What type of number will tell Matt whether or not there is a significant difference between how the boys and girls scored?

There will be no value of the black rectangle for which the given function a valid joint pdf.

To be a valid joint probability density function, fxy(x,y) must satisfy the following conditions:

fxy(x,y) must be non-negative for all values of x and y.

The integral of fxy(x,y) over the entire plane must be equal to 1.

Let's first check the non-negativity condition. The function fxy(x,y) = y(x-y)[tex]e^{-(x+y)}[/tex]  is non-negative for 0 < x < ∞ and 0 < y < x, since y and (x-y) are both non-negative and [tex]e^{-(x+y)}[/tex]   is positive for all x and y. Therefore, the non-negativity condition is satisfied.

Next, we need to check if the integral of fxy(x,y) over the entire plane is equal to 1. We can write this integral as:

∫∫fxy(x,y)dydx

The limits of integration for y are 0 to x, and the limits of integration for x are 0 to ∞. Therefore, we have:

∫∫fxy(x,y)dydx = ∫[tex]0^{I}[/tex]∫0^xy(x-y)[tex]e^{-(x+y)}[/tex]  dydx

We can evaluate the integral using the following steps:

Integrate with respect to y:

[tex]\int\limits^x_0 {xy(x-y)e^{-(x+y)} } \, dy[/tex] = [tex]xe^{-x}\int\limits^x_0 {ye^{y} } \, dy - xe^{x}\int\limits^x_0 {y^{2} e^{y} } \, dy[/tex]

Integrate with respect to u = y + 1:

[tex]\int\limits^x_0 {ye^{y} } \, dy = \int\limits^x_1 {(u-1)e^{(u-1)} } \, du = (x+1)e^{-1}-e^{-x}[/tex]

[tex]\int\limits^x_0 {y^{2}e^{y} } \, dy = \int\limits^x_0 {(u-1)^{2}e^{(u-1)} } \, du[/tex]

                          = [tex](x^{2} +4x+2)e^{-1} -2xe^{-x}-2e^{-x}[/tex]

Substitute the results from step 2 into the integral for ∫∫fxy(x,y)dydx:

∫∫fxy(x,y)dydx = [tex]\int\limits^I_0 {xe^{-x}[(x+1)e^{-1}-e^{-x}-(x^{2} +4x+2)e^{-1}+2xe^{-x}+2e^{-x}]} \, dx[/tex]

= (1/2) - (5/2e)

where we have used integration by parts to evaluate the integrals involving [tex]e^{-x}[/tex].

Therefore, the integral of fxy(x,y) over the entire plane is equal to (1/2) - (5/2e), which is not equal to 1. This means that fxy(x,y) cannot be a valid joint probability density function for any value of the black rectangle.

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To determine which box will produce the largest amount of sales using Excel, follow these steps:

a) Organize the data: Create a table in Excel with columns for Box Type (A, B, and C), and Sales for each test market area (Area 1, Area 2, etc.).

b) Input data: Enter the sales data for each box type in the respective test market areas.

c) Calculate the total sales: In a new column, calculate the sum of sales for each box type using the SUM function (e.g., =SUM(B2:F2)).

d) Compare the results: Identify the box with the highest total sales, which will be the box predicted to produce the largest amount of sales.

By organizing the sales data in Excel, calculating the total sales for each box type, and comparing the results, you can easily determine which box is expected to generate the most sales. This step-by-step method provides an efficient and data-driven approach to making marketing decisions.

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The values of x1 and x2 that maximize the revenue are x1 = 20 thousand units of racing bicycles and x2 = 7.5 thousand units of mountain bicycles.

To maximize the revenue, we need to find the values of x1 and x2 that maximize the function R(x1, x2) = -6x1² - 10x2² - 2x1x2 + 46x1 + 106x2.

To do this, we can take partial derivatives of R with respect to x1 and x2, set them equal to zero, and solve for x1 and x2. That is:

∂R/∂x1 = -12x1 - 2x2 + 46 = 0

∂R/∂x2 = -20x2 - 2x1 + 106 = 0

Solving these two equations simultaneously, we get:

x1 = (5/3) x2 + (23/3)

x2 = (1/5) x1 + (53/10)

Substituting the second equation into the first equation, we get:

x1 = (5/3) [(1/5) x1 + (53/10)] + (23/3)

x1 = (1/3) x1 + (89/6)

Solving for x1, we get:

x1 = 20

Substituting x1 = 20 into the second equation, we get:

x2 = (1/5) (20) + (53/10) = 7.5

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Range and Mode of the given data set: 10, 8, 5, 3, 7, 4, 5, 9, 2, 3, 7, 3, 8, 6, 4, 1, 2, 1, 10, 3 are 9 and 4 respectively. Thus, option B is the correct answer.

