Define a random variable x = number of cups of coffee consumed on an average day. Let x = 4 represent four or more cups. Round your answers to four decimal places.

Answer:

[tex]P(9 < x < 12)=P(X=10)+P(X=11)[/tex]

[tex]P(X=10)=(19C10)(0.2)^{10} (1-0.2)^{19-10}=0.00127[/tex]

[tex]P(X=11)=(19C11)(0.2)^{11} (1-0.2)^{19-11}=0.00026[/tex]

[tex]P(9 < X < 12)=0.00127 +0.00026=0.0015[/tex]

Step-by-step explanation:

Previous concepts

The binomial distribution is a "DISCRETE probability distribution that summarizes the probability that a value will take one of two independent values under a given set of parameters. The assumptions for the binomial distribution are that there is only one outcome for each trial, each trial has the same probability of success, and each trial is mutually exclusive, or independent of each other".

Solution to the problem

Let X the random variable of interest, on this case we now that:

[tex]X \sim Binom(n=19, p=0.2)[/tex]

The probability mass function for the Binomial distribution is given as:

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

Where (nCx) means combinatory and it's given by this formula:

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

And we want to find this probability:

[tex]P(9 < x < 12)=P(X=10)+P(X=11)[/tex]

[tex]P(X=10)=(19C10)(0.2)^{10} (1-0.2)^{19-10}=0.00127[/tex]

[tex]P(X=11)=(19C11)(0.2)^{11} (1-0.2)^{19-11}=0.00026[/tex]

[tex]P(9 < X < 12)=0.00127 +0.00026=0.0015[/tex]

The probability that less than 12 but more than 9 bulbs from the sample are defective is approximately 1.0230 (rounded to four decimal places).

We want to find the probability of having between 10 and 11 defective bulbs out of a sample of 19, with a 20% defect rate.

Probability of a single bulb being defective (p): 20% or 0.20.

Probability of a single bulb not being defective (q): 1 - p = 1 - 0.20 = 0.80.

Now, we'll calculate the probabilities for k = 10 and k = 11 using the binomial probability formula:

P(X = k) = (n choose k) * p^k * q^(n-k),

where:

n is the total number of trials (sample size), which is 19 in this case.

k is the number of successful trials (defective bulbs) we want.

For k = 10:

P(X = 10) = (19 choose 10) * (0.20)^10 * (0.80)^(19-10)

P(X = 10) = 92,378.49 * 0.0000001024 * 0.1073741824

P(X = 10) ≈ 0.9899 (rounded to four decimal places)

For k = 11:

P(X = 11) = (19 choose 11) * (0.20)^11 * (0.80)^(19-11)

P(X = 11) = 61,967.28 * 0.000002048 * 0.0262144

P(X = 11) ≈ 0.0331 (rounded to four decimal places)

Now, sum these probabilities to find the overall probability of having between 10 and 11 defective bulbs:

P(10 ≤ X ≤ 11) = P(X = 10) + P(X = 11)

P(10 ≤ X ≤ 11) ≈ 0.9899 + 0.0331

P(10 ≤ X ≤ 11) ≈ 1.0230 (rounded to four decimal places)

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Answer:

Probability that a randomly selected individual who was unemployed in 2000 would have been unemployed between 12 and 13 weeks is 0.83144 .

Step-by-step explanation:

We are given that according to the Bureau of Labor Statistics, in calendar year 2000, the average duration of unemployment was 12.7 weeks, with a standard deviation of 0.3 weeks.

Let X = randomly selected individual who was unemployed in 2000

Since distribution is approximately normal, so X ~ N([tex]\mu=12.7,\sigma^{2} = 0.3^{2}[/tex])

The z score probability distribution is given by;

                 Z = [tex]\frac{X-\mu}{\sigma}[/tex] ~ N(0,1)

where, [tex]\mu[/tex] = population mean = 12.7 weeks

            [tex]\sigma[/tex] = population standard deviation = 0.3 weeks

So, the probability that a randomly selected individual who was unemployed in 2000 would have been unemployed between 12 and 13 weeks is given by = P(12 < X < 13) = P(X < 13) - P(X <= 12)

P(X < 13) = P( [tex]\frac{X-\mu}{\sigma}[/tex] < [tex]\frac{13-12.7}{0.3}[/tex] ) = P(Z < 1) = 0.84134

P(X <= 12) = P( [tex]\frac{X-\mu}{\sigma}[/tex] < [tex]\frac{12-12.7}{0.3}[/tex] ) = P(Z < -2.33) = 1 - P(Z <= 2.33) = 1 - 0.99010

                                                                                                   = 0.0099

Therefore, P(12 < X < 13) = 0.84134 - 0.0099 = 0.83144 .

The equation of the quadratic function is [tex]y=(x-7)(x+3)[/tex]

Explanation:

The vertex form of the quadratic function is given by

[tex]y=a(x-h)^{2}+k[/tex]

It is given that the quadratic function has a vertex at [tex](2,-25)[/tex]

The vertex is represented by the coordinate [tex](h,k)[/tex]

Hence, substituting [tex](h,k)=(2,-25)[/tex] in the vertex form, we get,

[tex]y=a(x-2)^{2}-25[/tex]

Now, substituting the x - intercept [tex](7,0)[/tex] , we have,

[tex]0=a(7-2)^{2}-25[/tex]

[tex]0=a(5)^{2}-25[/tex]

[tex]25=a(25)[/tex]

 [tex]1=a[/tex]

Thus, the value of a is 1.

Hence, substituting [tex]a=1[/tex], [tex](h,k)=(2,-25)[/tex] in the vertex form [tex]y=a(x-h)^{2}+k[/tex] , we get,

[tex]y=1(x-2)^{2}-25[/tex]

[tex]y=(x-2)^{2}-25[/tex]

[tex]y=x^2-2x+4-25[/tex]

[tex]y=x^2-2x-21[/tex]

[tex]y=(x-7)(x+3)[/tex]

Thus, the equation of the quadratic function is [tex]y=(x-7)(x+3)[/tex]

Answer:

the answer is d

Step-by-step explanation:

Answer:

[tex]r=\frac{n(\sum xy)-(\sum x)(\sum y)}{\sqrt{[n\sum x^2 -(\sum x)^2][n\sum y^2 -(\sum y)^2]}}[/tex]

And for this case [tex] r =0.45[/tex]

The % of variation is given by the determination coefficient given by [tex]r^2[/tex] and on this case [tex]0.45^2 =0.2025[/tex], so then the % of variation explained is 20.25%.

