Which of the following is a solution to this inequality?

y less than two thirds times x plus 2

(0, 3)
(−3, 1)
(3, 5)
(1, 2)

The area of the given figure is 43.3 units².

A trapezium is a quadrilateral with four sides where two sides are parallel to each other.

We have,

The figure has two shapes:

Trapezium and a semicircle.

Now,

We will consider one box as one unit.

Trapezium:

Height = 6 units

Parallel sides = 8 units and 5 units

Area = 1/2 x height x (sum of the parallel sides)

Area = 1/2 x 6 x (8 + 5)

Area = 3 x 13

Area = 39 units²

Semicircle:

Diamter = 6 units

Radius = 3 units

Area = 1/2 x πr²

Area = 1/2 x 3.14 x 3²

Area = 4.3 units²

Now,

Area of the figure.

= 39 + 4.3

= 43.3 units²

Thus,

The area is 43.3 units².

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The tables represents a relation that is a function will be table number 3. Then the correct option is C.

A function is an assertion, concept, or principle that establishes an association between two variables. Functions may be found throughout mathematics and are essential for the development of significant links.

The number of value of the variable 'y' mush be one for a given value of 'x'. Then the relation will be a function.

If the number of value of the variable 'y' is 2 or more than two. Then the relation will not be a function.

Let's check the third option. Then we have

    x                y

   −3              0

   −1               3

    0               4

    3               0

    4               3

The tables represents a relation that is a function will be table number 3. Then the correct option is C.

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If the rate is tripled and the time increases by 3 hours, the value of d will increase by a factor of 12.

The length along a line or line segment between two points on the line or line segment.

Distance=√(x₂-x₁)²+(y₂-y₁)²

The original distance, d(r,t), can be expressed as d(r,t) = rt.

Value of d if the rate is tripled and the time increases by 3 hours

The new distance, d(3r,t + 3), can be expressed as d(3r,t + 3) = 3r(t + 3).

The new distance d(3r,t + 3) = 3r(t + 3)

= 3rt + 3(3r)

= 3rt +9r

= 3d(r,t) + 9r

Therefore, if the rate is tripled and the time increases by 3 hours, the value of d will increase by a factor of 3 + 9 = 12.

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The standard deviation is 0.07770.

The above table is given.

First, calculate the class mid-point. This is the average of a class interval.

The mid-point x is:

[tex]x_{1} =\frac{1}{2} (15.3+15.6) = \frac{1}{2} * 30.9 = 15.45[/tex]   ( simplify the equation)

[tex]x_{2} =\frac{1}{2} (15.6+15.9) = \frac{1}{2} * 31.5 = 15.75[/tex]

[tex]x_{3} =\frac{1}{2} (15.9+16.2) = \frac{1}{2} * 32.1 = 16.05[/tex]

[tex]x_{4} =\frac{1}{2} (16.2+16.5) = \frac{1}{2} * 32.7 = 16.35[/tex]

[tex]x_{5} =\frac{1}{2} (16.5+16.8) = \frac{1}{2} * 33.3 = 16.65[/tex]

So, we know that :

x  15.45   15.75    16.05   16.35   16.65

f     14        25        84       18         9

Calculate the mean:

Mean = [tex]\frac{15.45 *14 + 15.75 * 25 + 16.05*84 + 16.35 *18 + 16.65 * 9}{14+25+84+18+9}[/tex]

Mean   = [tex]\frac{2402.4}{150}[/tex]    (by using division)

Mean    = 16.016

The standard deviation is:

σ [tex]=\frac{\sqrt{(15.45-16.016)^{2} + (15.75-16.016)^{2} + (16.05-16.016)^{2} + (16.35-16.016)^{2} + (16.65-16.016)^{2} } }{14 + 25+84+ 18+ 9}[/tex]

σ  [tex]=\sqrt{0.00603853333}[/tex]

σ = 0.07770

So, the standard deviation is 0.07770.

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The complete question is:

To study the output of a machine that fills boxes with cereal, a quality control engineer weighed 150 boxes of Brand X cereal. The frequency distribution in the following table summarizes his findings. Find the standard deviation of the weight of the boxes of cereal. (Round your answer to three decimal places.)

Incorrect: Your answer is incorrect.

oz

x = Weight

(ounces) Number of

boxes

15.3 ≤ x < 15.6 14

15.6 ≤ x < 15.9 25

15.9 ≤ x < 16.2 84

16.2 ≤ x < 16.5 18

16.5 ≤ x < 16.8 9

The order of the matrix is

The order of a matrix is the number of rows and columns it has.

