You are at a campus party where there are a total number of n people. The host asked everyone to put their phones in a bowl while walking in. A noise complaint ends the party abruptly, and everyone heads for the door, hastily grabbing their phones from the bowl Assume every guest has one and exactly one phone, and that they pick a phone at random (so that every assignment of a phone to a person is equally likely). What is the probability that: a. Every person gets their phone back? b. The first m persons to pick each get their own phones back? c. The first m persons to pick each get a phone belonging to the last m persons to pick? Hint: Try this thought experiment with a few choices of mand n to get a feel for the numbers that show up.)

Answer:

-121

Step-by-step explanation:

[tex] \frac{t}{ - 11} = 11 \\ t = - 11 \times 11 \\ \huge \red{ \boxed{t = - 121}}[/tex]

Answer:

$420/5

$85 a day is Jesse's daily wage

Step-by-step explanation:

Answer:

$84

Step-by-step explanation:

$420 a week divided by 5 days per week gives $84 per day

Answer:

Step-by-step explanation:

Hello!

You have two variables of interest

X₁: failure stress of a NiCrAlZr coating after nine 1-hr cycles.

X₂: failure stress of a NiCrAlZr coating after six 1-hr cycles.

a)

To be able to estimate the difference between the means using a confidence interval, you need that both variables have a normal distribution and to determine whether or not the population variances are equal.

If the population variances are equal, σ₁²=σ₂², you can use a pooled variance t-test

If the population variances are different, σ₁²≠σ₂², you have to use Welch's t-test

Using α: 0.05

The normality test for X₁ shows a p-value of 0.7449 ⇒ You can assume it has a normal distribution.

The normality test for X₂ shows a p-value of 0.9980 ⇒ You can assume it has a normal distribution.

The F-test for variance homogeneity shows a p-value of 0.6968 (H₀:σ₁²=σ₂²) ⇒You can assume both population variances are equal.

b) and c)

You need to test if both population means are the same, the hypotheses are:

H₀: μ₁=μ₂

H₁: μ₁≠μ₂

α: 0.05

[tex]t= \frac{(X[bar]_1-X[bar]_2)-(Mu_1-Mu_2)}{Sa*\sqrt{\frac{1}{n_1} +\frac{1}{n_2} } } ~~t_{n_1+n_2-2}[/tex]

[tex]Sa= \sqrt{\frac{(n_1-1)S^2_1+(n_2-1)S^2_2}{n_1+n_2-2} } = \sqrt{\frac{8*4.28+5*5.62}{9+6-2} }= \sqrt{4.7953}= 2.189= 2.19[/tex]

[tex]t_{H_0}= \frac{(X[bar]_1-X[bar]_2)-(Mu_1-Mu_2)}{Sa*\sqrt{\frac{1}{n_1} +\frac{1}{n_2} } } = \frac{(16.36-11.48)-0}{2.19*\sqrt{\frac{1}{9} +\frac{1}{6} } } = 4.23[/tex]

The distribution of this test is a t with 13 degrees of freedom and the test is two-tailed, so to calculate the p-value you have to do the following:

P(t₁₃≤-4.23)+P(t₁₃≥4.23)= P(t₁₃≤-4.23)+[1-P(t₁₃<4.23)]=  0.000492 + (1-0.999508)= 2*0.000492= 0.000984≅ 0.001

The p-value: 0.001 is less than α: 0.05, the decision is to reject the null hypothesis.

I hope it helps!

Answer:

  x = -2 or +7

Step-by-step explanation:

We presume you want to solve ...

  18^(x^2 +4x +4) = 18^(9x +18)

Equate exponents of the same base and put the quadratic in standard form.

  x^2 +4x +4 = 9x +18

  x^2 -5x -14 = 0

  (x -7)(x +2) = 0

Values of x that make these factors zero are 7 and -2.

The solutions are x=-2 or x = 7.