The proportion of the variability seen in final grade performance that can be predicted by math ability scores is 20.25%.

Step-by-step explanation:

For this case we asume that we fit a linear model:

[tex] y = mx+b[/tex]

Where y represent the final grade and x the math ability scores

[tex]m=\frac{S_{xy}}{S_{xx}}[/tex]  

Where:  

[tex]S_{xy}=\sum_{i=1}^n x_i y_i -\frac{(\sum_{i=1}^n x_i)(\sum_{i=1}^n y_i)}{n}[/tex]  

[tex]S_{xx}=\sum_{i=1}^n x^2_i -\frac{(\sum_{i=1}^n x_i)^2}{n}[/tex]  

[tex]\bar x= \frac{\sum x_i}{n}[/tex]  

[tex]\bar y= \frac{\sum y_i}{n}[/tex]  

And we can find the intercept using this:  

[tex]b=\bar y -m \bar x[/tex]  

The correlation coeffcient is given by:

[tex]r=\frac{n(\sum xy)-(\sum x)(\sum y)}{\sqrt{[n\sum x^2 -(\sum x)^2][n\sum y^2 -(\sum y)^2]}}[/tex]

And for this case [tex] r =0.45[/tex]

The % of variation is given by the determination coefficient given by [tex]r^2[/tex] and on this case [tex]0.45^2 =0.2025[/tex], so then the % of variation explained is 20.25%.

The proportion of the variability seen in final grade performance that can be predicted by math ability scores is 20.25%.

Answer:

Her z-score is 2.03.

Step-by-step explanation:

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]

Michelle’s height is 1.758 meters. What is her z-score?

Michelle's is a woman.

The average height of women is [tex]\mu = 1.618[/tex] and the standard deviation is [tex]\sigma = 0.069[/tex]

This is Z when X = 1.758. So

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

[tex]Z = \frac{1.758 - 1.618}{0.069}[/tex]

[tex]Z = 2.03[/tex]

Her z-score is 2.03.

Michelle's z-score is approximately 2.03, which means her height is 2.03 standard deviations above the mean height for American women.

To calculate Michelle’s z-score for her height, we use the formula for the z-score:

Z = (X - μ) / σ

Where Z is the z-score, X is the value (Michelle's height), μ is the mean, and σ is the standard deviation. Since Michelle is a woman, we will use the mean and standard deviation for women. The mean height for American women, μ, is 1.618 meters, and the standard deviation, σ, is 0.069 meters.

Plugging in the values we get:

Z = (1.758 - 1.618) / 0.069

Z = 0.14 / 0.069

Z ≈ 2.029

Michelle's height is approximately 2.03 standard deviations above the mean height for American women.

Answer:

[tex]P(X<4)=P(\frac{X-\mu}{\sigma}<\frac{4-\mu}{\sigma})=0.2[/tex]  

[tex]P(z<\frac{4-\mu}{\sigma})=0.2[/tex]

But we know which value of z satisfy the previous equation so then we can do this:

[tex]z=-0.842<\frac{4-\mu}{1.1}[/tex]

And if we solve for the mean we got

[tex]\mu =4 +0.842*1.1=4.926[/tex]

Step-by-step explanation:

Assuming this question "Suppose that the lifetimes of TV tubes are normally distributed with a standard deviation of 1.1 years. Suppose also that exactly 20% of the tubes die before 4 years. Find the mean lifetime of TV tubes. Carry your intermediate computations to at least four decimal places. Round your answer to at least one decimal place. ?

Previous concepts

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 Z-score is "a numerical measurement used in statistics of a value's relationship to the mean (average) of a group of values, measured in terms of standard deviations from the mean".  

Solution to the problem

Let X the random variable that represent the lifetimes of TV tubes of a population, and for this case we know the distribution for X is given by:

[tex]X \sim N(\mu,1.1)[/tex]  

Where [tex]\mu[/tex] and [tex]\sigma=1.1[/tex]

And the best way to solve this problem is using the normal standard distribution and the z score given by:

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

For this part we know the following condition:

[tex]P(X>4)=0.8[/tex]   (a)

[tex]P(X<4)=0.2[/tex]   (b)

Both conditions are equivalent on this case. We can use the z score again in order to find the value a.  

As we can see on the figure attached the z value that satisfy the condition with 0.2 of the area on the left and 0.8 of the area on the right it's z=-0.842. On this case P(Z<-0.842)=0.2 and P(z>-0.842)=0.8

If we use condition (b) from previous we have this:

[tex]P(X<4)=P(\frac{X-\mu}{\sigma}<\frac{4-\mu}{\sigma})=0.2[/tex]  

[tex]P(z<\frac{4-\mu}{\sigma})=0.2[/tex]

But we know which value of z satisfy the previous equation so then we can do this:

[tex]z=-0.842<\frac{4-\mu}{1.1}[/tex]

And if we solve for the mean we got

[tex]\mu =4 +0.842*1.1=4.926[/tex]

Mean lifetime is 4.9 years. Found using z-score of -0.8416 with 20% dying before 4 years, [tex]\(\sigma = 1.1\)[/tex].

To find the mean lifetime of TV tubes, we will use the properties of the normal distribution and the given information. Here's the step-by-step solution:

1. Identify the given values:

  - Standard deviation [tex](\(\sigma\))[/tex]: 1.1 years

  - Percentage of tubes that die before 4 years: 20% (or 0.20)

2. Find the z-score corresponding to the given percentage:

  - Since the percentage is 20%, we need to find the z-score for which 20% of the area under the normal curve lies to the left.

  - Using a standard normal distribution table or a calculator, the z-score for 0.20 is approximately -0.8416.

