How does the product of 1/2 x 6/5 compare to the product of 1/2 x 5/6?

Answer:

Step-by-step explanation:

Scenarios are: Sample Distribution, sampling distribution, Population distribution

sample distribution: A sample distribution contains the data of individuals of sample collected from a population.

Sampling distribution: distribution of multiple sample statistics

Population distribution: Statistics of entire population without drawing sny sample

In scenario 1, only one saple is taken from a population and that sample is plotted

In Scenario 2, multiple samples are taken from a population and means of each sample is plotted

In Scenario 3, scores of entire population is considered.

Answer:

97.10% probability that five or more of the original 2000 components fail during the useful life of the product.

Step-by-step explanation:

For each component, there are only two possible outcomes. Either it works correctly, or it does not. The probability of a component falling is independent from other components. 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.

In this problem we have that:

[tex]n = 2000, p = 1-0.995 = 0.005[/tex]

Approximate the probability that five or more of the original 2000 components fail during the useful life of the product.

We know that either less than five compoenents fail, or at least five do. The sum of the probabilities of these events is decimal 1. So

[tex]P(X < 5) + P(X \geq 5) = 1[/tex]

We want [tex]P(X \geq 5)[/tex]

So

[tex]P(X \geq 5) = 1 - P(X < 5)[/tex]

In which

[tex]P(X < 5) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4)[/tex]

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

[tex]P(X = 0) = C_{2000,0}.(0.005)^{0}.(0.995)^{2000} = 0.000044[/tex]

[tex]P(X = 1) = C_{2000,1}.(0.005)^{1}.(0.995)^{1999} = 0.000445[/tex]

[tex]P(X = 2) = C_{2000,2}.(0.005)^{2}.(0.995)^{1998} = 0.002235[/tex]

[tex]P(X = 3) = C_{2000,3}.(0.005)^{3}.(0.995)^{1997} = 0.007480[/tex]

[tex]P(X = 4) = C_{2000,4}.(0.005)^{4}.(0.995)^{1996} = 0.018765[/tex]

[tex]P(X < 5) = P(X = 0) + `P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) = 0.000044 + 0.000445 + 0.002235 + 0.007480 + 0.018765 = 0.0290[/tex]

[tex]P(X \geq 5) = 1 - P(X < 5) = 1 - 0.0290 = 0.9710[/tex]

97.10% probability that five or more of the original 2000 components fail during the useful life of the product.

Answer and Step-by-step explanation:

The answer is attached below

In this exercise we have to use the knowledge of parameterization and calculate the direction and direction of the equation, so we have to:

A) Clockwise: [tex]t \in [ -\infty, 1/6][/tex]

B) Counter-clockwise: [tex]t \in [ 1/6, \infty][/tex]

C) [tex]\theta \in [ 0, 2 \pi][/tex]

D) [tex]t= 0 \ or \ t=1/3[/tex]

For this exercise, the following equations were informed:

[tex]x= cos(3t^2-t)\\y= sin(3t^2-t)\\t \in [ -\infty, \infty][/tex]

taking the parameterization we have that:

[tex]\phi = 3t^2 - t= t(3t-1)[/tex]

As t increases from [tex][ -\infty, \infty][/tex]  [tex]\phi[/tex] decreases, after 0 it becomes negative and after 1/3, goes on increasing. Also:

[tex]\frac{d\phi}{dt} = (6t-1)\\t= 1/6[/tex]

a) For clockwise begin [tex]\phi[/tex] must be decreasing, so:

[tex]t \in [ -\infty, 1/6][/tex]

b) For counter-clockwise  [tex]\phi[/tex] must be increasing, so:

[tex]t \in [ 1/6, \infty][/tex]

c) Entise circle gets traced out. For we know:

[tex]x= cos\theta\\y= sin\theta[/tex]

Circle gets traced out once for:

[tex]\theta \in [ 0, 2 \pi][/tex]

d) When point (1, 0) so:

[tex]1= cos(3t^2-t)\\0= sin(3t^2-t)\\t= 0 \or \ t=1/3[/tex]

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Follow below steps:

To solve for q in the equation √3q + 2 = √5, we first isolate the term with q by subtracting 2 from both sides of the equation.

√3q = √5 - 2

Then we square both sides of the equation to remove the square root:

(√3q)² = ( √5 - 2 )²

3q = ( √5 - 2 )²

Now, expand the right side of the equation:

3q = 5 - 2√5 * 2 + 2²

3q = 5 - 4√5 + 4

3q = 9 - 4√5

Then, divide both sides of the equation by 3 to solve for q:

q = (9 - 4√5) / 3

So, the value of q is (9 - 4√5) / 3.

The value of q is [tex]\frac{9 - 4\sqrt{5}}{3}[/tex].

To solve for q, we'll isolate it by performing operations to both sides of the equation to get q by itself.