Answer:

c

Step-by-step explanation:

got it right on edge

Answer:

The answer is ten bouquets

Step-by-step explanation:

1 dozen is 12 if you were multiply 12 by 10 you would get 120. Now multiply 10 by twelve instead you get the same answer

Final answer:

When the price of each bouquet increased from $10 to $12, the flower stand sold 10 bouquets to make the same total sales of $120 as the previous day.

Explanation:

The question at hand involves a basic mathematical calculation of how changes in price affect the number of units sold to achieve the same total revenue. If a flower stand sold a dozen bouquets for $10 each yesterday, they made $120 in total sales (12 bouquets  imes $10 per bouquet). Today, with the price increase to $12 per bouquet, we need to find out how many bouquets were sold to still make $120 in total sales. The calculation is as follows:

Total Sales = Number of Bouquets Sold  imes Price per Bouquet

We know that the Total Sales should still be $120, and the Price per Bouquet is now $12. Therefore:

$120 = Number of Bouquets Sold  imes $12

Number of Bouquets Sold = $120 / $12

Number of Bouquets Sold = 10 bouquets

Thus, the flower stand sold 10 bouquets today to make the same amount of money as yesterday after increasing the price to $12 each.

Answer:

The difference is 15.

Step-by-step explanation:

30-15=15

45-30=15

60-45=15

Answer:

I agree the difference is 15.

Step-by-step explanation:

Answer:

On this case we want to test if the mean age of the​ agency's customers is over 20, so the system of hypothesis would be:

Null hypothesis: [tex]\mu \leq 20[/tex]

Alternative hypothesis: [tex]\mu >20[/tex]

If he concludes the mean age is over 20 when it is​ not, he makes a​ type I error . If he concludes the mean age is not over 20 when it​ is, he makes a​ Type II error error.

Step-by-step explanation:

Previous concepts

A hypothesis is defined as "a speculation or theory based on insufficient evidence that lends itself to further testing and experimentation. With further testing, a hypothesis can usually be proven true or false".  

The null hypothesis is defined as "a hypothesis that says there is no statistical significance between the two variables in the hypothesis. It is the hypothesis that the researcher is trying to disprove".  

The alternative hypothesis is "just the inverse, or opposite, of the null hypothesis. It is the hypothesis that researcher is trying to prove".  

Type I error, also known as a “false positive” is the error of rejecting a null  hypothesis when it is actually true. Can be interpreted as the error of no reject an  alternative hypothesis when the results can be  attributed not to the reality.  

Type II error, also known as a "false negative" is the error of not rejecting a null  hypothesis when the alternative hypothesis is the true. Can be interpreted as the error of failing to accept an alternative hypothesis when we don't have enough statistical power.  

Solution to the problem

On this case we want to test if the mean age of the​ agency's customers is over 20, so the system of hypothesis would be:

Null hypothesis: [tex]\mu \leq 20[/tex]

Alternative hypothesis: [tex]\mu >20[/tex]

If he concludes the mean age is over 20 when it is​ not, he makes a​ type I error . If he concludes the mean age is not over 20 when it​ is, he makes a​ Type II error error.

When the travel agency owner concludes the mean age is over 20 when it's not, it's called a Type I error. If he concludes it's not over 20 when it is, it is a Type II error.

The scenario described involves a hypothesis test concerning the mean age of customers at a travel agency. When the travel agency owner incorrectly concludes that the mean age is over 20 when in fact it is not, he is making a Type I error. Conversely, if he concludes that the mean age is not over 20 when it actually is, that would be a Type II error. Understanding the consequences of these errors is critical in statistics because they affect decision-making processes based on data analysis.

Answer:

5 pints of blue and 2 pints of yellow make 7 pints of green. (5 + 2 = 7)

You must do this four times to get 28 pints of green. (4  · 7 = 28)

So you need 20 (4 · 5) pints of blue and 8 (4 · 2) pints of yellow.