3. Set up the z-score formula:

  The z-score formula is given by:

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

where [tex]\(X\)[/tex] is the value (4 years in this case), [tex]\(\mu\)[/tex] is the mean lifetime we want to find, and [tex]\(\sigma\)[/tex] is the standard deviation.

4. Substitute the known values into the z-score formula and solve for [tex]\(\mu\)[/tex]:

  [tex]\[ -0.8416 = \frac{4 - \mu}{1.1} \][/tex]

  Rearrange to solve for [tex]\(\mu\)[/tex]:

  [tex]\[ -0.8416 \times 1.1 = 4 - \mu \][/tex]

  [tex]\[ -0.92576 = 4 - \mu \][/tex]

  [tex]\[ \mu = 4 + 0.92576 \][/tex]

  [tex]\[ \mu \approx 4.9258 \][/tex]

5. Round the mean lifetime to one decimal place:

  [tex]\[ \mu \approx 4.9 \][/tex]

So, the mean lifetime of the TV tubes is approximately 4.9 years.

Answer:

0.344 = 34.4% probability that at least 1 of them have blue eyes.

Step-by-step explanation:

For each person, there are only two possible outcomes. Either they have blue eyes, or they have not. The probability of a person having blue eyes is independent of any other person. So we use the binomial probability distribution to solve this question.

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 combinations 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.

A survey reports that the probability a person has blue eyes is 0.10.

This means that [tex]p = 0.1[/tex]

4 people are randomly selected at Miramar College

This means that [tex]n = 4[/tex]

Find the probability that at least 1 of them have blue eyes.

Either none of them have blue eyes, or at least one do. The sum of the probabilities of these events is decimal 1. So

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

We want [tex]P(X \geq 1)[/tex]. So

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

In which

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

[tex]P(X = 0) = C_{4,0}.(0.1)^{0}.(0.9)^{4} = 0.656[/tex]

[tex]P(X \geq 1) = 1 - P(X = 0) = 1 - 0.656 = 0.344[/tex]

0.344 = 34.4% probability that at least 1 of them have blue eyes.

To find the probability that at least one out of four people has blue eyes at Miramar College, we use the complement rule and calculate that the probability is 0.344 after rounding to three decimal places.

The question asks us to calculate the probability that at least one person out of four randomly selected people at Miramar College will have blue eyes, given that the probability a person has blue eyes is 0.10. To find the probability of at least one person having blue eyes, we can use the complement rule. The complement of at least one person having blue eyes is that no person has blue eyes.

The probability that a single person does not have blue eyes is 1 - 0.10 = 0.90. For four independent selections, the probability that none of them have blue eyes is 0.90 raised to the fourth power. Therefore, the probability that at least one out of four people has blue eyes is 1 minus this result.

Calculation:

P(no one has blue eyes) = 0.90 ^ 4

P(no one has blue eyes) = 0.6561

P(at least one has blue eyes) = 1 - P(no one has blue eyes)

P(at least one has blue eyes) = 1 - 0.6561

P(at least one has blue eyes) = 0.3439

Thus, the probability that at least one person out of the four has blue eyes, rounded to three decimal places, is 0.344.

Answer:

[tex]z=\frac{0.529 -0.6}{\sqrt{\frac{0.6(1-0.6)}{240}}}=-2.24[/tex]  

[tex]p_v =P(z<-2.24)=0.0125[/tex]  

If we compare  the p value and the significance level given [tex]\alpha=0.01[/tex] we see that [tex]p_v>\alpha[/tex] so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, and we can said that at 1% of significance the proportion college students expressed an intent to vote is not higher than 0.6

Step-by-step explanation:

Assuming the following question: A presidential candidate's aide estimates that, among all college students, the proportion p who intend to vote in the upcoming election is at least 60% . If 127 out of a random sample of 240 college students expressed an intent to vote, can we reject the aide's estimate at the 0.1 level of significance?

Data given and notation

n=240 represent the random sample taken

X=127 represent the college students expressed an intent to vote

[tex]\hat p=\frac{127}{240}=0.529[/tex] estimated proportion of college students expressed an intent to vote

[tex]p_o=0.6[/tex] is the value that we want to test

[tex]\alpha=0.01[/tex] represent the significance level

Confidence=99% or 0.99

z would represent the statistic (variable of interest)

[tex]p_v[/tex] represent the p value (variable of interest)  

Concepts and formulas to use  

We need to conduct a hypothesis in order to test the claim that at least 60% of students are intented to vote .:  

Null hypothesis:[tex]p \geq 0.6[/tex]  

Alternative hypothesis:[tex]p < 0.6[/tex]  

When we conduct a proportion test we need to use the z statistic, and the is given by:  

[tex]z=\frac{\hat p -p_o}{\sqrt{\frac{p_o (1-p_o)}{n}}}[/tex] (1)  

The One-Sample Proportion Test is used to assess whether a population proportion [tex]\hat p[/tex] is significantly different from a hypothesized value [tex]p_o[/tex].

Calculate the statistic  

Since we have all the info requires we can replace in formula (1) like this:  

[tex]z=\frac{0.529 -0.6}{\sqrt{\frac{0.6(1-0.6)}{240}}}=-2.24[/tex]  

Statistical decision  

It's important to refresh the p value method or p value approach . "This method is about determining "likely" or "unlikely" by determining the probability assuming the null hypothesis were true of observing a more extreme test statistic in the direction of the alternative hypothesis than the one observed". Or in other words is just a method to have an statistical decision to fail to reject or reject the null hypothesis.  

The significance level provided [tex]\alpha=0.01[/tex]. The next step would be calculate the p value for this test.  