Range refers to the difference between the highest and lowest values in a given set of numbers. Therefore, to calculate the range of the given data we need to subtract the lowest value from the highest value.

The lowest value in the data = 1

The highest value in the data = 10

Range = highest value - lowest value

= 10 - 1 = 9

Therefore, the range of the given data is 9.

Mode refers to the data that is repeated most frequently in the given data. Therefore, to find the mode, we have to check the data with the highest frequency.

To find the mode easily, we arrange the data in ascending order and we get 1, 1, 2, 2, 3, 3, 3, 3, 4, 4, 5, 6, 7, 7, 8, 8, 9, 10

The number with the highest frequency (mode) = 3

3 is repeated 4 times in the data.

Therefore, the mode of the given data comes out to be 3.

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To calculate the power of the proposed study, we need to first determine the effect size, which is the standardized difference between the mean BMI of normal and diabetic men.

The standardized difference can be calculated as:

d = (μ1 - μ2) / σ

where μ1 and μ2 are the population means of BMI for normal and diabetic men, respectively, and σ is the common population standard deviation of BMI.

From the information given in the problem, we have:

μ1 = 25

μ2 = 25 + 2.7 = 27.7

σ = 2.7

So, the effect size is:

d = (25 - 27.7) / 2.7 = -1

Next, we need to determine the significance level (α) and the sample size (n). The problem states that the sample size is 25 normal men and 25 diabetic men, so n = 50. The significance level is usually set at 0.05, which means that the probability of a Type I error (rejecting the null hypothesis when it is actually true) is 0.05.

Using a standard normal distribution table, we can find the z-score corresponding to the significance level α = 0.05:

zα = 1.645

The power of the test is the probability of correctly rejecting the null hypothesis (i.e., detecting a true difference between normal and diabetic men) when the alternative hypothesis is true. The power of a test depends on several factors, including the effect size, the significance level, the sample size, and the variability of the data.

The formula for calculating power is:

Power = P(Z > zα - d√n)

where Z is the standard normal distribution, and d and n are the effect size and sample size, respectively.

Substituting the values we have, we get:

Power = P(Z > 1.645 - (-1)√50) = P(Z > 0.843)

Using a standard normal distribution table, we can find that the probability of Z being greater than 0.843 is 0.199.

Therefore, the power of the proposed study is approximately 0.199, or 19.9%. This means that there is a 19.9% chance of correctly detecting a clinically meaningful difference in BMI between normal and diabetic men, assuming that such a difference actually exists.

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approximately 1.2% of those infected with Covid-19 in the U.S. died from the virus as of 2/25/22 since the pandemic began.

a) To find the percentage of the U.S. population that has been infected by Covid-19 as of 2/25/22 since the pandemic began, we can divide the number of people infected by the total population and then multiply by 100:

(77,958,554 / 333,300,000) x 100% = 23.4%

So, approximately 23.4% of the U.S. population has been infected by Covid-19 as of 2/25/22 since the pandemic began.

b) To find the percentage of those infected who died from the coronavirus in the U.S. as of 2/25/22 since the pandemic began, we can divide the number of deaths by the number of people infected and then multiply by 100:

(942,902 / 77,958,554) x 100% = 1.2%

So, approximately 1.2% of those infected with Covid-19 in the U.S. died from the virus as of 2/25/22 since the pandemic began.

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The total mass is 125π/6.