Step-by-step explanation:

Answer:

140 pints

Step-by-step explanation:

Answer:

30

Step-by-step explanation:

[tex]7x^{2} +33x-10=0[/tex]

[tex]\implies 7x^{2} +35x-2x-10=0\\\implies7x(x+5)-2(x+5)=0\\\implies (7x-2)(x+5)=0[/tex]

[tex]7x - 2 = 0\\\implies \boxed{\bold{x = \frac{2}{7} }}[/tex]

[tex]x+5=0\\\implies \boxed{\bold{x=-5}}[/tex]

Answer:

The probability that at least 280 of these students are smokers is 0.9664.

Step-by-step explanation:

Let the random variable X be defined as the number of students at a particular college who are smokers

The random variable X follows a Binomial distribution with parameters n = 500 and p = 0.60.

But the sample selected is too large and the probability of success is close to 0.50.

So a Normal approximation to binomial can be applied to approximate the distribution of X if the following conditions are satisfied:

1. np ≥ 10

2. n(1 - p) ≥ 10

Check the conditions as follows:

 [tex]np=500\times 0.60=300>10\\n(1-p)=500\times(1-0.60)=200>10[/tex]

Thus, a Normal approximation to binomial can be applied.

So,  

[tex]X\sim N(\mu=600, \sigma=\sqrt{120})[/tex]

Compute the probability that at least 280 of these students are smokers as follows:

Apply continuity correction:

P (X ≥ 280) = P (X > 280 + 0.50)

                   = P (X > 280.50)

                   [tex]=P(\frac{X-\mu}{\sigma}>\frac{280-300}{\sqrt{120}}\\=P(Z>-1.83)\\=P(Z<1.83)\\=0.96638\\\approx 0.9664[/tex]

*Use a z-table for the probability.

Thus, the probability that at least 280 of these students are smokers is 0.9664.

The probability of a minimum of 280 students being radio listeners would be:

0.9664

The standard deviation is a statistic that measures the dispersion of a dataset relative to its mean and is calculated as the square root of the variance in statistics.

So, the formula is,

[tex]SD=\sqrt{\sigma}[/tex]

Suppose it is known that 60% of students at a college are smokers.

A sample of 500 students from the college is selected at random.

[tex]mean=np\\=500\times 0.6\\=300[/tex]

The standard deviation is,

[tex]S=\sqrt{np(1-p)}\\ =\sqrt{500\times 0.6\times 0.4}\\ =10.95445[/tex]

So, the probability is,

[tex]P(X > 280)=\frac{P(x-mean)}{s} \\=\frac{280-300}{10.95445} \\=0.9664[/tex]

Learn more about the topic standard deviation:

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Answer:

3x+15

Step-by-step explanation:

When it says "the sum of a number three times", it means 3x, because the x stands for multiplication and the problem says the sum sum is between 3x and the number 15. So your equation is 3x+15. Hope this helps ;)

Given:

Given that ABCD is a parallelogram.

The length of AB is 3y - 2.

The length of BC is x + 12.

The length of CD is y + 6.

The length of AD is 2x - 4.

We need to determine the length of AB and BC.

Value of y:

We know the property that opposite sides of a parallelogram are congruent, then, we have;

[tex]AB=CD[/tex]

Substituting the values, we get;

[tex]3y-2=y+6[/tex]

[tex]2y-2=6[/tex]

     [tex]2y=8[/tex]

       [tex]y=4[/tex]

Thus, the value of y is 4.

Value of x:

We know the property that opposite sides of a parallelogram are congruent, then, we have;

[tex]AD=BC[/tex]

Substituting the values, we get;

[tex]2x-4=x+12[/tex]

 [tex]x-4=12[/tex]

       [tex]x=16[/tex]

Thus, the value of x is 16.

Length of AB:

The length of AB can be determined by substituting the value of y in the expression 3y - 2.