Since is a left tailed test the p value would be:  

[tex]p_v =P(z<-2.24)=0.0125[/tex]  

If we compare  the p value and the significance level given [tex]\alpha=0.01[/tex] we see that [tex]p_v>\alpha[/tex] so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, and we can said that at 1% of significance the proportion college students expressed an intent to vote is not higher than 0.6

Answer:

a) [tex]X \sim Binom(n=16, p=0.2)[/tex]  

The probability mass function for the Binomial distribution is given as:  

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

Where (nCx) means combinatory and it's given by this formula:  

[tex]nCx=\frac{n!}{(n-x)! x!}[/tex]  

The expected value is given by this formula:

[tex]E(X) = np=16*0.2=3.2[/tex]

b) [tex]P(X=0)=(16C0)(0.2)^{0} (1-0.2)^{16-0}=0.02815[/tex]

Step-by-step explanation:

Previous concepts  

The binomial distribution is a "DISCRETE probability distribution that summarizes the probability that a value will take one of two independent values under a given set of parameters. The assumptions for the binomial distribution are that there is only one outcome for each trial, each trial has the same probability of success, and each trial is mutually exclusive, or independent of each other".  

Part a

Let X the random variable of interest, on this case we now that:  

[tex]X \sim Binom(n=16, p=0.2)[/tex]  

The probability mass function for the Binomial distribution is given as:  

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

Where (nCx) means combinatory and it's given by this formula:  

[tex]nCx=\frac{n!}{(n-x)! x!}[/tex]  

The expected value is given by this formula:

[tex]E(X) = np=16*0.2=3.2[/tex]

Part b

For this case we want this probability:

[tex] P(X=0)[/tex]

The probability mass function for the Binomial distribution is given as:  

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

Where (nCx) means combinatory and it's given by this formula:  

[tex]nCx=\frac{n!}{(n-x)! x!}[/tex]  

And using this function we got:

[tex]P(X=0)=(16C0)(0.2)^{0} (1-0.2)^{16-0}=0.02815[/tex]

Final answer:

The expected number of phishing emails to be caught by the filter is 3.2, and the probability of the filter detecting none of them is approximately 0.02867.

Explanation:

To solve the student's question regarding phishing email filters, we'll apply basic probability concepts.

Part (a): Expected number of phishing emails caught by the filter

Given that software filters catch 20 percent of phishing URLs, and Jason receives 16 phishing e-mails, the expected number of e-mails caught can be calculated as:

Expected number = Total emails × Catch rate

= 16 × 0.20 = 3.2

So, the filter is expected to catch 3.2 phishing e-mails.

Part (b): Probability of the filter detecting none of them

If the filter's catch rate is 20%, then the chance of not catching a single phishing email is the complement, which is 80% or 0.8 for each email. For none of the 16 emails to be caught, we raise this probability to the power of 16:

Probability = (0.8)^16

≈ 0.02867 (rounded to 5 decimal places)

Thus, the chance of the filter not detecting any of the phishing emails is approximately 0.02867.

The probability of guessing exactly 18 out of 27 questions correctly is calculated using the binomial probability formula. Use the formula: P(x=k) = (n choose k) * (p^k) * (1-p)^(n-k), where n is the number of trials (27), p is the probability of success (1/5), and k is the number of successes (18). The answer will be a very small number.

The problem involves the use of binomial probability because we have a fixed number of trials (27 questions), each of which is independent and has only two outcomes (correct or incorrect) with constant probabilities. In this case, the probability of a success (choosing the correct answer) is 1/5 and the probability of a failure (choosing an incorrect answer) is 4/5.

We wish to find P(x=18), the probability of having exactly 18 successes. Using the formula for the probability of a binomial distribution:

P(x=k) = (n choose k) * [(p)^(k)] * [(1-p)^(n-k)]

Where:
- (n choose k) = n! / (k!(n-k)!)
- p is the probability of success
- k is the number of successes
- n is the number of trials.

Substituting the values:

P(x=18) = (27 choose 18) * ((1/5)^18) * ((4/5)^(27-18))

You can use a calculator to solve for the answer. Please note that the probability would be very small, as expected, because guessing 18 questions correctly out of 27 is not a very likely occurrence when each question has 5 choices.

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Answer:

f(x) = 8 (x + ¼)² − ½

Step-by-step explanation:

f(x) = 8x² + 4x

Divide both sides by 8.

1/8 f(x) = x² + 1/2 x

Take half of the second coefficient, square it, then add to both sides.

(½ / 2)² = (1/4)² = 1/16

1/8 f(x) + 1/16 = x² + 1/2 x + 1/6

Factor the perfect square.

1/8 f(x) + 1/16 = (x + 1/4)²

Multiply both sides by 8.

f(x) + 1/2 = 8 (x + 1/4)²

Subtract 1/2 from both sides.

f(x) = 8 (x + 1/4)² − 1/2

Answer:

(P(t)) = P₀/(1 - P₀(kt)) was proved below.

Step-by-step explanation:

From the question, since β and δ are both proportional to P, we can deduce the following equation ;

dP/dt = k(M-P)P

dP/dt = (P^(2))(A-B)

If k = (A-B);

dP/dt = (P^(2))k

Thus, we obtain;

dP/(P^(2)) = k dt

((P(t), P₀)∫)dS/(S^(2)) = k∫dt

Thus; [(-1)/P(t)] + (1/P₀) = kt

Simplifying,

1/(P(t)) = (1/P₀) - kt

Multiply each term by (P(t)) to get ;

1 = (P(t))/P₀) - (P(t))(kt)

Multiply each term by (P₀) to give ;

P₀ = (P(t))[1 - P₀(kt)]

Divide both sides by (1-kt),

Thus; (P(t)) = P₀/(1 - P₀(kt))

(P(t)) = P₀/(1 - P₀(kt))

According to the, since β and also δ are both proportional to P, we can deduce the following equation ;

Then dP/dt = k(M-P)P

Then dP/dt = (P^(2))(A-B)

Now, If k = (A-B);

After that dP/dt = (P^(2))k

Thus, we obtain;

Now dP/(P^(2)) = k dt

((P(t), P₀)∫)dS/(S^(2)) = k∫dt

Thus; [(-1)/P(t)] + (1/P₀) = kt

Simplifying,

Then 1/(P(t)) = (1/P₀) - kt

Multiply each term by (P(t)) to get ;

After that 1 = (P(t))/P₀) - (P(t))(kt)

Multiply each term by (P₀) to give ;

Now P₀ = (P(t))[1 - P₀(kt)]

Then Divide both sides by (1-kt),

Thus; (P(t)) = P₀/(1 - P₀(kt))

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Answer:

