is same as that of the system one. Also their phase rotations is same like the system.

The negative sequence components have equal magnitudes with same 120° phase displacements. Their phase sequence is opposite to that of system but their phase rotation is same like the system.

Zero sequence components have equal magnitude with no phase displacement. They behave like negative sequence in terms of phase sequence and phase rotation.

Answer:

Equal, 120°, Equal, 120°, Equal, no

Step-by-step explanation:

Answer:

a) P ( A & B ) = 0.1995

b) P (A U B ) = 0.7905

c) P (A/B) = 0.2625

d) P(B/A')  = 0.194805

e) NOT mutually exclusive

f) NOT Independent

Step-by-step explanation:

Declare Events:

- buying gasoline = Event A

- buying groceries = Event B

Given:

- P(A) = 0.23

- P(B) = 0.76

- P(B/A) = 0.85

Find:

- a. Find the probability that a typical customer buys both gasoline and groceries.

- b. Find the probability that a typical customer buys gasoline or groceries.

- c. Find the conditional probability of buying gasoline given that the customer buys groceries.

- d. Find the conditional probability of buying groceries given that the customer did not buy gasoline.

- e Are these two events (groceries, gasoline) mutually exclusive?

- f  Are these two events independent?

Solution:

- a) P ( A & B ) ?

                     P ( A & B ) = P(B/A) * P(A) = 0.85*0.23 = 0.1995

- b) P (A U B ) ?

                    P (A U B ) = P(A) + P(B) - P(A&B)

                    P (A U B ) = 0.23 + 0.76 - 0.1995

                    P (A U B ) = 0.7905

- c) P ( A / B )?

                    P ( A / B ) = P(A&B) / P(B)

                                    = 0.1995 / 0.76

                                    = 0.2625

- d) P( B / A') ?

                   P( B / A') = P ( B & A') / P(A')

                   P ( B & A' ) = 1 - P( A / B) = 1 - 0.85 = 0.15

                   P ( B / A' ) = 0.15 / (1 - 0.23)

                                    = 0.194805

- e) Are the mutually exclusive ?

        The condition for mutually exclusive events is as follows:

                    P ( A & B ) = 0 for mutually exclusive events.

        In our case P ( A & B ) = 0.1995 is not zero.

        Hence, NOT MUTUALLY EXCLUSIVE

- f) Are the two events independent?

         The condition for independent events is as follows:

                    P ( A & B ) = P (A) * P(B) for mutually exclusive events.

        In our case,

                        0.1995 = 0.23*0.76

                        0.1995 = 0.1748 (NOT EQUAL)

        Hence, NOT INDEPENDENT

a. The probability a customer buys both gasoline and groceries is 0.23 * 0.85 = 0.1955.

b. The probability a customer buys either gasoline or groceries is 0.23 + 0.76 - 0.1955 = 0.7945.

c. Conditional probability of buying gasoline given groceries: 0.1955 / 0.76 ≈ 0.2572.

d. Conditional probability of buying groceries given no gasoline: 0.76 - 0.1955 / 0.77 ≈ 0.7331.

e. No, these events are not mutually exclusive.

f. No, these events are not independent.

Let's calculate the probabilities and answer each part of the question:

a. To find the probability that a typical customer buys both gasoline and groceries, you can use the formula for conditional probability:

P(Gasoline and Groceries) = P(Groceries | Gasoline) * P(Gasoline)

P(Gasoline and Groceries) = 0.85 * 0.23 = 0.1955

So, the probability that a typical customer buys both gasoline and groceries is 0.1955.

b. To find the probability that a typical customer buys gasoline or groceries, you can use the addition rule for probabilities:

P(Gasoline or Groceries) = P(Gasoline) + P(Groceries) - P(Gasoline and Groceries)

P(Gasoline or Groceries) = 0.23 + 0.76 - 0.1955 = 0.7945

So, the probability that a typical customer buys gasoline or groceries is 0.79

c. To find the conditional probability of buying gasoline given that the customer buys groceries, you can use the formula for conditional probability:

P(Gasoline | Groceries) = P(Gasoline and Groceries) / P(Groceries)

P(Gasoline | Groceries) = 0.1955 / 0.76 ≈ 0.2572

So, the conditional probability of buying gasoline given that the customer buys groceries is approximately 0.2572.