Given the equation:

[tex]\[ \sqrt{3q} + 2 = \sqrt{5} \][/tex]

Subtract 2 from both sides:

[tex]\[ \sqrt{3q} = \sqrt{5} - 2 \][/tex]

Now, to isolate q, we need to square both sides of the equation:

[tex]\[ (\sqrt{3q})^2 = (\sqrt{5} - 2)^2 \]\[ 3q = (\sqrt{5} - 2)^2 \]\[ 3q = 5 - 4\sqrt{5} + 4 \]\[ 3q = 9 - 4\sqrt{5} \][/tex]

Now, divide both sides by 3 to solve for q:

[tex]\[ q = \frac{9 - 4\sqrt{5}}{3} \][/tex]

So, [tex]\( q = \frac{9 - 4\sqrt{5}}{3} \).[/tex]

Answer:

[tex] P(X <9) [/tex]

And we can use the cumulative distribution function given by:

[tex] F(X) = \frac{x-a}{b-a}, a\leq X \leq b[/tex]

And using this we got:

[tex] P(X <9) = F(9) = \frac{9-3}{11-3}= 0.75[/tex]

Step-by-step explanation:

For this case we assume that X= represent the waiting times and for this case we have the following distribution:

[tex] X \sim Unif (a= 3, b =11)[/tex]

And the expected value is given by:

[tex]\mu= E(X) = \frac{a+b}{2}= \frac{3+11}{2}=7[/tex]

And the variance is given by:

[tex] Var(X) \sigma^2 = \frac{(b-a)^2}{12} = \frac{(11-3)^2}{12} = 5.333[/tex]

And we can find the deviation like this:

[tex] Sd(X) = \sqrt{5.333}= 2.309[/tex]

And we want to find this probability:

[tex] P(X <9) [/tex]

And we can use the cumulative distribution function given by:

[tex] F(X) = \frac{x-a}{b-a}, a\leq X \leq b[/tex]

And using this we got:

[tex] P(X <9) = F(9) = \frac{9-3}{11-3}= 0.75[/tex]

The probability that a randomly selected Starbucks drive-through customer in the US waits at most 9 minutes to receive their order is 0.75.

Uniform distributions are probability distributions in which all events are equally likely to occur.

Let's assume that X represents the waiting time.

As the distribution is the uniform distribution, we can write a =3 and b =11

Now, the expected value can be written as,

[tex]\mu = E(X) = \dfrac{a+b}{2} = \dfrac{3+11}{2}= 7[/tex]

the variance of the distribution can be written as,

[tex]Var(X) = \sigma^2 = \dfrac{(b-a)^2}{12} = \dfrac{(11-3)^2}{12} =5.34[/tex]

And, the standard deviation can be written as,

[tex]\sigma = \sqrt{5.34}=2.309[/tex]

In order to calculate the probability, we will use the cumulative distribution function. The cumulative function is given by the formula,

[tex]F(X) = \dfrac{x-a}{b-a}, a\leq X\leq b[/tex]

Using the function we can get the probability as,

[tex]F(X < 9) = F(9) = \dfrac{9-3}{11-3} = 0.75[/tex]

Hence,  the probability that a randomly selected Starbucks drive-through customer in the US waits at most 9 minutes to receive their order is 0.75.

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Answer:

0.000001 = 0.0001% probability that exactly 2 of them have never been married

Step-by-step explanation:

For each employed women, there are only two possible outcomes. Either they have already been married, or they have not. The women are chosen at random, which means that the probability of a woman having been already married is independent from other women. 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.

74% of employed women have never been married.

This means that [tex]p = 0.74[/tex]

15 employed women are randomly selected

This means that [tex]n = 15[/tex]

a. What is the probability that exactly 2 of them have never been married

This is P(X = 2).

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

[tex]P(X = 2) = C_{15,2}.(0.74)^{2}.(0.26)^{13} = 0.000001[/tex]

0.000001 = 0.0001% probability that exactly 2 of them have never been married

Answer:

a) Mean = 1.6, standard deviation = 1.21

b) 18.87% probability that zero online retail orders will turn out to be fraudulent.

c) 32.82% probability that one online retail order will turn out to be fraudulent.

d) 48.31% probability that two or more online retail orders will turn out to be fraudulent.

Step-by-step explanation:

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.

The mean of the binomial distribution is:

[tex]E(X) = np[/tex]

The standard deviation of the binomial distribution is:

[tex]\sqrt{V(X)} = \sqrt{np(1-p)}[/tex]

In this problem, we have that:

[tex]p = 0.08, n = 20[/tex]

a. What are the mean and standard deviation of the number of online retail orders that turn out to be fraudulent?

Mean

[tex]E(X) = np = 20*0.08 = 1.6[/tex]

Standard deviation

[tex]\sqrt{V(X)} = \sqrt{np(1-p)} = \sqrt{20*0.08*0.92} = 1.21[/tex]

b. What is the probability that zero online retail orders will turn out to be fraudulent?

This is P(X = 0).