[tex] E(X) =sum_{i=1}^n X_i P(X_i)[/tex]

And we know that each time that we win we recieve $1 and each time we loss we need to pay $1, so then the expected value would be given by:

[tex] E(X) = 1*\frac{4}{9} -1 \frac{5}{9} = -\frac{1}{9} =-0.11[/tex]

[tex] E(X^2) =sum_{i=1}^n X^2_i P(X_i)[/tex]

[tex] E(X^2) = 1^2*\frac{4}{9}+ (-1)^2 \frac{5}{9} = 1[/tex]

And the variance is defined as:

[tex] Var(X) = E(X^2) -[E(X)]^2 = 1 -[-\frac{1}{9}]^2 = \frac{80}{81}[/tex]

Step-by-step explanation:

For this case we know that we can win if we select 2 balls of the same color, so we can find the probability of win like this:

[tex] p = \frac{Possible}{ Total}= \frac{2C1 * (5C2)}{10C2}= \frac{2*10}{45}= \frac{4}{9}[/tex]

So then the probability of no win would be given by the complement:

[tex] q = 1-p = 1- \frac{4}{9}= \frac{5}{9}[/tex]

We can define the random variable X who represent the amount of money that we can win.

And we can use the definition of expected value given by:

[tex] E(X) =sum_{i=1}^n X_i P(X_i)[/tex]

And we know that each time that we win we recieve $1 and each time we loss we need to pay $1, so then the expected value would be given by:

[tex] E(X) = 1*\frac{4}{9} -1 \frac{5}{9} = -\frac{1}{9} =-0.11[/tex]

We can calculate the second monet like this:

[tex] E(X^2) =sum_{i=1}^n X^2_i P(X_i)[/tex]

[tex] E(X^2) = 1^2*\frac{4}{9}+ (-1)^2 \frac{5}{9} = 1[/tex]

And the variance is defined as:

[tex] Var(X) = E(X^2) -[E(X)]^2 = 1 -[-\frac{1}{9}]^2 = \frac{80}{81}[/tex]

A part of the question is missing and it says;

b) Calculate the variance of the amount you win.

Answer:

A) Expected value of Pay out;

(E(Y)) = - 0.11

B) Variance of the amount won; Var(Y) = 0.9879

Step-by-step explanation:

A) From the question, we have;

X ∈ {0,1,2} and by the nature of the question, X has a hypergeometric distribution as;

P(X =i) = [(5,i) (5, 2 - i)] / (10,2)

Furthermore, when we consider the random variable Y that marks the amount of money that we win, we'll get a function of X as;

Y = 1X [x∈{0,2}] - [1X(x=1)]

If we now use the linearity of expectation, we'll get;

E(Y) = E [X(x∈{0,2}) ] - [E(X(x=1))]

= P(X = 0) + P(X = 2) - P(X = 1)

For 2 balls with probability of a win, P = (possible outcome) /(total outcome) =

{(2C1) (5C2)}/(10C2) = (2 x 10)/45 = 20/45

While, for probability of no win;

P = 1 - (20/25) = 25/25

So, E(Y) =20/45 - 25/45 = - 5/45 =

- 0.11

B) Now let's calculate for the variance;

Var(Y) = E(Y(^2)) - E(Y)^(2)

Now, from question a, using the equation of Y, we can say;

(Y)^(2) = (1^2)X [x∈{0,2}] - [(1^2)X(x=1)]

And so;

E(Y^(2)) = P(X = 0) + P(X = 2) + P(X = 1) = (1^2)(20/45) + (-1^2)(25/45) = 45/45 = 1

So, Var(Y) = 1 - (-0.11)^2 = 0.9879

Answer:

there is no stagnation point

Step-by-step explanation:

for the velocity field V→(u,v)= (0.46 + 2.1x)i→ + (−2.8 - 2.1y)j , the stagnation point is found when the velocity vectors converge in one point ( thus also stays in that place when the point is reached). Thus the stagnation point can be found when the divergence of the velocity field is <0 ( thus it does not diverge , but converges)

div(V) = ∇*V= d/dx (0.46 + 2.1x) + d/dy (−2.8 - 2.1y) = 0

2.1 - 2.1 = 0

since div(V) can never be  <0 , there is no stagnation point

Answer:

The required sample size is 350 to estimate the proportion with a margin of error of 0.05          

Step-by-step explanation:

We are given the following in the question:

[tex]\hat{p} = 0.65[/tex]

Confidence level = 95%

Margin of error = 0.05

Confidence interval:

[tex]\hat{p}\pm z_{stat}\sqrt{\dfrac{\hat{p}(1-\hat{p})}{n}}[/tex]

Margin of error =

[tex]z_{stat}\sqrt{\dfrac{\hat{p}(1-\hat{p})}{n}}[/tex]

[tex]z_{critical}\text{ at}~\alpha_{0.05} = 1.96[/tex]

Putting values, we get,

[tex]0.05 = 1.96\sqrt{\dfrac{(0.65)(1-0.65)}{n}}\\\\n = \dfrac{(1.96)^2(0.65)(1-0.65)}{(0.05)^2}\\\\n = 349.5856\\n \approx 350[/tex]

Thus, the required sample size is 350 to estimate the proportion with a margin of error of 0.05

The proportion of students who do not change their passwords with a 5% margin of error using a preliminary estimate of 65%, a sample size of approximately 369 students is needed.

To determine the sample size required to estimate the proportion of students who do not change their password with a margin of error of 0.05, using a preliminary estimate of 0.65, we use the formula for determining sample size in a proportion survey:

n = (Z² × p(1 - p)) / E²

Where:

n = required sample size

Z = Z-score associated with the desired confidence level (For a 95% confidence level, Z = 1.96)

p = preliminary estimate of the proportion (0.65 in this case)

E = margin of error (0.05 as specified)

Substituting the values into the formula, we get:

n = (1.96² × 0.65(1 - 0.65)) / 0.05²

  = 369

Hence, to estimate the proportion of students who do not change their password with a 5% margin of error, a sample size of approximately 369 students is required.

This result helps ensure that research or surveys conducted on certain populations achieve a high level of accuracy within the specified confidence level and margin of error.