d. To find the conditional probability of buying groceries given that the customer did not buy gasoline, you can use the formula for conditional probability:

P(Groceries | No Gasoline) = P(Groceries and No Gasoline) / P(No Gasoline)

First, calculate P(No Gasoline):

P(No Gasoline) = 1 - P(Gasoline) = 1 - 0.23 = 0.77

Now, calculate P(Groceries and No Gasoline):

P(Groceries and No Gasoline) = P(Groceries) - P(Gasoline and Groceries) = 0.76 - 0.1955 = 0.5645

Now, find P(Groceries | No Gasoline):

P(Groceries | No Gasoline) = 0.5645 / 0.77 ≈ 0.7331

So, the conditional probability of buying groceries given that the customer did not buy gasoline is approximately 0.7331.

e. These two events (buying groceries and buying gasoline) are not mutually exclusive because it's possible for a customer to buy both groceries and gasoline, as we calculated in part (a).

f. To determine whether these two events are independent, we need to check if the conditional probabilities match the unconditional probabilities:

P(Gasoline | Groceries) = P(Gasoline) and P(Groceries | No Gasoline) = P(Groceries)

Let's check:

P(Gasoline | Groceries) ≈ 0.2572

P(Gasoline) = 0.23

P(Groceries | No Gasoline) ≈ 0.7331

P(Groceries) = 0.76

These conditional probabilities are not equal to the unconditional probabilities, so the events are not independent. In an independent event scenario, the conditional probabilities would be equal to the unconditional probabilities.

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Answer:

See below

Step-by-step explanation:

Remember some definitions about binary relations. If R⊆S×S then

a) R is not reflexive since (1,1)∉R.

R is irreflexive, since (a,a)∉R for all a=1,2,3,4

R is asymmetric: (2,3)∈R and (3,2)∉R (thus R is not symmetric).

R is antisymmetric, there are no cases to check. R is transitive, there are no cases to check.

b) R is reflexive, checking case by case, (a,a)∈R for all a=1,2,3,4. Hence R is not irreflexive.

R is not asymmetric: (1,2)∈R but (2,1)∈R. R is not symmetric, since (4,1)∈R but (1,4)∉R

R is not antisymmetric: (1,2)∈R and (2,1)∈R but 1≠2.

R is not transitive: (1,2)∈R and (2,4)∈R but (1,4)∉R.

Answer: False

Step-by-step explanation:

In Inferential statistics the sample data is analysed rather than analysing the whole population, analysing the whole population may sometimes be impossible, like analysing the population of a whole country. Therefore, for inferential statistics, conclusions about the whole population are drawn by analyzing the corresponding sample data. The sample data are selected from the population and then analysed, the results can then be used to conclude on the whole population.

a. To find the margin of error, we first calculate the standard error of the mean using the formula [tex]\(SE = \frac{s}{\sqrt{n}}\)[/tex], where s is the sample standard deviation and n is the sample size. Then, we use the formula for the margin of error [tex]\(ME = Z \times SE\)[/tex], where Z is the z-score corresponding to the desired level of confidence.

b. Once we have the margin of error, we can construct the confidence interval estimate of the population mean by adding and subtracting the margin of error from the sample mean.

c. To compare the prices for meals for two in mid-range restaurants in Hong Kong to those in Tokyo, we can use the confidence interval estimate of the population mean. If the confidence interval includes the mean price for Tokyo ($40), it suggests that there may not be a significant difference in prices between the two cities. However, if the confidence interval does not include $40, it suggests that there may be a significant difference in prices between the two cities.

Explanation:

a. Margin of error:

1. Calculate the sample mean [tex](\( \bar{x} \))[/tex] and sample standard deviation s from the given data.

2. Determine the sample size n.

3. Find the standard error of the mean SE using the formula [tex]\( SE = \frac{s}{\sqrt{n}} \)[/tex].

4. Look up the z-score corresponding to the desired level of confidence (e.g., 95%) from the standard normal distribution table.

5. Multiply the z-score by the standard error to find the margin of error [tex](\( ME \))[/tex].

b. Confidence interval estimate:

1. Calculate the margin of error ME.

2. Subtract the margin of error from the sample mean to find the lower bound of the confidence interval.

3. Add the margin of error to the sample mean to find the upper bound of the confidence interval.

c. Price comparison:

1. Check if the confidence interval estimate of the population mean includes the mean price for Tokyo ($40).

2. If the confidence interval includes $40, it suggests that there may not be a significant difference in prices between the two cities.