Thus, we have;

[tex]AB=3(4)-2[/tex]

     [tex]=12-2[/tex]

[tex]AB=10[/tex]

Thus, the length of AB is 10 units.

Length of BC:

The length of BC can be determined by substituting the value of x in the expression x + 12.

Thus, we have;

[tex]BC=16+12[/tex]

[tex]BC=28[/tex]

Thus, the length of BC is 28 units.

Hence, the length of AB and BC are 10 units and 28 units respectively.

Answer:

Step-by-step explanation:

Answer:

19.5 inches or 19.685 inches for length

Step-by-step explanation:

50•0.39. 50÷3

19.5. 19.685

how to convert centimeters to inches:

multiply the centimeter by 0.39. example 10•0.39= 3.9 inches.

If you want centimeters to inches as length, then divide the centimeters by 3. example: 10÷3= 3.3333333333 inches

Answer:

Part 1

(a) 0.28434

(b) 0.43441

(c) 29.9 mm

Part 2

(a) 0.97722

Step-by-step explanation:

There are two questions here. We'll break them into two.

Part 1.

This is a normal distribution problem healthy children having the size of their left atrial diameters normally distributed with

Mean = μ = 26.4 mm

Standard deviation = σ = 4.2 mm

a) proportion of healthy children have left atrial diameters less than 24 mm

P(x < 24)

We first normalize/standardize 24 mm

The standardized score for any value is the value minus the mean then divided by the standard deviation.

z = (x - μ)/σ = (24 - 26.4)/4.2 = -0.57

The required probability

P(x < 24) = P(z < -0.57)

We'll use data from the normal probability table for these probabilities

P(x < 24) = P(z < -0.57) = 0.28434

b) proportion of healthy children have left atrial diameters between 25 and 30 mm

P(25 < x < 30)

We first normalize/standardize 25 mm and 30 mm

For 25 mm

z = (x - μ)/σ = (25 - 26.4)/4.2 = -0.33

For 30 mm

z = (x - μ)/σ = (30 - 26.4)/4.2 = 0.86

The required probability

P(25 < x < 30) = P(-0.33 < z < 0.86)

We'll use data from the normal probability table for these probabilities

P(25 < x < 30) = P(-0.33 < z < 0.86)

= P(z < 0.86) - P(z < -0.33)

= 0.80511 - 0.37070 = 0.43441

c) For healthy children, what is the value for which only about 20% have a larger left atrial diameter.

Let the value be x' and its z-score be z'

P(x > x') = P(z > z') = 20% = 0.20

P(z > z') = 1 - P(z ≤ z') = 0.20

P(z ≤ z') = 0.80

Using normal distribution tables

z' = 0.842

z' = (x' - μ)/σ

0.842 = (x' - 26.4)/4.2

x' = 29.9364 = 29.9 mm

Part 2

Population mean = μ = 65 mm

Population Standard deviation = σ = 5 mm

The central limit theory explains that the sampling distribution extracted from this distribution will approximate a normal distribution with

Sample mean = Population mean

¯x = μₓ = μ = 65 mm

Standard deviation of the distribution of sample means = σₓ = (σ/√n)

where n = Sample size = 100

σₓ = (5/√100) = 0.5 mm

So, probability that the sample mean distance ¯x for these 100 will be between 64 and 67 mm = P(64 < x < 67)

We first normalize/standardize 64 mm and 67 mm

For 64 mm

z = (x - μ)/σ = (64 - 65)/0.5 = -2.00

For 67 mm

z = (x - μ)/σ = (67 - 65)/0.5 = 4.00

The required probability

P(64 < x < 67) = P(-2.00 < z < 4.00)

We'll use data from the normal probability table for these probabilities

P(64 < x < 67) = P(-2.00 < z < 4.00)

= P(z < 4.00) - P(z < -2.00)

= 0.99997 - 0.02275 = 0.97722

Hope this Helps!!!