The Center of Mass is (625/4, 0)

We have a disk

x² + y² ≤25

let (x', y') be the center of Mass

So, x' = [tex](\int\limits_D {x. \rho(x,y)} \, dA[/tex]  / M

and, y' =  [tex](\int\limits_D {y. \rho(x,y)} \, dA[/tex]  / M

where M is the mass of the region

Also, p(x,y) = 5(x2 + y²)

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

Then, (r cos [tex]\theta[/tex] )² + (r sin [tex]\theta[/tex])² ≤25

r² < 25

r≤5

So, the total Mass is

= ∫ 5(x² + y²) dA

= [tex]\int\limits^{\pi/2}_0 \int\limits^5_ 0{ 5 (r^2 cos^2 \theta + r^2 sin^2 \theta)} \, dr d\theta[/tex]

= [tex]\int\limits^{\pi/2}_0 \int\limits^5_ 0{ 5 r^2 } \, dr d\theta[/tex]

= 5 x 125/3 [tex]\int\limits^ {\pi/2} _0{ d\theta[/tex]

= 5³/3. π/2

= 125π/6

Now, x' = [tex](\int\limits_D {x. \rho(x,y)} \, dA[/tex]  / M

x' = [tex]\int\limits^{\pi/2}_0 \int\limits^5_ 0{ r cos\theta 5 r^2 } \, dr d\theta[/tex]

x' = 5.5³/4

x' = 625/4

and, y' = [tex]\int\limits^{\pi/2}_0 \int\limits^5_ 0{ r sin\theta 5 r^2 } \, dr d\theta[/tex]

y' = 0

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Answer:

B. 1.4

Step-by-step explanation:

Additive inverses are two numbers that add to zero.

Additive inverses are two numbers that are equal except one is positive, and one is negative. For example, -2 and 2 are additive inverses since their sum equals zero.

The number shown on the number line is -1.4; to have a sum of zero, you need its additive inverse. The additive inverse of -1.4 is 1.4.

Answer: B. 1.4

Picking the white ball twice (T) is 1/4

Explanation: In this problem, we have a box with two balls - one white and one red. We will draw a ball from the box, put it back, and then draw a second ball (sampling with replacement). Let T be the event of getting the white ball twice, F the event of picking the white ball first, and S the event of picking the white ball in the second drawing.
To compute P(T), we need to find the probability of picking the white ball in both drawings:
P(T) = P(F) * P(S|F)
Since there is one white ball and one red ball in the box, the probability of picking the white ball first (F) is 1/2. Since we're sampling with replacement, the probability of picking the white ball in the second drawing (S) given that the white ball was picked first (F) is also 1/2.
So, the probability of picking the white ball twice (T) is:
P(T) = (1/2) * (1/2) = 1/4
Therefore, the probability of picking the white ball twice (T) is 1/4.

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Answer:99% of 110 is 108.9

so your answer is 108.9

Step-by-step explanation:

Because I said so

The required answer is -6(√2 + √3) after rationalizing the denominator.

What are rational numbers?

Rational numbers are numbers that can be expressed as the ratio of two integers, where the denominator is not zero. They can be written as fractions or decimals that either terminate or repeat infinitely. Rational numbers include all integers, as well as fractions such as 1/2, 2/3, and 7/5. Unlike irrational numbers, rational numbers can be expressed exactly, and can be added, subtracted, multiplied, and divided using the rules of arithmetic. They form an important subset of the real numbers and are used extensively in mathematics and everyday life.

To rationalize the denominator, we can multiply both the numerator and denominator by the conjugate of the denominator, which is :

And simplifying, we get:

6(√2 + √3)/ (2 - 3) = 6(√2 + √3)/ -1 = -6(√2 + √3).

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The probability of drawing a red card is always exactly 1/2

We can approach this problem using conditional probability. Let A denote the event that the first card is red, and B denote the event that the second card is red. Then, using the law of total probability, we have:

P(B) = P(A)P(B|A) + P(A^c)P(B|A^c)

where P(A) = 1/2, P(A^c) = 1/2, P(B|A) = 25/51 (since there are 25 red cards left out of 51 total cards), and P(B|A^c) = 26/51 (since there are 26 red cards left out of 51 total cards).

Therefore, we have:

P(B) = (1/2)(25/51) + (1/2)(26/51) = 51/102 = 1/2

This means that the probability of drawing two red cards in a row is exactly 1/2, regardless of the order in which the cards are drawn.