3. If the confidence interval does not include $40, it suggests that there may be a significant difference in prices between the two cities.

Answer:

340 of the adults in the sample voted.

The 95% confidence interval estimate of the population percent of Americans that vote is (0.6391, 0.7209). This means that we are 95% sure that the true proportion of Americans that vote is between 0.6391 and 0.7209.

Step-by-step explanation:

In 2008, a random sample of 500 american adults was take and we found that 68% of them voted. How many of the 500 adults in the sample voted?

This is 68% of 500.

So 0.68*500 = 340.

340 of the adults in the sample voted.

Now construct a 95% confidence interval estimate of the population percent of Americans that vote and write a sentence to explain the confidence interval.

In a sample with a number n of people surveyed with a probability of a success of [tex]\pi[/tex], and a confidence interval [tex]1-\alpha[/tex], we have the following confidence interval of proportions.

[tex]\pi \pm z\sqrt{\frac{\pi(1-\pi)}{n}}[/tex]

In which

Z is the zscore that has a pvalue of [tex]1 - \frac{\alpha}{2}[/tex].

For this problem, we have that:

[tex]n = 500, p = 0.68[/tex]

95% confidence interval

So [tex]\alpha = 0.05[/tex], z is the value of Z that has a pvalue of [tex]1 - \frac{0.05}{2} = 0.975[/tex], so [tex]Z = 1.96[/tex].

The lower limit of this interval is:

[tex]\pi - z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.68 - 1.96\sqrt{\frac{0.68*0.32}{500}} = 0.6391[/tex]

The upper limit of this interval is:

[tex]\pi + z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.68 + 1.96\sqrt{\frac{0.68*0.32}{500}} = 0.7209[/tex]

The 95% confidence interval estimate of the population percent of Americans that vote is (0.6391, 0.7209). This means that we are 95% sure that the true proportion of Americans that vote is between 0.6391 and 0.7209.

Answer:

first quadrant

Step-by-step explanation:

To know this we have to take into account that for the breast we have to look at the y-axis and in the cosine we will look at the x-axis

 We have the positive and the negative part of the x-axis, now we notice that it says that the cosine is positive, then the quadrant will be with the positive x,  these quadrants can be the first or the fourth.

Now if we look at sine = 4 is positive, this means that the axis y will also be positive,  this means that it will be between the first and second quadrant.

Now if we look to meet the conditions of the sine and cosine, the angle has to be found in the first quadrant

Step-by-step explanation:

If x is her score on the final exam, then the average is:

(74 + 88 + 61 + 83 + 2x) / 6

(306 + 2x) / 6

51 + ⅓x

We want this to be greater than or equal to 76.

51 + ⅓x ≥ 76

⅓x ≥ 25

x ≥ 75

In order to get an average of 76 or higher, Shureka's score on the final exam must be greater than or equal to 75.

Answer:

The probability of selecting 5 female and 2 male students is 0.052.

Step-by-step explanation:

The class comprises of 7 female students and 10 male students.

Total number of students: 17.

Number of female students, 7.

Number of male students, 10.

The probability of an event E is:

[tex]P(E)=\frac{Favorable\ outcomes}{Total\ number\ of] outcomes}[/tex]

The number of ways to select 7 students from 17 is:

[tex]N ={17\choose 7}=\frac{17!}{7!(17-7)!}= 19448[/tex]

The number of ways to select 5 female students of 7 females is:

[tex]n(F) ={7\choose 5}=\frac{7!}{5!(7-5)!}= 21[/tex]

The number of ways to select 2 male students of 10 males is:

[tex]n(M) ={10\choose 2}=\frac{10!}{2!(10-2)!}= 45[/tex]

Compute the probability of selecting 5 female and 2 male students as follows:

P (5 F and 2 M) = [n (F) × n (M)] ÷ N

                         [tex]=\frac{21\times45}{19448} \\=0.05183\\\approx0.052[/tex]

Thus, the probability of selecting 5 female and 2 male students is 0.052.

