The following are distances (in miles) traveled to the workplace by 6 employees of a certain computer company. 11, 6, 36, 16, 5, 40 Send data to calculator Find the standard deviation of this sample o

The statement which correctly describes the data set include the following:

D. the data is bivariate and numerical.

In Mathematics, a bivariate data can be defined as a type of data set which comprises information that are based on two (2) variables, usually two types of related data.

In Mathematics and statistics, a numerical data can be defined as a type of data set that is primarily expressed in numbers only. This ultimately implies that, a numerical data simply refers to a type of data set consisting of numbers (numerals), rather than words or letters.

Thus, In conclusion, we can logically deduce that the given data set is both bivariate and numerical.

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Outliers can affect the performance of the classifier and make it difficult to find a linear decision boundary.

True. The level of significance in statistics refers to the probability of making a Type I error, which is rejecting a true null hypothesis. It is typically denoted as alpha and set by the analyst or researcher prior to conducting a hypothesis test. It is considered the amount of risk that an analyst is willing to take in rejecting a true null hypothesis.

True. Linear separability is the ability to classify data points in a dataset using a linear decision boundary. If a dataset is not linearly separable, it means that a linear boundary cannot accurately classify all the data points. One of the reasons why this may happen is due to experimental errors that lead to errant data points. These outliers can affect the performance of the classifier and make it difficult to find a linear decision boundary.

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The confidence interval:

Lower Endpoint: 6.418

Upper Endpoint: 6.782

To construct the confidence interval, we can use the formula:

CI = x ± z(σ/√n)

Where:

x = sample mean = 6.6

σ = population standard deviation = 2.3

n = sample size = 432

z = z-score for 90% confidence level = 1.645 (from the standard normal distribution table)

Plugging in the values, we get:

CI = 6.6 ± 1.645(2.3/√432)

CI = 6.6 ± 0.182

Therefore, the 90% confidence interval for the true mean number of reproductions per hour for the bacteria is:

Lower Endpoint: 6.418

Upper Endpoint: 6.782

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Answer:  is x=3

Step-by-step explanation:

Answer:

The awnser is 9

Step-by-step explanation:

1 × 9 = 9 9 dovided by 7 = -2 + 8 is 6 6 times 0 is 0 0 + 1 = 1 1 × 6 = 6

The solutions of the given expressions are as follows :

(a) (x^2+3x+9) / (x^2+9x+18) = x+2

(b) x ≈ 0.004 or x ≈ 0.056

(c)  we have two solutions: x = 0 or x = V8 (cube root of 8)

a. To divide *3-27 by 9-x^2, we can first factor both  expressions :
*3-27 = 3*(-9)
9-x^2 = (3-x)(3+x)
So we have:

(*3-27) / (9-x^2) = (3*(-9)) / ((3-x)(3+x))

To divide x^2+3x+9 by x^2+9x+18, we can use long division or synthetic division. Using long division, we have:

         x + 2
 -------------------
x^2 + 9x + 18 | x^2 + 3x + 9
         -x^2 - 2x
         ----------
                x + 9
                -x - 9
                -------
                      0

So we have:

(x^2+3x+9) / (x^2+9x+18) = x+2

b. To solve V3x + 2-27x = 0, we can first isolate the radical:

V3x = 27x - 2

Then we can square both sides:

3x = (27x - 2)^2

Expanding the right side and simplifying, we get:

3x = 729x^2 - 108x + 4

Bringing everything to one side, we have:

729x^2 - 111x + 4 = 0

Using the quadratic formula, we get:

x = (111 ± V(111^2 - 4*729*4)) / (2*729)

x ≈ 0.004 or x ≈ 0.056

c. To solve 3x^7 - 24x^4 = 0, we can factor out x^4:

3x^4(x^3 - 8) = 0

So we have two solutions:

x = 0 or x = V8 (cube root of 8)

Note that the equation has a total of seven roots (since it is a seventh-degree equation), but we only found two of them. The other five roots are complex numbers.

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First, we can simplify the equation by isolating the variable on one side:

2√x + 6 - 3 = 19

2√x + 3 = 19

2√x = 16

√x = 8

Now we can square both sides of the equation to isolate x:

(√x)² = 8²

x = 64

Therefore, the solution to the equation 2√x + 6 - 3 = 19 is x = 64, which corresponds to answer choice (e).