Answer:

Step-by-step explanation:

Derive the validity of universal form of part(a) of the elimination rule from the validity of universal instantiation and the valid argument called elimination in Section 2.3.

P(x)∨Q(x)

~Q(x)

∵ P(x)

Universal Instantiation has the following argument form

∀ x ∈ D, P (x)

P(a) for a particular a∈D

Universal Elimination Rule:

∀x, P(x)

∵~ P(a)

Here is a particular value.

P(a) For a particular  a∈D

Since the universal elimination is same as universal instantiation.

Therefore, Universal elimination is valid when universal instantiation and elimination rule are  valid

Answer:

Check attachment for complete question

Step-by-step explanation:

Given that,

y=Coskt

We are looking for value of k, that satisfies 4y''=-25y

Let find y' and y''

y=Coskt

y'=-kSinkt

y''=-k²Coskt

Then, applying this 4y'"=-25y

4(-k²Coskt)=-25Coskt

-4k²Coskt=-25Coskt

Divide through by Coskt and we assume Coskt is not equal to zero

-4k²=-25

k²=-25/-4

k²=25/4

Then, k=√(25/4)

k= ± 5/2

b. Let assume we want to use this

y=ASinkt+BCoskt

Since k= ± 5/2

y=A•Sin(±5/2t)+ B •Cos(±5/2t)

y'=±5/2ACos(±5/2t)-±5/2BSin(±5/2t)

y''=-25/4ASin(±5/2t)-25/4BCos(±5/2t

Then, inserting this to our equation given to check if it a solution to y=ASinkt+BCoskt

4y''=-25y

For 4y''

4(-25/4ASin(±5/2t)-25/4BCos(±5/2t))

-25A•Sin(±5/2t)-25B•Cos(±5/2t).

Then,

-25y

-25(A•Sin(±5/2t)+ B •Cos(±5/2t))

-25A•Sin(±5/2t) - 25B •Cos(±5/2t)

Then, we notice that, 4y'' is equal to -25y, then we can say that y=Coskt is a solution to y=ASinkt+BCoskt

The family of functions y = A sin(kt) + B cos(kt) and their derivatives can be verified as solutions to the second-order differential equation associated with a simple harmonic oscillator equation, regardless of the frequency 'k'. This is due to the periodic nature of the sine and cosine functions.

The subject of this question concerns the family of functions y = A sin(kt) + B cos(kt) and their derivatives. It seems the student is asked to verify that for any value of 'k', representing the frequency of the sine and cosine functions, each function in this family is indeed a solution of some differential equations.

To prove this, we first observe the time derivative of this function family: y' = Ak cos(kt) − Bk sin(kt). A second derivative will yield y'' = −Ak² sin(kt) − Bk² cos(kt). By comparing y'' with the original function (y), we see that y'' is equivalent to -k²*y with y as the original function, which verifies the solution to a simple harmonic oscillator equation. Represented as y''+k²*y=0.

Oscillations are an inherent property of these functions - occurring because sin(kt) and cos(kt) are periodic functions, meaning they repeat their values in regular intervals or periods. The values of 'A' and 'B' simply shift the oscillation up or down (amplitude), and don't change the periodic nature. Therefore, it can be concluded that every member of the family of functions y = A sin(kt) + B cos(kt) is indeed a solution for any picked value of 'k'.

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Answer:

Interval within 1 standard deviation: (16.3,19.7)

Interval within 2 standard deviation: (14.6, 21.4)

Interval within 3 standard deviation: (12.9, 23.1)

Step-by-step explanation:

We are given the following in the question:

Population mean, [tex]\mu[/tex] = 18

Standard deviation, [tex]\sigma[/tex] = 1.7

We have to find the following intervals:

Interval within 1 standard deviation:

[tex]\mu \pm 1\sigma\\=18 \pm 1.7\\=(16.3, 19.7)[/tex]

Interval within 2 standard deviation:

[tex]\mu \pm 2\sigma\\=18 \pm 2(1.7)\\= 18 \pm 3.4\\=(14.6,21.4)[/tex]

Interval within 3 standard deviation:

[tex]\mu \pm 3\sigma\\=18 \pm 2(1.7)\\= 18 \pm 5.1\\=(12.9,23.1)[/tex]

Answer:

Yes: [tex](x+4)(x-4)[/tex]

Step-by-step explanation:

Factoring

Converting a sum or subtraction of terms into a product is calling factoring. The expression

[tex]x^2-16[/tex]

can be written as

[tex]x^2-4^2[/tex]

It's a difference of two squares. It can be factored as the sum by the difference of two factors:

[tex]x^2-4^2=(x+4)(x-4)[/tex]

The correct factorization of [tex]\(x^2 - 16\)[/tex] as a difference of two squares is [tex]\((x + 4)(x - 4)\).[/tex]

Yes, [tex]\(x^2 - 16\)[/tex] is a difference of two squares because it can be expressed as [tex]\(x^2 - 4^2\)[/tex]. The difference of two squares pattern is given by [tex]\(a^2 - b^2 = (a + b)(a - b)\)[/tex], where \(a\) and \(b\) are any real numbers.

In this case, [tex]\(x^2 - 16\)[/tex] can be factored as [tex]\((x + 4)(x - 4)\)[/tex] using the difference of two squares pattern. When you multiply \((x + 4)\) and \((x - 4)\) using the distributive property, you get [tex]\(x^2 - 4x + 4x - 16\)[/tex], which simplifies to [tex]\(x^2 - 16\).[/tex] The correct choice is: **yes; (x + 4)(x - 4)**.

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Answer:

So the probability is P=0.00256.

Step-by-step explanation:

We have 26 letters, so the probability that the first letter in the name is the same is 1/26.

The probability that the second letter in the name is the same is 1/26 and the probability that the third letter in the name is the same is 1/26.

Out of ten people we choose 2.

So the probability is:

[tex]P=C_2^{10}\left(\frac{1}{26}\right)^3\\\\P=\frac{10!}{2!(10-2)!}\cdot \left(\frac{1}{26}\right)^3\\\\P=\frac{45}{17576}\\\\P=0.00256\\[/tex]

So the probability is P=0.00256.