Answer:

13/4260 tons

Step-by-step explanation:

We have the rate at which they remove tons of dirt per hour. We also know that total that needs to be removed. We can determine the time by dividing  the amount of tons that need to be removed by the rate:

[tex]=(13/6)/\cdot{710}=13/4260[/tex]

will take 13/4260 hours to remove the dirt

Answer:

This process is out of control.

Step-by-step explanation:

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.

A probability is said to be unusual if it's z-score has a pvalue of 0.05 or lower, or a pvalue of 0.95 or higher.

In this problem, we have that:

[tex]\mu = 10.2 \sigma = 1.04[/tex]

If a randomly selected minute has 13.9 hits, would the process be considered in control or out of control?

The process will be considered out of control if it's z-score(Z when X = 13.9) has a pvalue of 0.95 or higher. Otherwise(it will be positive, since 13.9 is above the mean), it will be considered in control.

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

[tex]Z = \frac{13.9 - 10.2}{1.04}[/tex]

[tex]Z = 3.56[/tex]

[tex]Z = 3.56[/tex] has a pvalue of 0.9999. So there is only a 1-0.9999 = 0.0001 = 0.01% probability of getting 13.9 hits a minute.

So this process is out of control.

If using the empirical rule, since 13.9 hits is more than two standard deviations above the mean of 10.2 hits, the process could be considered out of control. However, establishing specific control limits is necessary for a definitive answer.

To determine whether a process is in control or out of control, we assess whether a given measurement is within the expected range of a normal distribution, often using the empirical rule or control limits. Given that the process has a mean of 10.2 hits per minute and a standard deviation of 1.04 hits, under the empirical rule, approximately 95% of the data should fall within two standard deviations of the mean (that is, between roughly 8.12 and 12.28 hits).

With 13.9 hits in a randomly selected minute, this count is significantly more than two standard deviations above the mean, suggesting that the process might be out of control. However, to make a definitive statement about control status, specific control limits must be established, often based on the particular specifications of the process being monitored.

Answer:

Present value =  $4,122.4

Accumulated amount = $4,742

Step-by-step explanation:

Data provided in the question:

Amount at the Start of money flow = $1,000

Increase in amount is exponentially at the rate of 5% per year

Time = 4 years

Interest rate = 3.5%  compounded continuously

Now,

Accumulated Value of the money flow = [tex]1000e^{0.05t}[/tex]

The present value of the money flow = [tex]\int\limits^4_0 {1000e^{0.05t}(e^{-0.035t})} \, dt[/tex]

= [tex]1000\int\limits^4_0 {e^{0.015t}} \, dt[/tex]

= [tex]1000\left [\frac{e^{0.015t}}{0.015} \right ]_0^4[/tex]

= [tex]1000\times\left [\frac{e^{0.015(4)}}{0.015} -\frac{e^{0.015(0)}}{0.015} \right][/tex]

= 1000 × [70.7891 - 66.6667]

= $4,122.4

Accumulated interest = [tex]e^{rt}\int\limits^4_0 {1000e^{0.05t}(e^{-0.035t}} \, dt[/tex]

= [tex]e^{0.035\times4}\times4,122.4[/tex]

= $4,742

The present value and interest accumulated would be as follows:

Present Value = $ 4,122.4

Interest Accumulated = $ 4742

Given that,

Principal at the beginning of money flow = $1,000

Exponential interest rate = 5% per year

Time Period = 4 years

So,

The accumulated money flow's worth = [tex]1000e^{0.05t}[/tex]

The current value of the money can be determined by [tex]\int\limits^4_0 1000e^{0.05t}(e^{-0.035t}) {} \, dt[/tex]

On solving, we get

The present value = $ 4,122.4

Interest Accumulated = $4,742

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Answer:

a) left handed people

b) Binomial probability distribution with pdf

[tex]P(X=x)=15Cx0.1^{x} 0.9^{15-x}[/tex]

where x=0,1,2,...,15.

c) Histogram is attached

d) The shape of histogram depicts that distribution is rightly skewed.

e) 1.5

f) 1.35

g) 1.16

Step-by-step explanation:

a)

The random variable in the given scenario is " left handed people"

b)

The scenario represents the binomial probability distribution as the outcome is divided into one of two categories and experiment is repeated fixed number of times i.e. 15 and trails are independent. The pdf of binomial distribution is

[tex]P(X=x)=nCxp^{x} q^{n-x}[/tex]

Here n=15, p=0.1 and q=1-p=0.9.