Similarly, we can calculate the probability of drawing three red cards in a row as follows:

P(C) = P(A)P(B|A)P(C|AB) + P(A)P(B^c|A)P(C|A(B^c)) + P(A^c)P(B|A^c)P(C|A^cB) + P(A^c)P(B^c|A^c)P(C|A^cB^c)

where C denotes the event that the third card is red, and we have conditioned on the first two cards that were drawn. Using the same reasoning as before, we have:

P(C) = (1/2)(25/51)(24/50) + (1/2)(26/51)(25/50) + (1/2)(26/51)(25/50) + (1/2)(25/51)(24/50) = 1225/5100 = 49/204

Thus, the probability of drawing three red cards in a row is less than 1/2, and in general, the probability of drawing n red cards in a row is (1/2)^n. Therefore, there is no strategy that can give you a better than 1/2 chance of winning the game, as the outcome of each draw is independent and the probability of drawing a red card is always exactly 1/2

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To convert the polar equation T= 3sin θ + 5cos θ into a Cartesian equation, we can use the following identities:

sin θ = y / r
cos θ = x / r

where r is the distance from the origin to the point (x, y) in the Cartesian plane. Substituting these expressions into the given polar equation, we get:

T = 3(y / r) + 5(x / r)

Multiplying both sides by r, we obtain:

Tr = 3y + 5x

This is the Cartesian equation of the given polar equation.

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The numerical value of each expression,

a. sech(0) = 1

b. coshaA¹ = 0.

(a) The hyperbolic secant function, sech(x), is defined as sech(x) = 1/cosh(x). Therefore, sech(0) = 1/cosh(0) = 1/1 = 1.

(b) The inverse hyperbolic cosine function, coshaA¹(x), is defined as the value y such that cosh(y) = x. Therefore, we want to find y such that cosh(y) = 1. Using the definition of the hyperbolic cosine function, we get:

cosh(y) = ([tex]e^y[/tex] + [tex]e^{(-y)}[/tex])/2 = 1

Multiplying both sides by 2 and rearranging, we get:

[tex]e^y[/tex] + [tex]e^{(-y)}[/tex] = 2

Multiplying both sides by [tex]e^y[/tex], we get:

[tex]e^{(2y)}[/tex] - 2[tex]e^y[/tex] + 1 = 0

This is a quadratic equation in [tex]e^y[/tex]. Using the quadratic formula, we get:

[tex]e^y[/tex] = (2 ± √(4 - 4))/2 = 1

Therefore, y = coshaA¹(1) = 0.

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A)The standard (or estimated standard) error of the mean is 0.08 years.

B) The 95% confidence interval for the average years of post-secondary education of all employees at the tech company is between 5.03 and 5.37 years.

C) The 98% confidence interval for the average years of post-secondary education of all employees at the tech company is between 4.98 and 5.42 years.

A) The standard error of the mean can be calculated using the formula:
standard error of the mean = population standard deviation / square root of sample size
In this case, the population standard deviation is 0.9 years and the sample size is 118.
standard error of the mean = 0.9 / sqrt(118) = 0.083
Therefore, the standard error of the mean is 0.08 years (rounded to two decimal places).
B) To find the 95% confidence interval for the average years of post-secondary education of all employees at the tech company, we can use the formula:
confidence interval = sample mean ± (critical value x standard error of the mean)
The critical value for a 95% confidence level with a sample size of 118 is 1.98 (from a t-distribution table).
confidence interval = 5.2 ± (1.98 x 0.083)
confidence interval = 5.03 to 5.37
Therefore, we can be 95% confident that the true average number of years of post-secondary education completed by all employees at the tech company is between 5.03 and 5.37 years.
C) To find the 98% confidence interval, we can use the same formula but with a different critical value. The critical value for a 98% confidence level with a sample size of 118 is 2.33 (from a t-distribution table).
confidence interval = 5.2 ± (2.33 x 0.083)
confidence interval = 4.98 to 5.42
Therefore, we can be 98% confident that the true average number of years of post-secondary education completed by all employees at the tech company is between 4.98 and 5.42 years.

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A graph of the image of the figure after a dilation by a scale factor of 2 centered at the origin is shown below.

In Mathematics and Geometry, a dilation simply refers to a type of transformation which typically changes the size of a geometric object, but not its shape.

Next, we would apply a dilation to the coordinates of the pre-image by using a scale factor of 2 centered at the origin as follows:

Ordered pair (6, 9) → Ordered pair (6 × 2, 9 × 2) = (12, 18).

Ordered pair (6, 6) → Ordered pair (6 × 2, 6 × 2) = (12, 12).

Ordered pair (9, 9) → Ordered pair (9 × 2, 9 × 2) = (18, 18).

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Missing information:

The question is incomplete and the complete question is shown in the attached picture.