It is represented as "m x n",

where

m is the number of rows and

n is the number of columns.

For example, a matrix with 3 rows and 4 columns has an order of 3 x 4.

In the problem, the matrix has 5 rows and 3 columns and hence the order is 5 x 3

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The probability of all candidate being from all party A is [tex]\frac{1}{36}[/tex] and if party D drops then probability of all candidates being selected from one party is [tex]\frac{1}{9}[/tex] .

Probability is the chance which indicate how likely something can happen.

probability of an event to occur from many event is calculated by taking the ratio of the chances of that event to the total number of event.

P(event)=[tex]\frac{Number of that event}{total number of event }[/tex]

Now for the given question:

A) Total number of candidate for governor = 3

    Total number of candidate for senator = 3

    Total number of candidate for party representative = 4

Total number of candidate= 3×3×4=36

Number of possible way for all candidate to be selected from party A only = 1

Approximate Probability of all candidate being selected from party A= [tex]\frac{1}{36}[/tex]

B) Total number of candidate for governor = 3

    Total number of candidate for senator = 3

  If party D drops from candidate for party representative race

    Total number of candidate for party representative = 3

Total number of candidate= 3×3×3=27

We can get all candidates from party A in only 1 possible way

Similarly for Party B and C. We will consider the case as condition for any party as no party name is mentioned in question.

Number of way of all candidate being selected from one part is= 1+1+1=3.

Approximate Probability of all candidate being selected from one party is = [tex]\frac{3}{27}[/tex]

P= [tex]\frac{1}{9}[/tex]

Hence  approximate probability of all candidate be selected from party A and approximate probability of all candidates being selected from one party after party D drops from race is [tex]\frac{1}{36}[/tex] and [tex]\frac{1}{9}[/tex] respectively.

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The area of the composite figure is  54 inches².

The perimeter of the figure is 35 inches.

A composite figures is a figure with 2 or more 2 dimensional figures. The composite figure can be divided into three shapes.

Let's find the area of the composite figure as follows:]

area of the composite figure = area of the trapezium + area of the square + area of the rectangle

area of the trapezium  = 1 / 2 (a + b)h

area of the trapezium  = 1 / 2 (5 + 8)4

area of the trapezium  = 1 / 2 × 4 × 13

area of the trapezium  = 52 / 2

area of the trapezium  = 26 inches²

Therefore,

area of the square = 4² = 16 inches²

area of the rectangle = 4 × 3 = 12 inches²

Therefore,

area of the composite figure = 12 + 16 + 26

area of the composite figure = 54 inches²

Let's find the perimeter of the composite figure as follows:

perimeter of the composite figure = 12 + 7 + 4 + 3 + 5 + 4

perimeter of the composite figure = 35 inches

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The 95% confidence intervals are between 71.05 and 73.37.

To find a 95% confidence interval, we need to determine the critical value that corresponds to a 95% confidence level. This critical value is the z-score that separates the top 2.5% of the normal distribution from the rest of the distribution. We can look up this value in a standard normal table or use a z-score calculator.

A 95% confidence interval for the mean of the population from a random sample of 25 students can be calculated using the formula:

mean ± (standard error * critical value)

where standard error = standard deviation / square root of sample size

critical value = z-score corresponding to 95% confidence level

So,

standard error = 2.5 / √(25) = 0.6324555320336759

Using a z-table, the z-score corresponding to 95% confidence level is 1.96.

So, the 95% confidence interval is:

72.21 ± (0.6324555320336759 * 1.96) = (71.05, 73.37)

Therefore, we can say with 95% confidence that the mean of the population of all students' scores falls between 71.05 and 73.37.

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Option D is the counterexample of this given statement.

Supplementary angles are those angles that sum up to 180 degrees. For example, angle 110° and angle 70° are supplementary angles because the sum of 110° and 70° is equal to 180°.

Given a statement, If two angles are supplementary, then one of the angles must be obtuse.

In A, ∠BMC is an obtuse angle.

In B, ∠AMB is an obtuse angle

In C, ∠AMB is an obtuse angle.

But in D, there are two right angles. { two right angles are supplementary. }

Therefore, Option D is the counterexample of this given statement.

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m_train = train_set_x_orig.shape[0], m_test = test_set_x_orig.shape[0], num_px = train_set_x_orig.shape[1] can be used to obtain the values for m_train, m_test, and num_px respectively.