So, the pdf would be

[tex]P(X=x)=15Cx0.1^{x} 0.9^{n-x}[/tex]

where x=0,1,2,...,15.

c)

Histogram is constructed by first computing probabilities on all x points i.e. x=0, x=1 , .... ,x=15 and then plotting all probabilities with respective x values. Histogram is in attached image.

d)

The tail of histogram is to the right side and thus the histogram depicts that given probability distribution is rightly skewed.

e)

The mean of binomial probability distribution is computed by multiplying number of trails and probability of success.

mean=np=15*0.1=1.5

f)

The variance of binomial probability distribution is computed by multiplying number of trails and probability of success and probability of failure.

variance=npq=15*0.1*0.9=1.35

g)

The standard deviation can be calculated by simply taking square root of variance

S.D=√npq=√1.35=1.16

The proportion of left-handed people follows a binomial distribution

The given parameters are:

[tex]\mathbf{n = 15}[/tex] -- the sample size

[tex]\mathbf{p = 10\%}[/tex] --- the proportion of left-handed people

(a) The random variable

The distribution is about left-handed people.

Hence, the random variable is left-handed people

(b) The probability distribution

If the proportion of left-handed people is 10%, then the proportion of right-handed people is 90%.

So, the probability distribution function is:

[tex]\mathbf{P(x) = ^nC_x p^x (1 - p)^{n -x}}[/tex]

This gives

[tex]\mathbf{P(x) = ^nC_x (10\%)^x (1 - 10\%)^{n -x}}[/tex]

[tex]\mathbf{P(x) = ^nC_x 0.1^x 0.9^{n -x}}[/tex]

Hence, the probability distribution function is [tex]\mathbf{P(x) = ^nC_x 0.1^x 0.9^{n -x}}[/tex]

(c) The histogram

To do this, we calculate P(x) for x = 0 to 15

[tex]\mathbf{P(0) = ^{15}C_0 \times 0.1^0 \times 0.9^{15 -0} = 0.206}[/tex]

[tex]\mathbf{P(1) = ^{15}C_1 \times 0.1^1 \times 0.9^{15 -1} = 0.343}[/tex]

.....

..

[tex]\mathbf{P(15) = ^{15}C_{15} \times 0.1^{15} \times 0.9^{15 -15} = 10^{-15}}[/tex]

See attachment for the histogram

(d) The mean

This is calculated as:

[tex]\mathbf{\bar x = np}[/tex]

So, we have:

[tex]\mathbf{\bar x = 15 \times 10\% }[/tex]

[tex]\mathbf{\bar x= 1.5}[/tex]

Hence, the mean is 1.5

(e) The variance

This is calculated as:

[tex]\mathbf{Var = np(1 - p)}[/tex]

So, we have:

[tex]\mathbf{Var = 15 \times 10\% \times (1 - 10\%)}[/tex]

[tex]\mathbf{Var = 1.35}[/tex]

Hence, the variance is 1.35

(f) The standard deviation

This is calculated as:

[tex]\mathbf{\sigma = \sqrt{Var}}[/tex]

So, we have:

[tex]\mathbf{\sigma = \sqrt{1.35}}[/tex]

[tex]\mathbf{\sigma =1.16}[/tex]

Hence, the standard deviation is 1.16

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Answer:

If 40 percent of the English-speaking men at the hotel are English and 60 percent are American, the the probability that the writer is an Englishman

 is 40% or 0.4.

Step-by-step explanation:

i) If 40 percent of the English-speaking men at the hotel are English and 60 percent are American, the the probability that the writer is an Englishman

 is 40% or 0.4.

The probability that the writer is an Englishman given that a vowel was chosen from his spelling is approximately 45.45%. This was calculated using the respective probabilities of selecting a vowel from the words 'rigour' and 'rigor' and Bayes' Theorem.

Given the scenario where a man writes a word, and a letter taken at random is a vowel, we aim to determine the probability that the writer is an Englishman.