Hope this helped (:

If A is an m x n matrix, and let B be an n xp matrix such that AB = 0, then rank A + rank B ≤ n

To start, we can use the fact that for any matrix A, rank A + dim Nul A = n, where n is the number of columns in A. This is a result of the rank-nullity theorem.

Now, let's consider the subspaces associated with A and B. The column space of A, Col A, is a subspace of R^m, and the null space of A, Nul A, is a subspace of R^n. Similarly, the column space of B, Col B, is a subspace of R^n, and the null space of B, Nul B, is a subspace of R^p.

We want to show that rank A + rank B ≤ n, so let's consider the dimensions of the subspaces. We know that dim Col A = rank A and dim Col B = rank B. We also know that AB = 0, which means that every column of B is in the null space of A, Nul A. In other words, Col B is a subset of Nul A.

Using the fact that dim Col B + dim Nul B = n, we can write:

rank B + dim Nul B = n - dim Col B

Since Col B is a subset of Nul A, we know that dim Nul A ≥ dim Col B. Therefore:

dim Nul B ≥ dim Nul A

Substituting this inequality into the previous equation, we get:

rank B + dim Nul B ≥ rank B + dim Nul A = n - dim Col A

Finally, we can use the fact that rank A + dim Nul A = n to substitute for dim Nul A:

rank B + dim Nul B ≥ n - rank A

Adding rank A to both sides, we get:

rank A + rank B + dim Nul B ≥ n

But we know that dim Nul B ≥ 0, so:

rank A + rank B ≤ n

Therefore, we have shown that rank A + rank B ≤ n, as desired.

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In order to identify labeled points that do not lie on the x-y plane, several methods can be utilized.

The software tools MATLAB, Python's Matplotlib or Excel can be used to create a 3D plot of the labeled points. Through this approach, non-planar points become easily recognizable. It is possible to label points with different colors or symbols, based on their classification which would provide an added advantage in noticing any types of pattern or trends.

An alternate route involves having access to equation(s) of the fitted plane (e.g., by means of linear regression). Herein lies the ability to measure the distance between each point and the plane using the point-to-plane distance formula. Based upon fitting, if any point has substantial distance from the plane then it is likely to be situated off the plane.

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$1,560

To calculate simple interest, we use the formula:

Simple interest = Principal * Rate * Time

Given that Bianca invested $6,500 at a rate of 3%, the principal is $6,500 and the rate is 0.03 (since 3% is equivalent to 0.03 as a decimal).

We are asked to find the simple interest after 8 years, so the time is 8 years.

Using the formula, we get:

Simple interest = $6,500 * 0.03 * 8

Simple interest = $1,560

Therefore, the simple interest on Bianca's investment will be $1,560 after 8 years.

The car must achieve an acceleration of 120 mi/h² at some point between 2:00pm and 2:10 pm.

To find the acceleration the car must achieve using the Mean Value Theorem (MVT), we need to follow these steps:
1. Calculate the change in speed.
2. Calculate the change in time.
3. Apply the MVT to find the acceleration.
Step 1: The car's speedometer reads 50mph at 2:00 pm and 70mph at 2:00 pm. The change in speed is 70mph - 50mph = 20mph.
Step 2: The change in time is 10 minutes, which we need to convert to hours. To do this, divide 10 by 60 (since there are 60 minutes in an hour). So, 10/60 = 1/6 hour.
Step 3: Apply the MVT. The MVT states that there must be a point in time where the average acceleration equals the instantaneous acceleration. The average acceleration (a) can be found using the formula a = Δv/Δt. Here, Δv is the change in speed (20mph) and Δt is the change in time (1/6 hour).
So, a = (20mph) / (1/6 hour) = 20 * 6 = 120 mi/h².

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In other words, the system has infinitely many solutions, parameterized by the values of the free variables a and b.

The final step of the Gauss-Jordan elimination can be interpreted as the following system of equations:

x1 - 2x2 + 21x3 = 0

x4 + x5 = 1

x6 - 2x7 = 0

x8 + x9 = 1

From the second and fourth equations, we can conclude that x4 and x8 are free variables, which means they can take on any value. Let's set them to be a and b, respectively.