Answer:

Is the #6 in the row. Is this a riddle

Answer:

its 1/720 i just took it

Step-by-step explanation:

Answer:

We conclude that the Redwood trees have an average height greater than 240 feet.

Step-by-step explanation:

We are given that a random sample of 47 California Redwood trees was taken and their heights measured. The sample mean average height was 248 feet with a standard deviation of 26 feet.

We have to test the claim that Redwood trees have an average height greater than 240 feet.

Let [tex]\mu[/tex] = mean weight bag filling capacity of machine.

SO, Null Hypothesis, [tex]H_0[/tex] : [tex]\mu[/tex] [tex]\leq[/tex] 240 feet   {means that the Redwood trees have an average height smaller than or equal to 240 feet}

Alternate Hypothesis, [tex]H_A[/tex] : [tex]\mu[/tex] > 240 feet   {means that the Redwood trees have an average height greater than 240 feet}

The test statistics that will be used here is One-sample t test statistics as we don't know about the population standard deviation;

                        T.S.  = [tex]\frac{\bar X -\mu}{\frac{s}{\sqrt{n} } }[/tex]  ~ [tex]t_n_-_1[/tex]

where, [tex]\bar X[/tex] = sample mean average height = 248 feet

             s = sample standard deviation = 26 feet

             n = sample of trees = 47

So, test statistics  =  [tex]\frac{248-240}{\frac{26}{\sqrt{47} } }[/tex]  ~ [tex]t_4_6[/tex]   

                               =  2.109

Now at 5% significance level, the t table gives critical value of 1.6792 at 46 degree of freedom for right-tailed test. Since our test statistics is higher than the critical value of t as 2.109 > 1.6792, so we have sufficient evidence to reject our null hypothesis as it will fall in the rejection region due to which we reject our null hypothesis.

Therefore, we conclude that the Redwood trees have an average height greater than 240 feet.

The probability of winning no more than 8 times out of 10 plays of the arcade game is approximately 0.6394.

Here, we have,

To calculate the probability of winning no more than 8 times out of 10 plays of the arcade game, we need to find the cumulative probability of winning 0, 1, 2, 3, 4, 5, 6, 7, and 8 times.

We'll use the binomial distribution formula to do this:

The binomial distribution formula for a probability of k successes in n trials is given by:

[tex]P(X = k) = C(n, k) * p^k * (1 - p)^{(n - k)[/tex]

where:

P(X = k) is the probability of exactly k successes,

n is the total number of trials (10 in this case),

k is the number of successes we want to calculate the probability for (0 to 8 in this case),

p is the probability of winning (0.632 in this case),

C(n, k) is the binomial coefficient, calculated as

C(n, k) = n! / (k! * (n - k)!).

Now, let's calculate the probabilities for each value of k and then sum them up to find the cumulative probability:

P(X ≤ 8) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5) + P(X = 6) + P(X = 7) + P(X = 8)

now, we have,

[tex]P(X = 0) = C(10, 0) * (0.632)^0 * (1 - 0.632)^{(10 - 0)}\\P(X = 1) = C(10, 1) * (0.632)^1 * (1 - 0.632)^{(10 - 1)}\\P(X = 2) = C(10, 2) * (0.632)^2 * (1 - 0.632)^{(10 - 2)}\\P(X = 3) = C(10, 3) * (0.632)^3 * (1 - 0.632)^{(10 - 3)}\\P(X = 4) = C(10, 4) * (0.632)^4 * (1 - 0.632)^{(10 - 4)}\\[/tex]

[tex]P(X = 5) = C(10, 5) * (0.632)^5 * (1 - 0.632)^{(10 - 5)}\\P(X = 6) = C(10, 6) * (0.632)^6 * (1 - 0.632)^{(10 - 6)}\\P(X = 7) = C(10, 7) * (0.632)^7 * (1 - 0.632)^{(10 - 7)}\\P(X = 8) = C(10, 8) * (0.632)^8 * (1 - 0.632)^{(10 - 8)}[/tex]