The words are:

Letters 'i', 'o', and 'u' are the vowels. Let's calculate the likelihood of selecting a vowel from each spelling:

Using Bayes' Theorem, let's find the probability the writer is English (E), given a vowel (V) was selected:

P(E|V) = [P(V|E) * P(E)] / [P(V|E) * P(E) + P(V|A) * P(A)]

Where:

So:

P(E|V) = [0.5 * 0.4] / [0.5 * 0.4 + 0.4 * 0.6] = 0.2 / (0.2 + 0.24) = 0.2 / 0.44 ≈ 0.4545

Therefore, the probability that the writer is an Englishman is approximately 0.4545 or 45.45%.

Answer:

[tex]0.753 - 1.64\sqrt{\frac{0.753(1-0.753)}{1000}}=0.731[/tex]

[tex]0.753 + 1.64\sqrt{\frac{0.753(1-0.753)}{1000}}=0.775[/tex]

The 95% confidence interval would be given by (0.731;0.775)

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]

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 90% of confidence, our significance level would be given by [tex]\alpha=1-0.90=0.1[/tex] and [tex]\alpha/2 =0.005[/tex]. And the critical value would be given by:

[tex]z_{\alpha/2}=-1.64, z_{1-\alpha/2}=1.64[/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]

The estimated proportion for this case is:

[tex] \hat p = \frac{X}{n}= \frac{753}{1000}=0.753[/tex]

If we replace the values obtained we got:

[tex]0.753 - 1.64\sqrt{\frac{0.753(1-0.753)}{1000}}=0.731[/tex]

[tex]0.753 + 1.64\sqrt{\frac{0.753(1-0.753)}{1000}}=0.775[/tex]

The 95% confidence interval would be given by (0.731;0.775)

Answer:

Frist case: P=12/35

Second case: P=31/35

Step-by-step explanation:

An urn has 3 blue balls 4 red balls. 3 balls are drawn without replacement.

Frist case:

We calculate the number of possible combinations

{7}_C_{3}=\frac{7!}{3! · (7-3)!}=35

We calculate  the number of favorable combinations  

{3}_C_{2} · {4}_C_{1} =

=\frac{3!}{2! · (3-2)!} · \frac{4!}{1! · (4-1)!}

=3 · 4 = 12

Therefore, the probability is

P=12/35

Second case:

When we count on at least one ball to be blue, we go over the probability complement.

We calculate the probability that all the balls are red, then subtract this from 1.

We calculate the number of possible combinations

{7}_C_{3}=\frac{7!}{3! · (7-3)!}=35

We calculate  the number of favorable combinations  

{4}_C_{3}  = \frac{4!}{3! · (4-3)!=4

The probability is

P=4/35.

Therefore the probability on at least one ball to be blue

P=1-4/35

P=31/35

Answer:

The solution proved are in the attached file below. Also the explanation is in the attached file

Step-by-step explanation:

Answer:

Check below

Step-by-step explanation:

Vectors are quantities that possess both magnitude and direction. In engineering problems, it is best to think of vectors as arrows, and usually it is best to manipulate vectors using components. In this tutorial, we consider the addition of two vectors using both of these techniques. Consider two vectors [tex]\vec{A}[/tex]and [tex]\vec{B}[/tex] that have lengths A and B, respectively. Vector [tex]\vec{B}[/tex] makes an angle?

1) Vector [tex]\vec{B}[/tex] makes an angle?

Yes, it does. Vector [tex]\vec{B}[/tex] makes an angle with [tex]\vec{A}[/tex], since both have the same origin and different direction.

2) From the direction of A.(Figure 1)In vector notation, the sum is represented by  [tex]\vec{C}=\vec{A}+\vec{B}[/tex]  where [tex]\vec{C}[/tex] is a new vector that is the sum of [tex]\vec{A}[/tex] and [tex]\vec{B}[/tex].  Find C, the length of C, which is the sum of A and B.

C is the resultant vector of this sum of vectors([tex](\vec{C}=\vec{A}+\vec{B})[/tex]

The length of c is found through the law of cosines, after projecting, vector a.

(Check 3rd picture)

2.2) The other technique to add vectors is to write them. As C is the resultant vector then we have

[tex]\vec{A}=\left \langle a_{1},a_{2} \right \rangle \vec{B}=\left \langle b_{1},b_{2} \right \rangle \vec{C}=\left \langle a_{1}+b_{1}, a_{2}+b_{2}\right \rangle[/tex]

In physics, vectors are quantities possessing both magnitude and direction. They can be represented as arrows on a graph, with their length and direction corresponding to the vector's magnitude and direction. Vectors are added geometrically using the head-to-tail method or analytically, where they are broken down into components, and these components are added separately.