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

[tex]P(X = 0) = C_{20,0}.(0.08)^{0}.(0.92)^{20} = 0.1887[/tex]

18.87% probability that zero online retail orders will turn out to be fraudulent.

c. What is the probability that one online retail order will turn out to be fraudulent?

This is P(X = 1).

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

[tex]P(X = 1) = C_{20,1}.(0.08)^{1}.(0.92)^{19} = 0.3282[/tex]

32.82% probability that one online retail order will turn out to be fraudulent.

d. What is the probability that two or more online retail orders will turn out to be fraudulent?

Either one or less is fraudulent, or two or more are. The sum of the probabilities of these events is decimal 1. So

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

We want [tex]P(X \geq 2)[/tex]

So

[tex]P(X \geq 2) = 1 - P(X \leq 1)[/tex]

In which

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

From itens b and c

[tex]P(X \leq 1) = 0.1887 + 0.3282 = 0.5169[/tex]

[tex]P(X \geq 2) = 1 - P(X \leq 1) = 1 - 0.5169 = 0.4831[/tex]

48.31% probability that two or more online retail orders will turn out to be fraudulent.

The probability is an illustration of a binomial distribution.

The given parameters are:

n = 20

p = 0.08

The mean is calculated as:

[tex]\bar x = np[/tex]

So, we have:

[tex]\bar x = 20 * 0.08[/tex]

[tex]\bar x = 1.6[/tex]

The standard deviation is calculated as:

[tex]\sigma = \sqrt{\bar x * (1 - p)[/tex]

This gives

[tex]\sigma = \sqrt{1.6 * (1 - 0.08)[/tex]

[tex]\sigma = 1.21[/tex]

Hence, the mean is 1.6 and the standard deviation is 1.21

This is calculated as:

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

So, we have:

[tex]P(0) = ^{20}C_0 * 0.08^0 * (1 - 0.08)^{20 - 0}[/tex]

[tex]P(0) =0.1887[/tex]

The probability that zero online retail orders will turn out to be fraudulent is 0.1887

This is calculated as:

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

So, we have:

[tex]P(1) = ^{20}C_1 * 0.08^1 * (1 - 0.08)^{20 - 1}[/tex]

[tex]P(1) =0.3281[/tex]

The probability that one online retail orders will turn out to be fraudulent is 0.3281

This is calculated as:

[tex]P(x\ge 2) = 1 - P(0) - P(1)[/tex]

So, we have:

[tex]P(x\ge 2) = 1 - 0.1887 - 0.3281[/tex]

[tex]P(x\ge 2) = 0.4832[/tex]

The probability that two or more online retail orders will turn out to be fraudulent is 0.4832

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Answer:

0.2843 = 28.43% probability that no less than 92 out of 159 students will pass their college placement exams.

Step-by-step explanation:

I am going to use the binomial approximation to the normal to solve this question.

Binomial probability distribution

Probability of exactly x sucesses on n repeated trials, with p probability.

Can be approximated to a normal distribution, using the expected value and the standard deviation.

The expected value of the binomial distribution is:

[tex]E(X) = np[/tex]

The standard deviation of the binomial distribution is:

[tex]\sqrt{V(X)} = \sqrt{np(1-p)}[/tex]

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.

When we are approximating a binomial distribution to a normal one, we have that [tex]\mu = E(X)[/tex], [tex]\sigma = \sqrt{V(X)}[/tex].

In this problem, we have that:

[tex]p = 0.55, n = 159[/tex]. So

[tex]\mu = E(X) = 159*0.55 = 87.45[/tex]

[tex]\sigma = \sqrt{V(X)} = \sqrt{np(1-p)} = \sqrt{159*0.55*0.45} = 6.27[/tex]

Probability that no less than 92 out of 159 students will pass their college placement exams.

No less than 92 is more than 91, which is 1 subtracted by the pvalue of Z when X = 91. So

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

[tex]Z = \frac{91 - 87.45}{6.27}[/tex]

[tex]Z = 0.57[/tex]

[tex]Z = 0.57[/tex] has a pvalue of 0.7157

1 - 0.7157 = 0.2843

0.2843 = 28.43% probability that no less than 92 out of 159 students will pass their college placement exams.