Then, using the first and third equations, we can solve for x2, x3, x5, and x7 in terms of a and b:

x2 = (21/2)a - (1/2)b

x3 = a/2 - (21/4)b

x5 = 1 - a

x7 = b/2

Finally, substituting these values into the remaining equations, we can solve for x1 and x6:

x1 = 2x2 - 21x3 = -19a + (209/4)b

x6 = 2x7 = b

Therefore, the solution to the system of equations is:

x1 = -19a + (209/4)b

x2 = (21/2)a - (1/2)b

x3 = a/2 - (21/4)b

x4 = a

x5 = 1 - a

x6 = b

x7 = b/2

x8 = a

x9 = 1 - a

In other words, the system has infinitely many solutions, parameterized by the values of the free variables a and b.

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The limit of the ratio of the probabilities is[tex]$\frac{3}{4}$[/tex].

Given: [tex]$\$ P(Z=0)=p \$[/tex], [tex]\$ P(Z=1)=q \$[/tex], [tex]\$ P(Z=3)=r \$$[/tex], where[tex]$\$ p+q+r=1 \$$[/tex] and [tex]$\$ E[Z] < 1 \$$[/tex].

(a) To find the recursion formula for [tex]$\$ W_{-} a(u) \$$[/tex], we use the following formula:

[tex]$$W_a(u)=P(Z=a)+\sum_{k=0}^{u-1} P(Z=a+3 k) W_a(u-1-k)$$[/tex]

where [tex]$\$ \mathrm{a} \$$[/tex] is a non-negative integer and [tex]$\$ \mathrm{u} \$$[/tex] is a positive integer.

Using the given probabilities, we have:

[tex]$$\begin{aligned}& W_0(u)=p+\sum_{k=0}^{u-1} r W_0(u-1-k) \\& W_1(u)=q+\sum_{k=0}^{u-1} p W_1(u-1-k)+\sum_{k=0}^{u-1} r W_1(u-1-k-1) \\& W_3(u)=r+\sum_{k=0}^{u-1} q W_3(u-1-k)\end{aligned}$$[/tex]

(b) To find [tex]$\$ 19$[/tex], we use the fact that [tex]$\$ W_{-} O(u)+W_{-} 1(u)+W_{-} 3(u)=1 \$$[/tex] for all positive integers [tex]$\$[/tex] u [tex]\$$[/tex].

We take the limit as [tex]$\$$[/tex] u [tex]$\$$[/tex] approaches infinity:

[tex]$$\lim _{u \rightarrow \infty} W_0(u)+\lim _{u \rightarrow \infty} W_1(u)+\lim _{u \rightarrow \infty} W_3(u)=1$$[/tex]

Since [tex]$\$ E[Z] < 1 \$$[/tex], we have. Also, from part (a), we have:

[tex]$$\begin{aligned}& \lim _{u \rightarrow \infty} W_0(u)=\lim _{u \rightarrow \infty} W_0(u-1) \\& \lim _{u \rightarrow \infty} W_1(u)=p \lim _{u \rightarrow \infty} W_1(u-1)+\lim _{u \rightarrow \infty} W_1(u-2)\end{aligned}$$[/tex]

solve for the limits as:

[tex]$$\begin{aligned}& \lim _{u \rightarrow \infty} W_0(u)=\frac{p}{1-r} \\& \lim _{u \rightarrow \infty} W_1(u)=\frac{q p}{1-p-r}\end{aligned}$$[/tex]

Therefore, we have:

[tex]$$\lim _{u \rightarrow \infty} f(u)=\lim _{u \rightarrow \infty} \frac{W_1(u)}{W_0(u)+W_1(u)}=\frac{q p}{p+(1-r)}=\frac{3}{4}$$[/tex]

Thus, the limit of the ratio of the probabilities is[tex]$\frac{3}{4}$[/tex].

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a. There are 10 kg salt in the tank after t minutes

b.  After 30 minutes, the amount of salt in the tank is still 10 kg.

a) After t minutes, the amount of salt in the tank can be found by the formula:

Amount of salt = initial amount of salt + (rate of salt in - rate of salt out) x time

The initial amount of salt is 10 kg, and the rate of salt in is 0.01 kg/L x 15 L/min = 0.15 kg/min. The rate of salt out is also 0.01 kg/L x 15 L/min = 0.15 kg/min, because the solution is kept thoroughly mixed. Therefore, the amount of salt in the tank after t minutes is:

Amount of salt = 10 + (0.15 - 0.15) x t = 10 kg

b) After 30 minutes, the amount of salt in the tank is still 10 kg. This is because the rate of salt in and the rate of salt out are equal, and so the amount of salt in the tank remains constant. Therefore, the answer is the same as part (a), which is 10 kg

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Nine cans must be chosen to guarantee that we have three cans of the same type of soft drink.