This can be visualized as an arrow, where its length corresponds to the magnitude, and its direction is represented by the way the arrow points. Some examples of vectors include displacement, velocity, and force.

When adding vectors, there are two methods to consider: geometric or component wise method. The geometric method involves representing the vectors as arrows on a graph and adding them using the head-to-tail method. The component-wise method involves breaking down the vector into its x and y components, and adding these components separately.

For instance, if you have two vectors A and B, vector A can be broken down into its x and y components: A_x and A_y. Likewise, vector B can be broken down into B_x and B_y. Simply add corresponding components to yield the resultant vector, R_x = (A_x + B_x) and R_y = (A_y + B_y)

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Answer:

20

Step-by-step explanation:

At equilibrium quantity demanded Qd equals quantity supply Qs

                              Qd = Qs      (At equilibrium)

     ⇒                   150 - 2P = 50 + 3P

solving for P in the equation,

                             3P + 2P = 150 - 50

                              5P = 100

                                [tex]P = \frac{100}{5}[/tex]

                                P = 20

Answer:

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Step-by-step explanation:

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Answer:

Present Value = $1666666.67

Step-by-step explanation:

Present Value of a Growing Perpuity is calculated using the following formula

PV =D/(r - g)

Where D = Dividend

r = Discount Rate

g = Growth rate

D = $50,000

r = 7%

r = 7/100

r = 0.07

g = 4%

g = 4/100

g = 0.04

PV = D/(r-g)

Becomes

PV = $50,000/(0.07-0.04)

PV = $50,000/0.03

PV = $1,666,666.67

So the Present Value of the perpuity is $1,666,666.67

Answer: There is a 90% chance that the error range will be with in 3.29 ft² .

Step-by-step explanation:

Given : The  area of a parking lot is calculated to be 5,474 ft² .

Estimated standard deviation = 2 ft²

The critical z-value for 90% confidence interval is 1.645 (from z-table)

Then, the Maximum Anticipated Error = ( critical z-value ) x ( standard deviation )

= 1.645 (2) =3.29 ft²

i.e.  Maximum Anticipated Error = 3.29 ft²

Hence, there is a 90% chance that the error range will be with in 3.29 ft² .

Answer:

The possible number of ways to select distinct (a, b) such that (a + b) is even is 534.

Step-by-step explanation:

The range 1 - 99 has 99 numbers, since 1 and 99 are inclusive.

Of these 50 numbers are odd and 49 are even.

The two distinct numbers a and b must have an even sum and a should be a multiple of 9.

The sum of two numbers is even only when both are odd or both are even.

The possible values that a can assume are,

a = {9, 18, 27, 36, 45, 54, 63, 72, 81, 90 and 99}

Thus, a can assume 6 odd values and 5 even values.

Total number of ways to select a and b such that both are odd and their sum is even is:

[tex]n(Odd\ a\ and\ b)=n(Odd\ value\ of\ a)\times n(Odd\ value\ of\ b)=6\times49=294[/tex]

Total number of ways to select a and b such that both are even and their sum is even is

[tex]n(Even\ a\ and\ b)=n(E\ value\ of\ a)\times n(Even\ value\ of\ b)=5\times48=240[/tex]

Total number of ways to select distinct (a, b) such that (a + b) is even is =

[tex]=n(Odd\ a\ and\ b)+n(Even\ a\ and\ b)=294+240=534[/tex]

Thus, the possible number of ways to select distinct (a, b) such that (a + b) is even is 534.

Answer:

D) An observational study based on available data.

Step-by-step explanation:

This is an observational study based on available data.

If it had been on a sample, he would take a sample of a few pitchers, and not studied the statistics of all major league pitchers during those seasons.

It is not an anecdotal evidence, because an anecdotal evidence is something without study, just an impression.

It is not an experiment, because he just studies(observes, that is why it is an observational study) the data, he does not change anything about the pitchers.

So the correct answer is:

D) An observational study based on available data.

The baseball enthusiast's decision is based on an observational study based on available data.