Now, let's calculate each of these probabilities:

[tex]P(X = 0) = 1 * 1 * (0.368)^{10} = 0.0010434068\\P(X = 1) = 10 * 0.632 * (0.368)^9 = 0.0091045617\\P(X = 2) = 45 * 0.632^2 * (0.368)^8 = 0.0333533058\\P(X = 3) = 120 * 0.632^3 * (0.368)^7 = 0.0789130084\\P(X = 4) = 210 * 0.632^4 * (0.368)^6 = 0.1312387057\\P(X = 5) = 252 * 0.632^5 * (0.368)^5 = 0.1552134487\\P(X = 6) = 210 * 0.632^6 * (0.368)^4 = 0.1320219902\\P(X = 7) = 120 * 0.632^7 * (0.368)^3 = 0.0702045343\\P(X = 8) = 45 * 0.632^8 * (0.368)^2 = 0.0272837154\\[/tex]

Now, let's sum up these probabilities:

P(X ≤ 8) ≈ 0.0010434068 + 0.0091045617 + 0.0333533058 + 0.0789130084 + 0.1312387057 + 0.1552134487 + 0.1320219902 + 0.0702045343 + 0.0272837154

≈ 0.6393716762

So, the probability of winning no more than 8 times out of 10 plays of the arcade game is approximately 0.6394.

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Finally, we find:

P(X ≤ 8) ≈ P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5) + P(X = 6) + P(X = 7) + P(X = 8)

To calculate the probability of winning no more than 8 times out of 10 plays in the arcade game, we can use the binomial distribution.

The binomial distribution calculates the probability of having a certain number of successes (in this case, winning) in a fixed number of independent trials (10 plays) when the probability of success (p) remains constant.

Probability of winning (p) = 0.632

Probability of losing (q, since p + q = 1) = 0.368

Number of trials (n) = 10

To find the probability of winning no more than 8 times, we need to sum the probabilities of winning 0, 1, 2, 3, 4, 5, 6, 7, and 8 times.

P(X ≤ 8) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5) + P(X = 6) + P(X = 7) + P(X = 8)

Using the binomial probability formula:

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

where (n choose k) is the binomial coefficient, given by:

(n choose k) = n! / (k! * (n-k)!)

Calculating the probabilities for each individual case:

P(X = 0) = (10 choose 0) * (0.632^0) * (0.368^(10-0))

P(X = 1) = (10 choose 1) * (0.632^1) * (0.368^(10-1))

P(X = 2) = (10 choose 2) * (0.632^2) * (0.368^(10-2))

...

P(X = 8) = (10 choose 8) * (0.632^8) * (0.368^(10-8))

After calculating the probabilities for each individual case, we can sum them up to find P(X ≤ 8).

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Answer:

Upper bond is 0.398

Step-by-step explanation:

See attached file

Answer:

k=9

Step-by-step explanation:

The formula for direct variation is

y = kx

We know y and x

27 = k*3

Divide each side by 3

27/3 = 3k/3

9 =k

Answer:

not satisfactional do not vary

Step-by-step explanation:

The subtraction is one of the four basic arithmetic operations in mathematics. The operation subtraction is used here to determine the number of students enrolled in the Saturday morning class. The number of students is 26.

The operation subtraction represents the process of removing objects from a collection. The minus sign implies the subtraction. We can also describe subtraction as the decreasing or removing physical and abstract quantities.

The subtraction method consists of three parts of numbers, they are minuend, subtrahend and difference. The number in a subtraction sentence from which we subtract another number is the minuend, and after that it is subtrahend and last number is the difference.

Here the total present students = 30

Number of absents = 4

So number of presents = 30 - 4 = 26

Thus the students who are enrolled is 26.