To approximate the probability that at least 92 out of 159 students will pass their college placement exams using the normal distribution, we will follow these steps:
Step 1: Determine the parameters of the binomial distribution.
In a binomial distribution the parameters are the number of trials (n) and the probability of success in a single trial (p). For this particular case:
n = 159 (the number of students)
p = 0.55 (the probability that a given student will pass)
Step 2: Calculate the mean (μ) and standard deviation (σ) of the binomial distribution.
The mean (μ) of a binomial distribution is given by:
μ = n * p
The standard deviation (σ) is given by:
σ = sqrt(n * p * (1 - p))
For our case:
μ = 159 * 0.55 ≈ 87.45
σ = sqrt(159 * 0.55 * (1 - 0.55)) ≈ sqrt(159 * 0.55 * 0.45) ≈ sqrt(71.775) ≈ 8.472 (rounded to three decimal places for intermediate calculation)
Step 3: Apply the continuity correction.
Because we're approximating a discrete distribution with a continuous one, we use a continuity correction. To find the probability of at least 92 students passing, we'll look for the probability of X > 91.5 (since X is a discrete random variable, X ≥ 92 is equivalent to X > 91.5 when using a continuous approximation).
Step 4: Calculate the z-score for the corrected value.
The z-score is calculated by taking the value of interest, applying the continuity correction, subtracting the mean, and then dividing by the standard deviation:
z = (X - μ) / σ
Using the corrected value:
z = (91.5 - 87.45) / 8.472 ≈ 0.4774 (rounded to four decimal places for the calculation)
Step 5: Find the corresponding probability.
The z-score tells us how many standard deviations away from the mean our value of interest is. To find the probability that at least 92 students pass (i.e., P(X ≥ 92), we need to find 1 - P(Z < z) because the normal distribution table or a calculator gives us P(Z < z), which is the probability of being less than a certain z value.
We are looking for the probability that z is greater than our calculated value, which is the upper tail of the distribution.
Step 6: Consult the standard normal distribution table or use a calculator.
The z-score of approximately 0.4774 corresponds to a percentage in the standard normal distribution. Since we want P(Z > z), we need to subtract P(Z < z) from 1. If we look up the probability for z=0.4774 in standard normal distribution tables or use a calculator, we find that P(Z < z) is approximately 0.6832.
Therefore, P(Z > z) = 1 - P(Z < z) = 1 - 0.6832 = 0.3168.
Step 7: Round the answer.
Rounding P(Z > z) = 0.3168 to four decimal places, the answer remains 0.3168.
So, the approximate probability that at least 92 out of 159 students will pass the exam, using the normal distribution approximation, is 0.3168 when rounded to four decimal places.

Answer:

11.51% probability of observing an 83 degree F temperature or higher LA during a randomly chosen day in June

The coldest 10% of the days during June in LA have high temperatures of 70.6F or lower.

Step-by-step explanation:

Problems of normally distributed samples are 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.

In this problem, we have that:

[tex]\mu = 77, \sigma = 5[/tex]

What is the probability of observing an 83 degree F temperature or higher LA during a randomly chosen day in June?

This probability is 1 subtracted by the pvalue of Z when X = 83. So

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

[tex]Z = \frac{83 - 77}{5}[/tex]

[tex]Z = 1.2[/tex]

[tex]Z = 1.2[/tex] has a pvalue of 0.8849.

1 - 0.8849 = 0.1151

11.51% probability of observing an 83 degree F temperature or higher LA during a randomly chosen day in June

How cold are the coldest 10% of the days during June in LA?

High temperatures of X or lower, in which X is found when Z has a pvalue of 0.1, so whn Z = -1.28

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

[tex]-1.28 = \frac{X - 77}{5}[/tex]

[tex]X - 77 = -1.28*5[/tex]

[tex]X = 70.6[/tex]

The coldest 10% of the days during June in LA have high temperatures of 70.6F or lower.

A) The probability of observing a temperature ≥ 83°F in LA during a randomly chosen day in June is;

p(observing a temperature ≥ 83°F) = 11.507%

B) The coldest 10% of the days during June in LA have temperatures;

Less than or equal to 70.592 °F

This question involves z-distribution which is given by the formula;

z = (x' - μ)/σ

We are given;

Average daily temperature; μ = 77 °F

Standard deviation; σ = 5 °F

Since the temperatures follow a normal distribution, then if we want to find the probability of observing a temperature ≥ 83°F, then;

x' = 83 °F

Thus;

z = (83 - 77)/5

z = 6/5

z = 1.2

Thus;

from online z-score calculator, p-value = 0.11507

Thus, p(observing a temperature ≥ 83°F) = 11.507%

B) We want to find out how cold the coldest 10% of the days during June in LA;

Thus, it means that p = 10% = 0.1

z-score at p = 0.1 from z-score tables is;

z = -1.28155

Thus;

-1.28155 = (x' - 77)/5

-1.28155*5 = x' - 77

-6.40775 = x' - 77

x' = 77 - 6.40775

x' ≈ 70.592 °F

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Answer: 31

Step-by-step explanation:

This is solved Using the theory of sets.

Teachers in Albers school = 44

Teachers in Bothel school = 74

Let the number of teachers that work in both schools be denoted as "x"

This implies that:

Number of teachers in Albers only = 44 - x

Number of teachers in Bothel only = 74 - x

And total number of teachers in both schools = 87,

then

x + (44-x) + (74-x) = 87

118 -x = 87

31 = x

This means the number of teachers that teach in both schools = 31.

Answer: 31 teachers teach at both schools.

Let:

A =  Albers Elementary School

B = Bothel Elementary School

According to the question:

n(A) = 44, n(B) = 74, n(A U B) = 87.

Using the union formula we get:

[tex]n(A \cup B) = n(A)+n(B)-n(A \cap B)\\87 = 44+74-n(A \cap B)\\n(A \cap B)=44+74-87\\n(A \cap B)=31[/tex]

So, 31 teachers teach at both schools.