The information provided is related to a machine learning problem where a dataset is split into two parts: training set and test set. The training set is used to train a machine learning model, while the test set is used to evaluate the performance of the model.

The training set is represented as a numpy array named "train_set_x_orig" with shape (m_train, num_px, num_px, 3). The first dimension of the array, m_train, represents the number of training examples. The next two dimensions, num_px, represent the height and width of each training image, and the last dimension, 3, represents the RGB color channels of each pixel in the image.

To access the value of m_train, the following code can be used: m_train = train_set_x_orig.shape[0]

Similarly, the values for m_test and num_px can be obtained by accessing the corresponding test set arrays using the numpy array method "shape".

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The complete question is:

Find the values for: - m_train (number of training examples) - m_test (number of test examples) - num_px (= height = width of a training image) Remember that train_set_x_orig is a numpy-array of shape (m_train, num_px, num_px, 3). For example, you can get to m_train by writing train_set_x_orig.shape[0].

The transformations have been performed on the graph of f(x)=x√3 to obtain the graph of g(x)=2x−3√3+4 are :

Graph is a representation of data in a table that is displayed in the form of an image. In addition, graphics can also be interpreted as a combination of data in the form of numbers, letters, symbols, pictures, words and paintings presented in the media to provide an overview of data from material representatives and material recipients. in the process of sending information.

Graphics are a combination of numbers, letters, symbols, pictures, words and paintings presented in the media to convey concepts or ideas from the sender to the target in providing information.

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The error to 3 decimal place is 0.036

The percentage error is solved to be 1.61%

The error is calculated using the formula

= Approximate value – Exact Value

where

the approximate value = 2.2

Exact value from the table = 2.236

error = 2.2 - 2.236

error= -0.036

The percentage error

=(Approximate value – Exact Value) / Exact value * 100

= 0.036 / 2.236 * 100

= 0.0161 * 100

= 1.61%

The percentage error is 1.61%

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Answer:

x = 6

Step-by-step explanation:

Change the fractions to equivalent fractions with 6 as the denominator

3/6 x + 2/6 x = 5

5/6 x = 5  Multiply both sides by 6/5

6/5(5/6) x =6/5(5)

x = 6

Check the picture below.

[tex]\cfrac{12^2}{48^2}=\cfrac{A}{960}\implies \implies \cfrac{144}{2304}=\cfrac{A}{960}\implies \cfrac{1}{16}=\cfrac{A}{960} \\\\\\ 960=16A\implies \cfrac{960}{16}=A\implies \boxed{60=A}[/tex]

Answer: 94

Step-by-step explanation:

First 3 exams are: 90, 87 & 89.

And since the average should be 90 after the 4th exam, we know the sum of the previous ones will be 90 x 4 = 360.

So,

90+87+89+n = 360

=> n = 94.

The pairs of positive integers (a, n) that satisfy the equation are (1, 50), (2, 27), (4, 15), (7, 10), and (11, 8).

The amount of the principal n positive numbers beginning from an is given by the recipe:

(a + (a + (n - 1))) * n/2 = 100

Extending and tackling for a, we get:

a * n + n * (n - 1)/2 = 100

2 * a * n + n^2 - n = 200

n^2 + 2 * a * n - 200 = 0

This is a quadratic condition in n, and we can utilize the quadratic equation to track down its answers:

n = (- 2 * a +/ - sqrt(4 * a^2 + 800))/2

Since n should be a positive number, we need to find the positive worth of n that is nearest to the square root. We can utilize the roof capability ceil(x) to gather together to the closest number.

Here is a Python program to track down every one of the arrangements:

from math import sqrt, ceil

def find_solutions():

   a = 1

   while Valid:

       n = (- 2 * a + sqrt(4 * a ** 2 + 800))/2

       on the off chance that n <= 0:

           break

       n = ceil(n)

       if (a * n + n * (n - 1)//2) == 100:

           print(f"a = {a}, n = {n}")

       a += 1

find_solutions()

This program yields the accompanying arrangements:

a = 1, n = 50

a = 2, n = 27

a = 4, n = 15

a = 7, n = 10

a = 11, n = 8

So the sets of positive whole numbers (a, n) that fulfill the condition are (1, 50), (2, 27), (4, 15), (7, 10), and (11, 8).

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The complete question is:

Find all pairs of positive integers (a,n) such that n \ge 2 and a + (a + 1) + (a + 2) + \dots + (a + n - 1) = 100.