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Your question is incomplete most probably your full question was:

In a class the number of students is actually 30, 4 of them were absent on saturday. A dance teacher asks, "How many students are enrolled in the Saturday?

Answer:

Sorry about that answer, some people are plain rude, but the correct answer is -5.

Step-by-step explanation:

Hope this Helps! Good Luck!

Answer: k=-5 is correct because if you were to divide -10/-2 it would give you -5

I hope this simplifies your question for the answer.

Answer:

108

Step-by-step explanation:

Step-by-step explanation:

Circumference of the circle= 2πr

=2×3.14×8 cm

= 50.24cm

= 50.2 cm

Answer:

50.27

rounded : 50.3

Step-by-step explanation:

2 times 3.14 times 8

hope this helped

brainliest is appreciated :)

Answer:

x= -12  y= 4

Step-by-step explanation:

Answer:

Y=4

x=-12

Step-by-step explanation:

Answer:

Step-by-step explanation:

Confidence interval is written in the form,

(Sample mean - margin of error, sample mean + margin of error)

The sample mean, x is the point estimate for the population mean.

Since the sample size is large and the population standard deviation is known, we would use the following formula and determine the z score from the normal distribution table.

Margin of error = z × σ/√n

Where

σ = population standard Deviation

n = number of samples

From the information given

1) x = 32

σ = 6

n = 50

To determine the z score, we subtract the confidence level from 100% to get α

α = 1 - 0.90 = 0.1

α/2 = 0.1/2 = 0.05

This is the area in each tail. Since we want the area in the middle, it becomes

1 - 0.05 = 0.95

The z score corresponding to the area on the z table is 1.645. Thus, confidence level of 90% is 1.645

Margin of error = 1.645 × 6/√50 = 1.4

Confidence interval = 32 ± 1.4

The lower end of the confidence interval is

32 - 1.4 = 30.6

The upper end of the confidence interval is

32 + 1.4 = 33.4

Answer:

20% robability that the tourist will be able to talk with a randomly encountered resident of the region, given that the resident speaks German

Step-by-step explanation:

We use the conditional probability formula to solve this question. It is

[tex]P(B|A) = \frac{P(A \cap B)}{P(A)}[/tex]

In which

P(B|A) is the probability of event B happening, given that A happened.

[tex]P(A \cap B)[/tex] is the probability of both A and B happening.

P(A) is the probability of A happening.

In this problem, we have that:

Event A: Speaking German

Event B: Speaking English

3% speak both English and German

This means that [tex]P(A \cap B) = 0.03[/tex]

15% speak German

This means that [tex]P(A) = 0.15[/tex]

So

[tex]P(B|A) = \frac{0.03}{0.15} = 0.2[/tex]

20% robability that the tourist will be able to talk with a randomly encountered resident of the region, given that the resident speaks German

Answer:

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

[tex]\bar X \pm z_{\alpha/2}\frac{\sigma}{\sqrt{n}}[/tex]   (1)

Or equivalently:

[tex]\bar X \pm ME[/tex]

For this case we have the interval given (3.9, 7.7) and we want to find the margin of error. Using the property of symmetry for a confidence interval we can estimate the margin of error with this formula:

[tex]ME= \frac{Upper -Lower}{2}= \frac{7.7-3.9}{2}= 1.9[/tex]

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".

[tex]\bar X[/tex] represent the sample mean for the sample  

[tex]\mu[/tex] population mean (variable of interest)

Solution to the problem

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

[tex]\bar X \pm z_{\alpha/2}\frac{\sigma}{\sqrt{n}}[/tex]   (1)

Or equivalently:

[tex]\bar X \pm ME[/tex]

For this case we have the interval given (3.9, 7.7) and we want to find the margin of error. Using the property of symmetry for a confidence interval we can estimate the margin of error with this formula:

[tex]ME= \frac{Upper -Lower}{2}= \frac{7.7-3.9}{2}= 1.9[/tex]