To guarantee that we have three cans chosen from the same type of soft drink, we need to consider the worst-case scenario, which is that we choose two cans from each type of soft drink (a total of eight cans) and none of them is the same type. In this case, we would need to choose at least nine cans to guarantee that we have three cans chosen from the same type of soft drink.

To see why this is the case, imagine choosing eight cans from the four different types of soft drinks. There are two possibilities:

1. We choose two cans from each type of soft drink, and none of them is the same type. In this case, we would need to choose at least one more can from any of the types of soft drinks to guarantee that we have three cans chosen from the same type.

2. We choose three cans from at least one type of soft drink. In this case, we already have three cans chosen from the same type.

Therefore, we need to choose at least nine cans to guarantee that we have three cans chosen from the same type of soft drink, regardless of which cans we choose.

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Note that this prompt examines the given data using scatter plot whose details is analyzed below.

1) The scatter plot showing the relationship between the numbers of points scored and the numbers of rebounds is attached.

2) The association between the numbers of points scored and the numbers of rebounds is a positive one. This means that generally, there is a tendency to get more points when the number of rebounds is high.

3) No, we cannot conclude that greater numbers of points scored cause

greater or lesser numbers of rebounds. This would be an inverse relationship which contradicts our findings above.

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Answer: x = 4.

Step-by-step explanation:

Taking the logarithm of both sides with base 3, we get:

x = log3(81)

x = log3(3^4)  [Since 81 is equal to 3 raised to 4th power]

x = 4

Therefore, the solution is x = 4.

Answer:

Step-by-step explanatiON

X=81/3

X=4

Fill in each blank so that the resulting statement is true: A function f has an inverse that is a function if there is no vertical line that intersects the graph of f at more than one point. Such a function is called a/an injective function or a one-to-one function.

A function is injective if every distinct input produces a distinct output. Geometrically, this means that the function does not repeat any output values . If there is a vertical line that intersects the graph of f at more than one point, then the function fails to be injective, since two distinct input values will produce the same output value. In this case, the function does not have an inverse that is a function.

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Full Question ;

Fill in each blank so that the resulting statement is true. A function f has an inverse that is a function if there is no ____ line that intersects the graph of f at more than one point. Such a function is called a/an ____ function.

Yes, Cyrus can cover 6 miles by running 40 times.

Given that, Cyrus plans to run at least 6 miles each week, Cyrus has a circular route in the neighborhood to run, having a circumference of 340 yards,

We need to find if Cyrus runs that route 40 times during the week, will he cover at least 6 miles or not,

So,

1 mile = 1760 yards

Therefore,

6 miles = 1760 × 6 = 10560 yards

The circumference of the route = 340 yards

He took 40 rounds, so the total distance covered = 340 × 40 = 13600 yards.

Since, 6 miles = 10560 yards and he covered 13600 yards

Hence, yes, he can cover 6 miles by running 40 times.

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The statements that are supported by the graph are:

Given that the equation of the function is

f(x) = (x - 2)(x - 6)

From the equation of the graph, we can see that

Minimum = (4, -4)

This means that the vertex is at (4, -4)

The x coordinate of the vertex is the axis of symmetry

So, we have

x = 4

Next, we set the function to 0 to determine the zeros

So, we have

(x - 2)(x - 6) = 0

Solve for x

x = 2 and x = 6

This means that the zeros are at (2, 0) and (6, 0).

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Answer:

Step-by-step explanation:

You want the common ratio and first 3 terms of a decreasing geometric progression with the sum of the first three terms being 7, and their product being 8.

Let the first term be represented by x, and let r represent the common ratio. Then the first three terms are ...

  x, xr, xr²

Their sum is ...

  7 = x +xr +xr²

Their product is ...

  8 = (x)(xr)(xr²) = (xr)³

Taking the cube root of the product equation, we have ...

  2 = xr

Substituting this into the first equation, we have ...

  7 = x +2 + 2r

  5 = x +2r   ⇒   x = 5 -2r

And substituting back into the above, we get ...