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Answer:

The 95% confidence interval for the average heights of the population is between 1.7108m and 1.7892m.

Step-by-step explanation:

We have that to find our [tex]\alpha[/tex] level, that is the subtraction of 1 by the confidence interval divided by 2. So:

[tex]\alpha = \frac{1-0.95}{2} = 0.025[/tex]

Now, we have to find z in the Ztable as such z has a pvalue of [tex]1-\alpha[/tex].

So it is z with a pvalue of [tex]1-0.025 = 0.975[/tex], so [tex]z = 1.96[/tex]

Now, find M as such

[tex]M = z*\frac{\sigma}{\sqrt{n}}[/tex]

In which [tex]\sigma[/tex] is the standard deviation of the population and n is the size of the sample.

The standard deviation is the square root of the variance. So

[tex]\sigma = \sqrt{0.16} = 0.4[/tex]

Then

[tex]M = 1.96*\frac{0.4}{\sqrt{400}} = 0.0392[/tex]

The lower end of the interval is the mean subtracted by M. So it is 1.75 - 0.0392 = 1.7108m

The upper end of the interval is the mean added to M. So it is 1.75 + 0.0392 = 1.7892m

The 95% confidence interval for the average heights of the population is between 1.7108m and 1.7892m.

Answer:

Step-by-step explanation:

mean x = [tex]1.75[/tex]

Variance [tex]\rho^2 = 0.16[/tex]

standard deviation [tex](\rho) = \sqrt{0.16} = 0.4[/tex]

n = 400

[tex]95\%[/tex] confidence :

[tex]\alpha = 100\% - 95\% = 5\%\\\\\frac{\alpha}{2} = 2.5\% = 0.025[/tex]

From standard normal distribution table,

[tex]Z_\frac{\alpha}{2} = Z_{0.025} = 1.96[/tex]

Margin of error, [tex]E = Z_\frac{\alpha}{2} * \frac{\rho}{\sqrt{n}}[/tex]

[tex]E = 1.96 * \frac{0.4}{\sqrt{400}}\\\\E = 0.0392[/tex]

Lower limit: x - E

[tex]= 1.75 - 0.0392\\\\= 1.7108[/tex]

Upper limit: x + E

[tex]= 1.75 + 0.0392\\\\= 1.7892[/tex]

[tex]Limits : (1.7108, 1.7892)[/tex]

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Answer:

There is enough evidence to support the claim that the average number of sheets recycled per bin was more than 59.3 sheets.          

Step-by-step explanation:

We are given the following in the question:  

Population mean, μ = 59.3

Sample mean, [tex]\bar{x}[/tex] = 62.4

Sample size, n = 79

Alpha, α = 0.05

Sample standard deviation, s = 9.86

First, we design the null and the alternate hypothesis

[tex]H_{0}: \mu = 59.3\text{ sheets}\\H_A: \mu > 59.3\text{ sheets}[/tex]

We use one-tailed t test to perform this hypothesis.

Formula:

[tex]t_{stat} = \displaystyle\frac{\bar{x} - \mu}{\frac{s}{\sqrt{n}} }[/tex]

Putting all the values, we have

[tex]t_{stat} = \displaystyle\frac{62.4 - 59.3}{\frac{9.86}{\sqrt{79}} } = 2.7945[/tex]

Degree of freedom = n - 1 = 78

Now, [tex]t_{critical} \text{ at 0.05 level of significance, 78 degree of freedom } = 1.6646[/tex]

Since,                        

[tex]t_{stat} > t_{critical}[/tex]

We fail to accept the null hypothesis and reject it. We accept the alternate hypothesis.

Conclusion:

Thus, there is enough evidence to support the claim that the average number of sheets recycled per bin was more than 59.3 sheets.

Answer:

(a)f(t)=1090-5t

(b)f(0)=1090metres

(c)g(t)=1600-9t

(d)g(0)=1600metres

(e)The Hare

Step-by-step explanation:

Total Distance =1600 metres

(a)The tortoise has a 510m headstart and a speed of 5m/s

Distance=Speed X Time

Distance at 5m/s = 5t

Total Distance covered by the tortoise at any time t= 510+5t

Therefore, The tortoise's distance from the finish line (in meters) in terms of the number of seconds t since the start of the race is given as:

f(t)=1600-(510+5t)

f(t)=1090-5t

(b)f(0)=1090-(5X0)

=1090metres

(c)The hare has a speed of 9m/s

Distance=Speed X Time

Distance at 9m/s = 9t

Total Distance covered by the hare at any time t= 9t

Therefore, The hare's distance from the finish line (in meters) in terms of the number of seconds t since the start of the race is given as:

g(t)=1600-9t

(d)

g(0)=1600-(9X0)=1600metres

(e)The race is finished when the distance from the finish line=0

For the Tortoise

f(t)=1090-5t=0

1090=5t

t=218seconds

For the Hare

g(t)=1600-9t=0

9t=1600

t=177.8seconds

The hare takes a shorter time to reach the finish line so he won the race.