The baseball enthusiast's decision is based on A) an observational study based on available data. In this case, the enthusiast examined the records of all major league pitchers between 1990 and 1995. This observational study involved collecting and analyzing data that was already available, without manipulating any variables or conducting an experiment.

Answer:  

120 ways

Step-by-step explanation: If there are three different categories of components, that is, component A, comprising four types; component B comprising, five types and component C comprising, six types. Three types are to be selected, that is, one type from each category. The number of ways one component of each type be selected is:

Component A * Component B * Component C = 4 X 5 X 6 = 120

Answer:

There is a 90.46% probability that it was repaired by a competent shop, given that it was repaired correctly.

Step-by-step explanation:

We have these following probabilities:

An 87% probability that an air conditioner repair shop is competent.

A 13% probability that an air conditioner repair shop is incompetent.

An 85% probability that an compotent shop can repair the air.

A 60% probability than an incompetent shop can repair the air.

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 this problem we have that:

Probability that it was repaired by a competent shop, given that it was repaired correctly.

P(B) is the probability that it was repaired by a competent shop. 87% of the shops are competent, so [tex]P(B) = 0.87[/tex]

P(A/B) is the probability that it was repaired correctly, given that it was repaired by a competent shop. There is an 85% probability that an compotent shop can repair the air. So [tex]P(A/B) = 0.85[/tex]

P(A) is the probability that an air was repaired correctly.

This is 85% of 87% and 60% of 13%. So

[tex]P(A) = 0.85*0.87 + 0.60*0.13 = 0.8175[/tex]

Finally

[tex]P = \frac{P(B).P(A/B)}{P(A)} = \frac{0.87*0.85}{0.8175} = 0.9046[/tex]

There is a 90.46% probability that it was repaired by a competent shop, given that it was repaired correctly.

Answer:

Factory A should be operated 6.11 days

Factory B should be operated 6.88 days

The minimum cost for factory A is $6,110

The minimum cost for factory B is $5,504

Step-by-step explanation:

Factory A

Daily production: 16 3-speed and 20 10-speed bikes

Total daily production = 16+20 = 36 speed bikes

Order: 80 3-speed and 140 10-speed bikes

Total order = 80+140 = 220 speed bikes

Number of days = 220/36 = 6.11 days

Cost per day = $1,000

Minimum cost for 6.11 days = $1000 × 6.11 = $6,110

Factory B

Daily production: 12 3-speed and 20 10-speed bikes

Total daily production = 12+20 = 32 speed bikes

Total order = 220 speed bikes

Number of days = 220/36 = 6.88 days

Cost per day = $800

Cost for 6.88 days = $800 × 6.88 = $5,504

The business should operate Factory A for about 2.92 days and Factory B for about 4.17 days to fill the order at a minimum cost of $7148.

This problem can be solved using Linear Programming, a mathematical model used in optimization problems. Suppose we let A be the number of days factory A operates and B be the number of days factory B operates. The cost to operate the factories is given by the equation C = 1000A + 800B.

The constraints are as follows:

By solving these equations, you find that Factory A should be operated for about 2.92 days and Factory B should be operated for 4.17 days, giving a minimum cost of about $7148.

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Quota sampling is most commonly used in surveys. It is a non-probability sampling technique where researchers select individuals who meet certain criteria to be included in the sample.

Quota sampling is most commonly used in surveys. It is a non-probability sampling technique where researchers select individuals who meet certain criteria to be included in the sample.

In quota sampling, the researcher sets quotas or targets for each subgroup they want to include in the sample based on certain characteristics. For example, if the researcher wants equal representation of males and females in the sample, they would set quotas for each gender and continue selecting individuals until the quotas are met.

Quota sampling is often used when it is not possible or practical to obtain a random sample, but the researcher still wants to ensure representation of different subgroups within the sample.

Learn more about Quota sampling here:

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Answer:

la familia invierte 8.33% de los ingresos totales en ocio

Step-by-step explanation:

Representando los ingresos totales por I:

- Gastos de funcionamiento = 2/3*I

- Ahorro  = 1/4*I

- En ocio : lo que resta = I - 2/3*I - 1/4*I = I - 11/12*I = 1/12*I (8.33% de I)

por lo tanto la familia invierte 8.33% de los ingresos totales en ocio