  2 = (5 -2r)(r)

  2r² -5r +2 = 0

  (2r -1)(r -2) = 0

  r = 2 or 1/2

We want r < 1, so r = 1/2.

  x = 5 -2(1/2) = 4

For x = 4, r = 1/2, the first three terms are ...

  x, xr, xr² = 4, 2, 1

__

Additional comment

The equations are nicely solved by a graphing calculator. In the attached, we used y instead of r. We want the solution with y<1.

The two solutions give rise to terms 4, 2, 1 (decreasing) or 1, 2, 4 (increasing).

The approximate probability that between 12 and 20 patients, inclusive will experience weight gain as a side effect is 0.4147.

Normal approximation can be used to approximate the binomial distribution when the sample size is large enough (n >= 30) and the probability of success (p) and failure (q=1-p) are not too small or too large. In this case, we have a sample size of 500, which is sufficiently large, and the probability of success (p=0.03) and failure (q=0.97) are not too small or too large.

To approximate the probability that 15 or fewer users will experience weight gain as a side effect, we can use the normal approximation to the binomial distribution with mean (μ) = np = 500 x 0.03 = 15 and standard deviation (σ) = sqrt(npq) = sqrt(500 x 0.03 x 0.97) = 3.814. Then, we can use the normal cumulative distribution function (normalcdf) to calculate the probability that X ≤ 15, where X is the number of users who experience weight gain.

normalcdf(–1E99,15.5,15, 3.814) = 0.5522

Therefore, the approximate probability that 15 or fewer users will experience weight gain as a side effect is 0.5522.

To approximate the probability that 24 or more users experience weight gain as a side effect, we can use the normal approximation to the binomial distribution with the same mean and standard deviation as before. Then, we can use the normal complementary cumulative distribution function (normalccdf) to calculate the probability that X ≥ 24.

normalccdf(23.5,15,3.814) = 0.0097

Therefore, the approximate probability that 24 or more users experience weight gain as a side effect is 0.0097.

To approximate the probability that between 12 and 20 patients, inclusive will experience weight gain as a side effect, we can use the normal approximation to the binomial distribution with the same mean and standard deviation as before. Then, we can use the normal cumulative distribution function (normalcdf) to calculate the probability that 12 ≤ X ≤ 20.

normalcdf(11.5,20.5,15,3.814) = 0.6081 - 0.1934 = 0.4147

Therefore, the approximate probability that between 12 and 20 patients, inclusive will experience weight gain as a side effect is 0.4147.

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From mass unit conversion where using a numeric constant ( 0.001), the total weight of vegitables ( carrots and beans) in a bag is equals to the 3.420 kilograms.

A unit conversion is used to expresses the same property as a different unit of measurement. Unit conversion is a process with serval steps that involves multiplication or division by a numerical factor called conversion factor. So, there are different unit conversions charts like mass unit conversation, length unit conversion etc. Now, we have, a bag contains carrots and beans.

Weight of carrots in bag = 3 kg

weight of beans in bag = 420 grams

We have to determine total weight that bag contains. As we see both weights are present in different units ( i.e, kg and grams). Using unit conversion, 1 kilogram = 1000 grams

=>[tex] 1 gram= \frac{1}{1000}=0.001 kg[/tex]

So, 420 grams = 420× 0.001 = 0.420 kg

Total weight of vegitables in bag = weight of carrots + weight of beans

= 3 kg + 0.420 kg

= 3.420 kg

Hence, required value is 3.420 kilograms.

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The appropriate answer would be option B: two-tailed dependent samples t-test.

Since we are comparing scores from the same group of participants at two different points in time (first separation vs second separation), we would use a dependent samples t-test.

Therefore, the options are A and B. We cannot determine whether the test would be one-tailed or two-tailed based on the information given.

A one-tailed test would be appropriate if we had a specific directional hypothesis (e.g., we expect the scores to be higher on the first separation compared to the second separation). A two-tailed test would be appropriate if we had a non-directional hypothesis (e.g., we expect there to be a difference between the scores, but we do not have a specific expectation about the direction of the difference).

Since we do not have information about the directional hypothesis, the appropriate answer would be option B: two-tailed dependent samples t-test.

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The measure of the selected angles from the attached figure of coordinate plane equals to 44° are ∠2 and ∠4.