Answer:

[tex]f(t)=1804(1.04)^{t}\\f(25)=4809.17[/tex]

Step-by-step explanation:

1. Since the increasing rate is 0.04 or (4%) per day, then the factor is (1+0.04) raised to t days, and we have and exponential growth therefore we can write:

[tex]\\f(t)=c(1.04)^t\\f(t)=1,804(1.04)^t\\[/tex]

2. To estimate the number of cases, 25 days later following that exponential model

[tex]f(25)=1804(1.04)^{25}\\f(25)=4809.17[/tex]

Answer:

hope it helps you see the attachment for further information

Answer:

81

Step-by-step explanation:

Answer:

You earned an A.

Step-by-step explanation:

Problems of normally distributed samples are 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.

In this problem, we have that:

[tex]\mu = 79, \sigma = 6[/tex]

The top 15% of all scores have been designated As.

This means that if Z for the score has a pvalue of 1-0.15 = 0.85 or higher, the score is designated as A.

Your score is 89. Did you earn an A?

We have to find the pvalue of Z when X = 89. So

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

[tex]Z = \frac{89 - 79}{6}[/tex]

[tex]Z = 1.67[/tex]

[tex]Z = 1.67[/tex] has a pvalue of 0.9525. So yes, you earned an A.

Answer:

The function is provided below.

Step-by-step explanation:

Running the program in Python, the code for this problem is as follows:

def func(n):

       count = 0                                           **initializing the sum with 0**

       for i in range (1, n):

                   if i%3 == 0 or i%5 == 0:        

                              count = count + 1

      return (count)

When the program is executed enter value 1 to 10 one by one and the result will be 23, the sum of all the multiples of 3 and 5.

Answer:

[tex]r=\frac{19M}{14h}[/tex]

Step-by-step explanation:

The equation is given as:

[tex]M=\frac{7r2h}{19}[/tex]

Assuming all the unknown variables are positive, we can make [tex]r[/tex] the subject of the formula to obtain it in terms of M & h:

[tex]M=\frac{7r2h}{19}\\M\times19=7r2h\\\\\frac{19M}{2h}=7r\\\\r=\frac{19M}{2h\times7}\\\\r=\frac{19M}{14h}[/tex]

or [tex]r=1.3571M/h[/tex]

Hence, r as in terms of M& H is given as

[tex]r=\frac{19M}{14h} \ or \ 1.3571M/h[/tex]

Answer:

Step-by-step explanation:

A quadrilateral inscribed within a circle is known as a cyclic quadrilateral. Property of the cyclic quadrilateral is that, sum of its opposite angles is 180°.

∴ ∠U + ∠K = 180°

∴ ∠K = 180 - 85 = 95°

Answer:

The answer is a population parameter.

Step-by-step explanation:

Population can include people, but other examples include objects, event, businesses, and so on. Population is the entire pool from which statistical sample is drawn.

A parameter is a value that describes a characteristics of an entire population, such as population mean, because you can almost never measure an entire population, you usually don't know the real value of a parameter.

Consider all possible sample of size N that can be drawn from a given population (either with or without replacement). For example, we can compute a statistics (such as the mean and the standard deviation ) that will vary from sample to sample. In this manner we obtain a distribution of statistics that is called Sampling distribution.

In statistics, a sampling distribution is the theoretical distribution of a sample statistic that arises from drawing all possible samples of a specific size from a population. It helps to quantify the variability and predictability of sample statistics when used as estimates for population parameters.

A sampling distribution refers to the "distribution of a sample statistic". This is option C from your list. This term describes the probability distribution of a statistic based on a random sample. For example, if we study random samples of a certain size from any population, the mean score will form a distribution. This is the sampling distribution of the mean. Similarly, variance, standard deviations and other statistics also have sampling distributions. The purpose of a sampling distribution is to quantify the variation and uncertainty that arises when we use sample statistics (like the mean) to estimate population parameters (like the population mean).

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Answer:

Correct option: B. Less between group variability

Step-by-step explanation:

The Analysis of Variance (ANOVA) test is performed to determine whether there is a significant difference between the different group mean.

The hypothesis is defined as:

H₀: There is no difference between the group means, i.e. μ₁ = μ₂ = ... = μₙ

Hₐ: At least one of the mean is different from the others, i.e. μ[tex]_{i}[/tex] ≠ 0.

The test statistic is defined as:

[tex]F=\frac{SS_{between}}{SS_{within}}[/tex]

If the null hypothesis is true then the test statistic will be small and if it is false then the test statistic will be large.

In this case it is provided that the null hypothesis is true.

This implies that:

[tex]SS_{between}<SS_{within}[/tex]

Implying that the sum of squares for between group variability is less than within group variability.

Thus, if the null hypothesis is true there will be less between group variability.