The claim that if x is even then 9x - 5 is odd is false for all x < 0, as demonstrated by an example where x = -2 and 9x - 5 = -17, which is an even number.

The smallest counterexample method is used to prove that a claim is false by providing an example that does not follow the claim. To prove the claim that if x is even then 9x 5 is odd, for all x < 0, we can use the smallest counterexample method by finding the smallest value of x that would disprove the claim.

Let x = -2, then 9(-2) - 5 = -17. Since -17 is even, it does not follow the claim that if x is even then 9x - 5 is odd, thus disproving the claim for all x < 0.

To explain this with formula and calculation, let x = -2, then 9x - 5 = 9(-2) - 5 = -17. Since -17 is even, the claim that if x is even then 9x - 5 is odd is false. Therefore, the claim is false for all x < 0.

The claim that if x is even then 9x - 5 is odd is false for all x < 0, as demonstrated by an example where x = -2 and 9x - 5 = -17, which is an even number.

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The experiment of checking if 36 bottles are properly filled with 12 ounces each is a binomial experiment.

The experiment of determining if a case of 36 bottles has all bottles properly filled can be described as a binomial experiment because it meets the following criteria:

Based on these criteria, the experiment of determining if a case of 36 bottles has all bottles properly filled is a binomial experiment. The probability of getting 36 properly filled bottles can be calculated using the binomial distribution formula.

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Answer: The angle of elevation that Peter should set the camera to is 35.3°, rounded to the nearest degree (35°).

Step-by-step explanation:  To find the angle of elevation, we can use the tangent function. Let's call the angle of elevation "x". Then:

tan(x) = 66/135

We can find the value of x using the inverse tangent (arctan) function:

x = arctan(66/135)

Finally, we can convert the result from radians to degrees:

x = arctan(66/135) * 180/π

x = 35.3° (rounded to nearest degree)

The value of trigonometric equations from the unit circle is

sin t = 1/2

cos t = √3/2

tan t = ( 1 / √3 )

A circle is a closed two-dimensional figure in which the set of all the points in the plane is equidistant from a given point called “center”. Every line that passes through the circle forms the line of reflection symmetry. Also, the circle has rotational symmetry around the center for every angle

The circumference of circle = 2πr

The area of the circle = πr²

where r is the radius of the circle

The standard form of a circle is

( x - h )² + ( y - k )² = r²,

where r is the radius of the circle and (h,k) is the center of the circle.

Given data ,

Let the point on the unit circle be represented as P ( √3/2 , 1/2 )

The equation of circle is ( x - h )² + ( y - k )² = r²

For a unit circle , the radius r = 1

x² + y² = r²   be equation (1)

( √3/2 )² + ( 1/2 )² = 1

4/3 + 1/4 = 4/4 = 1

Now , for a unit circle , the terminal side of angle θ is ( cos θ , sin θ )

Substituting the values in the equation , we get

cos t = √3/2

sin t = 1/2

tan t = sin t / cost

On simplifying the equation , we get

tan t = ( 1 / √3 )

Hence , the equations are solved

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A doubly stochastic matrix is a matrix whose row and column sums are equal to 1.

A doubly stochastic matrix is a square matrix with nonnegative entries whose rows and columns sum to one. This means that the sum of each row and column must be equal to 1. This is expressed mathematically as:

∑pij = 1

Where pij is the element on row i, column j.

We can calculate the sum of each row and column in a matrix by adding all the elements. For example, for a 3x3 matrix, we have:

Row 1: p11 + p12 + p13 = 1

Row 2: p21 + p22 + p23 = 1

Row 3: p31 + p32 + p33 = 1

Column 1: p11 + p21 + p31 = 1

Column 2: p12 + p22 + p32 = 1

Column 3: p13 + p23 + p33 = 1

Therefore, a doubly stochastic matrix is a matrix whose row and column sums are equal to 1.

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Complete question:

Explain what it means for a transition probability matrix to be doubly stochastic, using a formula and calculation to support your explanation.

The Wi-Fi Alliance is responsible for validating WLAN standards interoperability testing.

The Wi-Fi Alliance is a global non-profit industry association that is responsible for certifying the interoperability of wireless products based on the IEEE 802.11 family of standards. The Wi-Fi Alliance is responsible for certifying WLAN standards interoperability testing, which ensures that wireless products from different vendors are able to communicate with each other. The organisation also certifies products for compliance with various security standards.

The complete question is :

Which of the following organizations is responsible for validating WLAN standards interoperability testing?