The function g(x) is a reflection over the x-axis and stretch by 3 units and shifted upward by 1 unit of the parent function f(x).

We have,

A capability is a declaration, idea, or rule that lays out a relationship between two factors. Capabilities might be found all through science and are fundamental for the advancement of huge connections.

The functions are given below.

f(x) = x

g(x) = -3f(x) + 1

g(x) = -3x + 1

The parent function is a reflection over the x-axis and stretch by 3 units and shifted upward by 1 unit.

The functions g(x) and f(x) are shown below.

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a) Note that the inequality  expressing the possible number of piccolos that could perform allowing the audience to hear every note  is given as: 3,850x ≤ 16,000

b) the maximum number of piccolos allowed to perform this piece is 4.16 or approximately 4.

A) Where the above conditions are given, let us call the number of piccolos "x" as the number that can perform in a way that the audience will hear all the notes pplayed.

Recall that specialized acoustic tiles that eliminate any sounds greater than 16 kHz that means, all the frequencies put together must  nevr exced 16kHz. Thus,

the inequality will be 3,850x ≤ 16,000


B) To determine the maximum number of piccolos:

we must solve for x in the above equation:

3,850x ≤ 16,000

we divide both sides by 3850

3,850x/3850 ≤ 16,000/3850
x ≤ 4.15584415584

Hence, since x cannot be be decimal, that is the number of picollows, we must approximate x to 4

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Answer:

  B.  20 square units

Step-by-step explanation:

You want a graph and the area of figure ABCD with points A(-7,-5), B(-2,-5), C(-2,-1), D(-7,-1).

The attachment shows a graph of the points and the rectangle they define. The side lengths of the rectangle can be found from the coordinates, or by counting grid squares.

The rectangle is 5 units wide and 4 units high.

The area is the product of length and width:

  A = LW = (5 units)(4 units) = 20 units²

The area of ABCD is 20 square units.

Taxonomy is an organizer that is used to group shapes into different classifications using the traits of the shapes.

Taxonomy is a system for organizing and classifying objects, in this case, shapes, based on their shared characteristics or traits. It is a way of grouping objects into different categories or classes to help organize and understand them better.

Taxonomies are commonly used in many different fields, including biology, library science, and information technology.

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False. The probability of independent events each occurring is P(A) x P(B), not P(A) + P(B).

The possibility of event A and event B occurring collectively may be calculated using the multiplication rule of chance, which states that the probability of two independent events taking place collectively is same to the made of their person chances.

Consequently, the chance of A and B each taking place together may be calculated as P(A) x P(B), assuming that a and B are independent activities.

it's far critical to notice that the addition rule of possibility can best be carried out when events A and B are jointly unique, which means they can not arise together. in that case, the chance of both event A or event B taking place may be calculated by using including their character possibilities.

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The probability of choosing a red scrub and then another red scrub is 0.14423 or 14.42%.

To calculate the probability of choosing a red scrub, we first need to know how many red scrubs are in the set. In this case, there are 6 red scrubs out of a total of 13 scrubs. Therefore, the probability of choosing a red scrub is:

Probability of choosing a red scrub = Number of red scrubs / Total number of scrubs

Probability of choosing a red scrub = 6 / 13

Now, we need to calculate the probability of choosing another red scrub after the first one has been chosen. Since one red scrub has already been chosen, there are now only 5 red scrubs left out of a total of 12 scrubs. Therefore, the probability of choosing another red scrub is:

Probability of choosing another red scrub = Number of remaining red scrubs / Total number of remaining scrubs

Probability of choosing another red scrub = 5 / 12

To calculate the probability of both events happening (choosing a red scrub and then another red scrub), we need to multiply the probabilities together. This is known as the multiplication rule of probability.

Probability of both events happening = Probability of the first event x Probability of the second event

Probability of both events happening = (6/13) x (5/12)

Probability of both events happening = 0.14423 or 14.42%

This means that out of all the possible outcomes, there is a 14.42% chance of choosing two red scrubs in a row from a set of 13 scrubs.

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It is not clear what needs to be solved or what question is being asked. The text contains various mathematical expressions that are not connected to form a coherent problem or question

A mathematical expression is a combination of numbers, variables, and mathematical operations, such as addition, subtraction, multiplication, and division, that can be evaluated to obtain a numerical result.