In the figure of the coordinate plane,

Measure of angle ABD = 90 degrees

measure of angle ABD = 44° + measure of angle 1

⇒  measure of angle 1 =  measure of angle ABD - 44°

⇒  measure of angle 1 =  90° - 44°

⇒  measure of angle 1 =  46°

Angles formed in the second quadrant.

measure of angle 1 + measure of angle 2 = 90°

⇒measure of angle 2 = 90° - 46°

⇒measure of angle 2 = 44°

From the figure,

Measure of ∠ABC = 44°

Using the result of vertically opposite angle we have,

Measure of angle 4 =Measure of ∠ABC

⇒Measure of angle 4 =  44°

Therefore, the measure of the angles in the figure equals to 44° are ∠2 and ∠4.

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The probability that more than one flat screen television will be produced in 2 hours is 0.2642.

(a) To find the probability that no flat screen television was produced on a particular day, we can use the Poisson distribution formula:

P(X = 0) = (e^(-λ) * λ^0) / 0!

where λ is the average number of flat screen televisions produced per day, which is 3 in this case.

P(X = 0) = (e^(-3) * 3^0) / 0! = 0.0498 (rounded to four decimal places)

Therefore, the probability that no flat screen television was produced on a particular day is 0.0498.

(b) To find the probability that there are at most nine flat screen televisions produced in 3 days, we can use the cumulative Poisson distribution formula:

P(X ≤ 9) = ∑(k=0 to 9) [(e^(-λ) * λ^k) / k!]

where λ is the average number of flat screen televisions produced per day, which is 3 in this case. To find the probability for 3 days, we need to multiply λ by 3.

P(X ≤ 9) = ∑(k=0 to 9) [(e^(-9) * 9^k) / k!] = 0.5874 (rounded to four decimal places)

Therefore, the probability that there are at most nine flat screen televisions produced in 3 days is 0.5874.

(c) If the production line only functions 8 hours a day, we can adjust λ accordingly. Since there are 24 hours in a day and the production line is functioning for 8 hours, the average number of flat screen televisions produced in 2 hours would be λ/3.

So, the new λ would be 3/3 = 1.

To find the probability that more than one flat screen television will be produced in 2 hours, we can use the Poisson distribution formula:

P(X > 1) = 1 - P(X ≤ 1)

P(X ≤ 1) = (e^(-1) * 1^0) / 0! + (e^(-1) * 1^1) / 1! = 0.7358 (rounded to four decimal places)

P(X > 1) = 1 - 0.7358 = 0.2642 (rounded to four decimal places)

Therefore, the probability that more than one flat screen television will be produced in 2 hours is 0.2642.

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The value of x is 11.6 ( option C).

Similar triangles are triangles that have the same shape, but their sizes may vary. The angles of similar triangles are congruent.

Also the ratio corresponding sides of similar triangles are equal. This means for two triangles to be similar , the corresponding angles must be equal and the ratio of corresponding sides are equal.

Therefore,

29/50 = x/20

29×20 = 50x

580 = 50x

divide both sides by 50

580/50 = x

x = 580/50

x = 58/5

x = 11.6

therefore the value of x 11.6

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The value of the missing ratio = 25. That is 5/25.

Ratio is an expression that shows the quantity of a variable that is found in another variable. It can be represented as a:b or can be written as a numerator all over a denominator. That is a/b.

To determine the missing part of the ratio, the following is carried out.

if 1 = 5, 2 = 10 then 5 = 25 = 1/5

Therefore the value of the missing part of the ratio is 5 and the complete ratio is written as 5/25.

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The prime factorization for 168 is 2 x 2 x 2 x 3 x 7.

(a) The prime factorization for 294 is 2 x 3 x 7 x 7.
(b) The prime factorization for 1,584 is 2 x 2 x 2 x 2 x 3 x 3 x 7.
(c) The prime factorization for 187 is 11 x 17.
(d) The prime factorization for 51 is 3 x 17.

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The probability that a randomly selected adult has an IQ greater than 134.4 is 0.025 or 2.5%

To find the probability that a randomly selected adult has an IQ greater than 134.4, we need to calculate the z-score and then find the corresponding area under the standard normal distribution curve.

The z-score is calculated as: [tex]z= \frac{x-μ}{σ}[/tex]

where x is the IQ score, μ is the mean IQ score, and σ is the standard deviation of IQ scores.

Substituting the given values, we get:

[tex]z = \frac{(134.4 - 101.1)}{17}[/tex]

z = 1.96

Using a standard normal distribution table, we find that the area to the right of z = 1.96 is approximately 0.025. Therefore, the probability that a randomly selected adult has an IQ greater than 134.4 is 0.025 or 2.5%.

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