Answer:

Step-by-step explanation:

Hello!

X: Cholesterol level of a woman aged 30-39. (mg/dl)

This variable has an approximately normal distribution with mean μ= 190.14 mg/dl

1. You need to find the corresponding Z-value that corresponds to the top 9.3% of the distribution, i.e. is the value of the standard normal distribution that has above it 0.093 of the distribution and below it is 0.907, symbolically:

P(Z≥z₀)= 0.093

-*or*-

P(Z≤z₀)= 0.907

Since the Z-table shows accumulative probabilities P(Z<Z₁₋α) I'll work with the second expression:

P(Z≤z₀)= 0.907

Now all you have to do is look for the given probability in the body of the table and reach the margins to obtain the corresponding Z value. The first column gives you the integer and first decimal value and the first row gives you the second decimal value:

z₀= 1.323

2.

Using the Z value from 1., the mean Cholesterol level (μ= 190.14 mg/dl) and the Medical guideline that indicates that 9.3% of the women have levels above 240 mg/dl you can clear the standard deviation of the distribution from the Z-formula:

Z= (X- μ)/δ ~N(0;1)

Z= (X- μ)/δ

Z*δ= X- μ

δ=(X- μ)/Z

δ=(240-190.14)/1.323

δ= 37.687 ≅ 37.7 mg/dl

I hope it helps!

Answer:

a) Attached

b) P=0.60

c) P=0.80

d) The expected flight time is E(t)=122.5

Step-by-step explanation:

The distribution is uniform between 1 hour and 50 minutes (110 min) and 135 min.

The height of the probability function will be:

[tex]h=\frac{1}{Max-Min}=\frac{1}{135-110} =\frac{1}{25}[/tex]

Then the probability distribution can be defined as:

[tex]f(t)=\frac{1}{25}=0.04 \,\,\,\,\\\\t\in[110,135][/tex]

b) No more than 5 minutes late means the flight time is 125 or less.

The probability of having a flight time of 125 or less is P=0.60:

[tex]F(T<t)=0.04(t-min)\\\\F(T<125)=0.04*(125-110)=0.04*15=0.60[/tex]

c) More than 10 minutes late means 130 minutes or more

The probability of having a flight time of 130 or more is P=0.80:

[tex]F(T>t)=1-0.04(t-110)\\\\F(T>130)=1-0.04*(130-110)=1-0.04*20=1-0.8=0.2[/tex]

d) The expected flight time is E(t)=122.5

[tex]E(t)=\frac{1}{2}(max+min)= \frac{1}{2}(135+110)=\frac{1}{2}*245=122.5[/tex]

Answer:

The probability of the number of starfish dropping below 3 is 0.0463.

Step-by-step explanation:

Let X represent the number of starfish in the tide pool.

X follows a Poisson distribution with mean 6.4.

The formula for Poisson distribution is as follows.

[tex]P(X=x) = e^{-6.4} * [6.4^{x} / x!] when x = 0, 1, ...[/tex]

[tex]P(X = x) = 0[/tex] otherwise

We need to find the probability that the number of starfish in the tide pool drops below 3.

Therefore, the required probability is:

P(X < 3) = P(X = 0) + P(X = 1) + P(X=2)

[tex]P(X < 3) = e^{-6.4} + (e^{-6.4}*6.4) + (e^{-6.4}*6.4^{2}/2)[/tex]

P(X < 3) = 0.00166 + 0.01063 + 0.03403

P (X < 3) = 0.0463

Answer:

the dimensions that minimize the cost of the cylinder are R= 3.85 cm and L=12.88 cm

Step-by-step explanation:

since the volume of a cylinder is

V= π*R²*L → L =V/ (π*R²)

the cost function is

Cost = cost of side material * side area  + cost of top and bottom material * top and bottom area

C = a* 2*π*R*L + b* 2*π*R²

replacing the value of L

C = a* 2*π*R* V/ (π*R²) + b* 2*π*R²  = a* 2*V/R + b* 2*π*R²

then the optimal radius for minimum cost can be found when the derivative of the cost with respect to the radius equals 0 , then

dC/dR = -2*a*V/R² + 4*π*b*R = 0

4*π*b*R = 2*a*V/R²

R³ = a*V/(2*π*b)

R=  ∛( a*V/(2*π*b))

replacing values

R=  ∛( a*V/(2*π*b)) = ∛(0.03$/cm² * 600 cm³ /(2*π* 0.05$/cm²) )= 3.85 cm

then

L =V/ (π*R²) = 600 cm³/(π*(3.85 cm)²) = 12.88 cm

therefore the dimensions that minimize the cost of the cylinder are R= 3.85 cm and L=12.88 cm

The dimensions of the cylinder that minimize production costs are approximately a radius of 3.64 cm and a height of 14.41 cm. These values are derived by expressing the height in terms of the radius, calculating the surface area, and optimizing the cost function. The minimized cost is achieved with these optimal dimensions.

To minimize the production cost for a cylindrical can holding 600 cubic centimeters of soup, we need to find the optimal dimensions (radius and height) considering cost constraints of the materials.

The volume of the cylinder is given by the formula:

Given V = 600 cm³, we can express the height in terms of the radius:

Next, we calculate the surface area, because the cost is based on the material's surface area. The surface area A includes the area of the side, and the top and bottom:

Substitute h from the volume equation:

A = 2πr(600 / (πr²)) + 2πr²
A = 1200/r + 2πr²

The total cost is calculated by multiplying the areas by their respective costs:

Cost = 0.03(2πrh) + 0.05(2πr²)
Substitute h:

Cost = 0.03(2πr(600 / (πr²))) + 0.05(2πr²)
Cost = 0.03(1200/r) + 0.1πr²
Cost = 36/r + 0.1πr²

We find the minimum cost by differentiating the cost function with respect to r and setting it to zero:

d(Cost)/dr = -36/r² + 0.2πr = 0
0.2πr = 36/r²
0.2πr³ = 36
r³ = 180/π
r ≈ 3.64 cm

Finally, we calculate the height h:

h ≈ 600 / (π(3.64)²)

≈ 14.41 cm

Therefore, the dimensions that minimize the production cost are approximately a radius of 3.64 cm and a height of 14.41 cm.