The probability that a randomly selected Canadian baby is a large baby (weighing more than 4,000 grams) is approximately 0.187 or 18.7%.

To find the probability that a randomly selected Canadian baby is a large baby, we need to calculate the area under the normal distribution curve to the right of 4,000 grams. First, we calculate the z-score using the formula: z = (x - mean) / standard deviation. Plugging in the values, we get z = (4000 - 3500) / 560 = 0.8929.

Next, we need to find the area under the curve to the right of this z-score using a standard normal distribution table or a calculator. The cumulative probability from the table or calculator is approximately 0.187. This means that the probability of a randomly selected Canadian baby being a large baby (weighing more than 4,000 grams) is approximately 0.187 or 18.7%.

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Answer:

a) 0.9772 = 97.72% probability that the mean weight of the sample is less than 3.10 pounds

b) 2.88 pounds

Step-by-step explanation:

To solve this question, we have to understand the normal probability distribution and the central limit theorem.

Normal probability distribution:

Problems of normally distributed samples are 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]s = \frac{\sigma}{\sqrt{n}}[/tex]

In this problem, we have that:

[tex]\mu = 3, \sigma = 0.25, n = 25, s = \frac{0.25}{\sqrt{25}} = 0.05[/tex]

a) What is the probability that the mean weight of the sample is less than 3.10 pounds? Round your answer to four decimal places

This is the pvalue of Z when X = 3.10. So

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

By the Central Limit Theorem

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

[tex]Z = \frac{3.1 - 3}{0.05}[/tex]

[tex]Z = 2[/tex]

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

0.9772 = 97.72% probability that the mean weight of the sample is less than 3.10 pounds

b) What value will the mean weight exceed with probability 0.99? Round your answer to two decimal places.

This is the value of X when Z has a pvalue of 1-0.99 = 0.01. So X when Z = -2.33.

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

[tex]-2.33 = \frac{X - 3}{0.05}[/tex]

[tex]X - 3 = -2.33*0.05[/tex]

[tex]X = 2.88[/tex]

Answer:

[tex]ydv = (v +2)dy\\[/tex]            

Step-by-step explanation:

We are given the following differential equation:

[tex]y dx = 2(x + y) dy[/tex]

We have to substitute

[tex]x = vy[/tex]

Differentiating we get,

[tex]\dfrac{dx}{dy} = v + y\dfrac{dv}{dy}[/tex]

Putting value in differential equation, we get,

[tex]y dx = 2(x + y) dy\\\\y\dfrac{dx}{dy}=2(x+y)\\\\y(v+y\dfrac{dv}{dy}) = 2(vy + y)\\\\vy + y^2\dfrac{dv}{dy} = 2vy +2y\\\\y^2\dfrac{dv}{dy}=vy +2y\\\\y^2dv = y(v+2)dy\\ydv = (v +2)dy\\[/tex]

is the differential equation after substitution.

The given homogeneous differential equation y dx = 2(x + y) dy can be rewritten in terms of y and v using the substitution x = vy. The result is the differential equation y dv/dy = v.

The given differential equation is y dx = 2(x + y) dy. To write his equation in terms of y and v using the substitution x = vy, we must first differentiate both sides of x = vy with respect to x to get 1 = v + y dv/dx. We rearrange this to get dx/dy = 1 / (v + y dv/dy). The original equation can now be rewritten after substituting these values, you will get y / (v + y dv/dy) = 2(v + y), simplifying, we get v = 2v + 2y, and after rearranging, we get y dv/dy = v. This is the differential equation in terms of v and y.

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Answer:

[tex] P(X=1)[/tex]

And using the probability mass function we got:

[tex] P(X=1) = (5C1) (0.25)^1 (1-0.25)^{5-1}= 0.396[/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  

For this cae that one buggy whip would be defective is [tex] p = \frac{5}{20}=0.25[/tex]

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

[tex]X \sim Binom(n=5, p=0.25)[/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(X=1)[/tex]

And using the probability mass function we got:

[tex] P(X=1) = (5C1) (0.25)^1 (1-0.25)^{5-1}= 0.396[/tex]

Answer:

True for n = 18, 19, 20, 21

Step-by-step explanation:

[tex]P(n) =[/tex] a postage of [tex]n[/tex] cents; where [tex]P(n) = 4x + 7y[/tex]. ( [tex]x[/tex] are the number of 4-cent stamps and [tex]y[/tex] are the number of 7-cent stamps)

For [tex]n=18, P(18)[/tex] is true.

This is a possibility, if [tex]x= 1 \ and \ y=2[/tex]

[tex]P(18) = 4(1) + 7(2) = 4 + 14 = 18[/tex]

Similarly for [tex]P(19)[/tex]:

[tex]P(19) = 4(3) + 7(0) = 19[/tex]

[tex]P(20) = 4(5) + 7(0) = 20\\P(21) = 4(0) + 7(3) = 21[/tex]

Answer:

4.76% probability of between 2 and 4 phone calls received (endpoints included) in a given day.