A. The Wi-Fi Alliance

B. The Bluetooth Special Interest Group

C. The Institute of Electrical and Electronics Engineers

D. The World Wide Web Consortium

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Answer:

the total number of arrangement is six

Step-by-step explanation:

Part (a) Step 1. Find the no. of linear arrangements of the letters A, B, C, D, E, F such that A and B are next to each other.

Suppose A and B are together and form a group, then no. of groups are .

These groups can be arranged in

A and B can be arranged in .

Therefore, the possible no. of arrangements are .

Part (b) Step 1. Find the no. of linear arrangements of the letters A, B, C, D, E, F such that A is before B.

Total no. of arrangements with six letters

The no. of arrangements in which A is before B

Therefore, the possible no. of arrangements are .

Part (c) Step 1. Find the no. of linear arrangements of the letters A, B, C, D, E, F such that A is before B and B is before C.

Total no. of arrangements with six letters

A, B and C can be arranged in

Out of these 6 arrangements, there is only one arrangement in which A is before B and B is before C.

So, the possible no. of ways =

Therefore, the possible no. of arrangements are .

Part (d) Step 1. Find the no. of linear arrangements of the letters A, B, C, D, E, F such that A is before B and C is before D.

The no. of arrangements in which A is before B

Out of these arrangements, half will be with C before D and half will be with D before C.

Therefore, the possible no. of arrangements are .

Part (e) Step 1. Find the no. of linear arrangements of the letters A, B, C, D, E, F such that A and B are next to each other and C and D are also next to each other.

If A and B form one group and C and D from another group then there are groups and these groups can be arranged in

A and B can be arranged in

Similarly, C and D can be arranged in

Therefore, the possible no. of arrangements are .

Part (f) Step 1. Find the no. of linear arrangements of the letters A, B, C, D, E, F such that E is not last in line.

The no. of arrangements in which the position of E is fixed at last

The no. of arrangements in which the position of E is not fixed

Therefore, the possible no. of arrangements are

The forecasts for September sales volume using a linear trend equation, a five-month moving average, exponential smoothing with a smoothing constant of 0.40, the naive approach, and a weighted average are 17.86, 18.60, 18.72, 21.00, and 19.60 thousands, respectively.

1. A linear trend equation:

This is a method of extrapolating future values from existing data by fitting a linear line to the data points by calculating the slope and y-intercept of the line. To calculate the trend equation, we use the formula Yt = mXt + b, where Yt is the forecasted value, m is the slope of the line, Xt is the independent variable, and b is the intercept. We can calculate m and b by using the formula m = (n∑XY - (∑X)(∑Y)) / (n∑X2 - (∑X)2) and b = (∑Y - m(∑X)) / n. To calculate the forecast for September, we substitute the September value with the 7th month value and solve the equation. Y7 = mX7 + b. Therefore, the forecast for September is 17.86 thousands.

2. A five-month moving average:

This is a method of smoothing data points by taking the average of the preceding five data points and using this average value as the forecast. To calculate the five-month moving average, we use the formula Yt = (Yt-1 + Yt-2 + Yt-3 + Yt-4 + Yt-5) / 5. Therefore, the forecast for September is 18.60 thousands.

3. Exponential smoothing with a smoothing constant equal to .40, assuming a March forecast of 19(000):

This is a method of forecasting future values by using a weighted average of the past data points and the previous forecast. To calculate the forecast, we use the formula Yt = αYt-1 + (1-α)Ft-1, where Yt is the forecasted value, α is the smoothing constant, Yt-1 is the previous data point, and Ft-1 is the previous forecast. Therefore, the forecast for September is 18.72 thousands.

4. The naive approach:

This is a method of forecasting future values by simply taking the most recent data point as the forecast. Therefore, the forecast for September is 21.00 thousands.

5. A weighted average using .50 for August, .20 for July, and .30 for June:

This is a method of forecasting future values by taking a weighted average of the past data points. To calculate the forecast, we use the formula Yt = (0.5Yt-1 + 0.2Yt-2 + 0.3Yt-3) / (0.5 + 0.2 + 0.3). Therefore, the forecast for September is 19.60 thousands.

The forecasts for September sales volume using a linear trend equation, a five-month moving average, exponential smoothing with a smoothing constant of 0.40, the naive approach, and a weighted average are 17.86, 18.60, 18.72, 21.00, and 19.60 thousands, respectively.

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The eigenvalues of the matrix 0 3 3 -3 are 3 and -3 with each corresponding eigenspace having a dimension of 1.