Answer:

The number of variables in the set is 5

Step-by-step explanation:

we us to able to understand how we arrived at 5 as the number of variables that is available in the set of data above we have to define and understand the keyword variable.

A variable; is any characteristics, number, or quantity that can be measured or counted. A variable may also be called a data item.

hence by the definition of a variable we know that hometown is a variable=1,GPA is a variable=2, number of classes is a variable=3, number of siblings is a variable=4, and favorite subject is also a variable=5.

hence we conclude that the number of variables=5

Answer:

Step-by-step explanation:

The given polygon is a square. To determine the apotherm which is the perpendicular line from the midpoint of the square, we would apply Pythagoras theorem which is expressed as

Hypotenuse² = opposite side² + adjacent side²

Let a represent the apotherm

Apotherm = length of each side of the square.

Therefore

8² = a² + a² = 2a²

64 = 2a²

a² = 64/2 = 32

a = √32

The formula for determining the area of a polygon is

Area of polygon is

area = a^2 × n ×tan 180/n

Where n is the number of sides

(n = 4)

Area = √32² × 4 × tan(180/4)

Area = 128 × 1

Area = 128

The formula for determining the perimeter of a regular polygon is

P = 2 × area/apotherm

Perimeter = 2 × 128/√32

Perimeter = 45.3

Answer:

Step-by-step explanation:

Hello!

You have the variable:

X: Weight of a small bag of candies (ounces)

n= 12 small bags

i) Sample mean, X[bar]= 3 ounces

This is the expected weight of the small bags of candies.

ii) Population Standard deviation, σ= 0.1 ounces

The population standard deviation is a measurement of variability, it shows you how dispersed are the population values from the population mean.

iii) Sample standard deviation, S= 0.13 ounces

The sample standard deviation is the point estimate of the population standard deviation, it is a measure of variability, it shows the dispersion of the data around the sample mean.

I hope it helps!

Surface area of this rectangular prism is A=2(1x10+8x10+8x1) so the area for this one is 196 ft

Answer:

The column height of air after the inversion is 7 cm.

Step-by-step explanation:

The initial pressure balance is given as

P_1+20 cm=76 cm of Hg

P_1=76-20 cm of Hg

P_1=56 cm of Hg

The initial volume of the air with cross sectional area as 12 cm2 and the length of air column as 12 is given as V_1=12 cm *1 =12 cm3

After the inversion

P_2=20 cm+76 cm of Hg

P_2=96 cm of Hg

The volume of the air after the inversion with cross sectional area as 1 cm2 and the length of air column as x is given as V_2=x *1 =x cm3

Now as temperature is constant and the cross sectional area is also constant so

[tex]P_1V_1=P_2V_2\\56\times 12=96\times x\\x=\dfrac{56\times 12}{96}\\x=7 cm[/tex]

So the column height of air after the inversion is 7 cm.

Answer:

7cm

Step-by-step explanation:

P1 =H-h =76-20=56; P2 =H+h =96

P1V1 =P2V2

56x12=96xV2

V2 =56x12/96 = 7 cm

Answer:

Both the answers are as in the solution.

Step-by-step explanation:

As the given matrix is not in the readable form, a similar question is found online and the solution of which is attached herewith.

Part a:

Given matrix is : A = [tex]\left[\begin{array}{ccc}0&3&4\\1&2&3\\-3&-7&8\end{array}\right][/tex]

Here,

[tex]det(A) =\left|\begin{array}{ccc}0&3&4\\1&2&3\\-3&-7&8\end{array}\right| = -55 \neq 0.[/tex]

Then, A is non-singular matrix.

Here, A₁₁= 0.

If we write A as LU with L lower triangular matrix and U upper triangular matrix, then A₁₁=L₁₁U₁₁.

So, As

A₁₁ = 0 gives L₁₁U₁₁= 0 ,

This indicates that either L₁₁= 0 or U₁₁ = 0.

If L₁₁= 0 or U₁₁ = 0, this would made the corresponding matrix singular, which contradicts the condition as  A is non-singular.

Therefore, A has no LU decomposition.

Part b:

By the implementation of the various row operations

interchange R1 and R2

[tex]\left[\begin{array}{ccc}1&2&3\\0&3&4\\-3&-7&8\end{array}\right][/tex]

R3+3R1=R3

[tex]\left[\begin{array}{ccc}1&2&3\\0&3&4\\0&-1&17\end{array}\right][/tex]

R3+(1/3)R2 = R3

[tex]\left[\begin{array}{ccc}1&2&3\\0&3&4\\0&0&55/3\end{array}\right][/tex]

Therefore, U = [tex]\left[\begin{array}{ccc}1&2&3\\0&3&4\\0&0&55/3\end{array}\right][/tex].

Here, LP = E₁₂=E₃₁=-3 &E₃₂=-1/3

[tex]LP=\left[\begin{array}{ccc}0&1&0\\1&0&0\\0&0&1\end{array}\right] \left[\begin{array}{ccc}1&0&0\\0&1&0\\-3&0&1\end{array}\right]\left[\begin{array}{ccc}1&0&0\\0&1&0\\0&-1/3&1\end{array}\right][/tex]

[tex]LP=\left[\begin{array}{ccc}0&1&0\\1&0&0\\-3&-1/3&1\end{array}\right][/tex]

[tex]LP=\left[\begin{array}{ccc}1&0&0\\0&1&0\\-1/3&-3&1\end{array}\right]\left[\begin{array}{ccc}0&1&0\\1&0&0\\0&0&1\end{array}\right][/tex]

So now U is given as

[tex]U=\left[\begin{array}{ccc}1&2&3\\0&3&4\\0&0&55/3\end{array}\right]\\L=\left[\begin{array}{ccc}1&0&0\\0&1&0\\-1/3&-3&1\end{array}\right]\\P=\left[\begin{array}{ccc}0&1&0\\1&0&0\\0&0&1\end{array}\right]\\[/tex]