Step-by-step explanation:

In a Poisson distribution, the probability that X represents the number of successes of a random variable is given by the following formula:

[tex]P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}[/tex]

In which

x is the number of sucesses

e = 2.71828 is the Euler number

[tex]\mu[/tex] is the mean in the given interval.

Records show that the average number of phone calls received per day is 9.2.

This means that [tex]\mu = 9.2[/tex].

Find the probability of between 2 and 4 phone calls received (endpoints included) in a given day.

[tex](2 \leq X \leq 4) = P(X = 2) + P(X = 3) + P(X = 4)[/tex]

[tex]P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}[/tex]

[tex]P(X = 2) = \frac{e^{-9.2}*(9.2)^{2}}{(2)!} = 0.0043[/tex]

[tex]P(X = 3) = \frac{e^{-9.2}*(9.2)^{3}}{(3)!} = 0.0131[/tex]

[tex]P(X = 4) = \frac{e^{-9.2}*(9.2)^{4}}{(4)!} = 0.0302[/tex]

[tex](2 \leq X \leq 4) = P(X = 2) + P(X = 3) + P(X = 4) = 0.0043 + 0.0131 + 0.0302 = 0.0476[/tex]

4.76% probability of between 2 and 4 phone calls received (endpoints included) in a given day.

Number of possible ways to arrange letters of word "possession"  = 75600.

Step-by-step explanation:

A classic counting problem is to determine the number of different ways that the letters of "possession" can be arranged.  We have, word "POSSESSION" which have 10 letters! as P,O,S,S,E,S,S,I,O,N . We have to find the number of ways letters of word "possession" can be arranged , in order to do that we will use simple logic as:

At first, we have a count of 10 letters and 10 places vacant to fill letters.

For first place we have a choice of 10 letters , after putting some letter from all 10 letters in first place now we are left with 9 places and 9 letters having 9 choices , similarly we'll be having 8 places , 8 letters and 8 choices and so on...... Therefore, Number of possible ways to arrange letters of word "possession"  = [tex]10.9.8.7.6.5.4.3.2.1[/tex] = [tex]10![/tex] ( 10 factorial ) , but there are 4 letters "s" are repeated and 2 letters "o" are repeated so we need to eliminate the similar combinations ∴ Number of possible ways = [tex]\frac{10!}{4!(2!)}[/tex] = 75600.

∴Number of possible ways to arrange letters of word "possession" = 75600

Answer:

(A) Assume m is an odd integer.

Therefore m is of the m=2n-1 type where n is any number.

We have m + 6 = 2n-1 + 6 = 2n+5=2n+4 + 1 = 2(n+2) + 1, in which n + 2 is an integer, adding 6 to both ends.

Because m+6 is of the 2x + 1 type, where x =  n + 2; then m + 6 is an odd integer too.

(B) Provided that mn is an integer also. For some integer y and m, n are integers, this means mn is of the form mn = 2y.

And y = mn/2.

Therefore either m is divided by 2 or n is divided by 2 since y, m, n are all integers. To put it another way, either m is a multiple of 2 or n is a multiple of 2, which means that m is even or n is even.

Final answer:

The proposition regarding odd integers is proven by correcting the initial flawed proof, demonstrating that adding 6 to an odd integer results in another odd integer by using the correct form of an odd integer in the calculation.

Explanation:

The given proposition states that adding 6 to an odd integer results in another odd integer. The proof starts with the assumption that there exists an integer n such that m + 6 = 2n + 1.

This equation is supposed to define m + 6 as an odd integer. By isolating m, the proof continues to show that m itself is expressed in the form of 2(n - 3) + 1, indicating that m is also an odd integer.

To correct the proof, we should start with an odd integer m such that m = 2k + 1, where k is an integer. Adding 6 to m gives m + 6 = 2k + 7, which can be written as 2(k + 3) + 1, clearly showing that m + 6 is an odd integer.

Therefore, this proves the original proposition correctly.

Answer:

There are 21 comparisons to be made.

Step-by-step explanation:

The number of species of Odonata monitored every year is, n = 7.

It is provided that the density of each species is compared with each other.

The number of ways to compare the species (N) without repetition is:

[tex]N=\frac{n(n-1)}{2}\\=\frac{7(7-1)}{2}\\=\frac{7\times6}{2}\\=21[/tex]

Thus, there are 21 comparisons.

The 21 comparisons are as follows:

Specie 1 is compared with the remaining 6.

Specie 2 has already with he 1st so it is compared with the remaining 5.

Specie 3 has already with he 1st and 2nd so it is compared with the remaining 4.

Specie 4 has already with he 1st, 2nd and 3rd so it is compared with the remaining 3.

Specie 5 has already with he 1st, 2nd, 3rd and 4th so it is compared with the remaining 2.

Specie 6 has already with he 1st, 2nd, 3rd, 4th and 5th so it is compared with the remaining 1.

And the specie 7 has already been compared with the others.

Total number of comparisons = 6 + 5 + 4 + 3 + 2 + 1 = 21.