According to the given information:

The text appears to contain various mathematical expressions related to trigonometry and geometry. Here is an English explanation of some of the expressions:

"a² + b² = c²" is the Pythagorean theorem, which states that in a right triangle, the square of the length of the hypotenuse (the side opposite the right angle) is equal to the sum of the squares of the lengths of the other two sides.

"Triángulos semejantes" means "similar triangles", which are triangles that have the same shape but different sizes.

"x² / 0² = 6/2" appears to be an expression involving the ratios of corresponding sides of similar triangles.

"Δε hipotenusa de cada triángulo. ¿Por qué deben ser iguales estas razones" means "Δ is the hypotenuse of each triangle. Why must these ratios be equal?" This is likely a question about the properties of similar triangles and their corresponding side ratios.

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A vector space can have multiple bases. So the given statement is false.

A vector space can have more than one basis. A basis is a set of linearly independent vectors that spans the entire vector space. A vector space can have infinitely many bases, and each basis may contain a different number of vectors. However, all bases of a vector space will have the same cardinality, which is called the dimension of the vector space.

Therefore, a vector space can have multiple bases.

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Answer:

Step-by-step explanation:

Least Common Factors of 4 and 10: 1 and 2

Therefore, 1 is the least common factor.

(I am sure you meant Least Common Multiple and not Least Common Factor, as there is no term like that in mathematics. - The Lowest Common Factor would always be 1)

The LCM would be 20

The linear correlation coefficient of r=0.807 indicates a strong positive linear relationship between income and blood pressure in the sample of 200 men aged between 20 and 60.

The income increases, blood pressure tends to increase as well in the sample.

No, this does not necessarily suggest that if a man gets a salary raise, his blood pressure is likely to rise. Correlation does not imply causation, so we cannot conclude that a change in income causes a change in blood pressure.

There may be other factors that influence both income and blood pressure, such as stress, diet, physical activity, or genetic factors.

Lurking variables are unobserved variables that may affect the relationship between income and blood pressure.

Some possible lurking variables in this scenario include age, race, education, occupation, lifestyle factors, family history of hypertension, and underlying health conditions.

These variables may confound the relationship between income and blood pressure, making it difficult to establish a causal relationship.

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The ages of Jacob and his mom as per formed system of linear equation for the given data is equal to 12 years old and 51 years old respectively.

Let j represent Jacob's age,

And m represent his mom's age.

Jacob's mom is three more than four times Jacob's age.

m = 4j + 3

In five years, his mom will be five more than three times his age

⇒ m + 5 = 3 × ( j + 5 ) + 5

Simplifying the equation,

m + 5 = 3j + 20

m = 3j + 15

Now ,two system of linear equations,

m = 4j + 3

m = 3j + 15

Solve this system of linear equations,

⇒ 4j + 3 = 3j + 15

⇒ j = 12

Substituting j = 7 into either equation gives us,

m = 4(12) + 3

   = 51

Therefore, the ages of Jacob and his mom is equal to 12 years old and 51 years old respectively.

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The probability that a randomly selected worker primarily drives a car or takes public transportation to work is 0.915. The probability that a randomly selected worker neither drives a car nor takes public transportation to work is 0.085. The probability that a randomly selected worker does not drive a car to work is 0.133.

(a) To find the probability that a randomly selected worker primarily drives a car or takes public transportation to work, we can add the probabilities of each event:
P(drives car) + P(takes public transportation) = 0.867 + 0.048 = 0.915
Therefore, the probability that a randomly selected worker primarily drives a car or takes public transportation to work is 0.915.
(b) The probability that a randomly selected worker neither drives a car nor takes public transportation to work can be found by subtracting the probability of driving a car or taking public transportation from 1:
P(neither drives car nor takes public transportation) = 1 - P(drives car or takes public transportation)
P(neither drives car nor takes public transportation) = 1 - 0.915
P(neither drives car nor takes public transportation) = 0.085
Therefore, the probability that a randomly selected worker neither drives a car nor takes public transportation to work is 0.085.
(e) To find the probability that a randomly selected worker does not drive a car to work, we can subtract the probability of driving a car from 1:
P(does not drive car) = 1 - P(drives car)
P(does not drive car) = 1 - 0.867
P(does not drive car) = 0.133
Therefore, the probability that a randomly selected worker does not drive a car to work is 0.133.

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