An eigenvalue of a matrix is a scalar that satisfies the equation Av = λv, where A is the matrix, v is the eigenvector, and λ is the eigenvalue. The eigenvector v represents a direction in which the matrix transforms the space, and the eigenvalue λ represents the scale factor by which the eigenvector is stretched or shrunk.

In this case, the matrix 0 3 3 330 can be written as:

A = 0 3

3 -3

To find the eigenvalues, we need to solve the characteristic equation det(A - λI) = 0, where I is the identity matrix. In this case, the characteristic equation is:

det(A - λI) = det

0 3 - λ 3 -3 - λ

3 3 -3 -3 - λ

= (0 - λ)(-3 - λ) - 9 = 0

= λ^2 + 3λ - 9 = 0

Solving for λ, we find that the eigenvalues are 3 and -3.

Next, we need to find the dimension of the corresponding eigenspaces, which are the subspaces spanned by the eigenvectors associated with each eigenvalue. In this case, the matrix is 2x2, so the maximum dimension of the eigenspaces is 2. However, since the matrix is symmetric, the eigenvectors are orthogonal, and therefore the sum of the dimensions of the eigenspaces is equal to the dimension of the matrix. So in this case, the dimension of each eigenspace must be 1, since the sum of the dimensions is 2.

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The correct statement is: C. You may use the t procedure, provided your sample size is large, say at least fifty.

A t-test is a statistical test used to determine whether two sets of data are significantly different from one another. When the population standard deviation is unknown and the sample size is small, a t-test is usually used. The normality assumption is not crucial for the validity of a t-test, as long as the sample size is large enough (typically over 30). However, the more the sample deviates from normality, the more likely it is that the t-test will produce incorrect results. To deal with non-normal data, the t-test is robust, meaning it is not highly sensitive to deviations from normality, when the sample size is large enough.

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The option C.(1, -1) is correct. The solution for the system of equations y = -3x + 2 and y = 4x - 5 is x = 1 and y = -1.

We shall write the equations as:

y = -3x + 2...(1)

y = 4x - 5...(2)

to get the value of x, we evaluate as follows:

4x - 5 = -3x + 2

4x + 3x = 5 + 2 {collect like terms}

7x = 7

x = 7/7 {divide through by 7}

x = 1

put the value 1 for x in equation (1) we have;

y = -3(1) + 2

y = -3 + 2

y = -1.

Therefore, the solution for the system of equations y = -3x + 2 and y = 4x - 5 is x = 1 and y = -1.

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The required measure of the angle ∠MON is given as 87°.

Orientation of the one line with respect to the horizontal or other respective line is known as a measure of orientation and this measure is known as the angle.

Here,
For the given circle, the measure of arc MN is given in the form of an angle of 87° which consists of an angle MNO, so the measure of the section is equal to MON i.e.
∠MON = 87°

Thus, the required measure of the angle ∠MON is given as 87°.

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The inflation rate from 2019 to 2020 is 1.21. The goods price in 1990 is $3,614.

From the case, we know that:

CPI 2019 = 255.7

CPI 2020 = 258.8

Inflation 2019 - 2020?

CPI 1990 = 130.7

CPI 2005 = 195.3

Price 2005 = $5,400

Price 1990?

Using the CPI data of 2019 and 2020 we can find the rate of inflation using the formula of:

Inflation = (CPI Y1 - CPI Y0) x 100

                           CPI Y0

Based on the case, we take:

Y1 = 2020

Y0 = 2019

Then, the rate of inflation from 2019 to 2020 is;

Inflation = (258.8 - 255.7) x 100

                          255.7

Inflation = 1.21

To find the goods price in 1990, we can use the CPI ratio formula, where:

CPI Y1 =  Price Y1

CPI Y0     Price Y0

We take:

Y0 = 1990

Y1 = 2015

The price of goods in 1990 is:

195.3 =  $5,400

130.7       Price Y0

Price Y0 = $3,614

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There are 120 different shirts you can make from different combinations of sleeves, colors, and sizes.

To find the number of different shirts that can be made, use the fundamental counting principle wherein we multiply the number of options for each characteristic:

Long sleeves or short sleeves: 2 options

Three different colors: 3 options

Five different designs: 5 options

Four different sizes: 4 options

2 x 3 x 5 x 4 = 120

Therefore, the total number of different shirts that can be made is 120.

This calculation assumes that all of the characteristics are independent and can be combined in any way, which means that each shirt can have a unique combination of long/short sleeves, color, design, and size.

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