FILL IN THE BLANK ______ can creep into a study if subjects are picked (intentionally or not) according to some criterion and not randomly.

If we total up the dots plot for 3, 4, and 5 pets, we find that 3 people have 2 pets at home, 10 individuals have at least 3 pets at home.

The fact that dogs are the most common pet in the world shouldn't be shocking. There is a reason why there are tens of millions of dogs living in the United States alone, which is why some people say that dogs are a man's greatest friend. Around the world, at least one dog is kept in one-third of all households.

A fully domesticated animal kept constitutes a "household pet." a pet kept by you for personal company, like a dog, cat, reptile, bird, or mouse. Any kind of horse, cow, pig, sheep, goat, chicken, turkey, other captive fur-bearing animal is not considered a household pet, nor is any animal that is typically kept for food or profit.

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The total number of people with pets at home is 11, which is the sum of the heights of the columns.

What is equation?

A math equation is a method that links two claims and represents equivalence using the equals sign (=). An equation is a mathematical statement that establishes the equivalence of two mathematical expressions in algebra.

Based on the given dot plot, we can answer the following questions:

How many people have 2 pets at home?

Answer: Two people have 2 pets at home, as indicated by the two dots in the second column.

How many people have at least 3 pets at home?

Answer: Six people have at least 3 pets at home, as indicated by the dots in the third column and beyond.

How many more people have 2 pets than 5 pets?

Answer: There are no dots in the last column, which represents 5 pets. Therefore, the difference between the number of people with 2 pets and those with 5 pets is 2 - 0 = 2.

How many people have less than 3 pets at home?

Answer: Three people have less than 3 pets at home, as indicated by the dots in the first two columns.

Therefore, the total number of people with pets at home is 11, which is the sum of the heights of the columns.

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The returned value would be 70 which is the missing value in the data set. Hence, option D is correct. We have some X values; we called these numeric inputs and some Y value that we are trying to predict.

This set of data would yield a result of 70 if a median imputer were run on it. In regression, we have some X values that are referred to as independent variables and some Y values that are referred to as dependent variables (this is the variable we are trying to predict). Several Y values are possible, but they are uncommon.

Learning a function that can predict Y given X is the fundamental concept behind a regression. Depending on the data, the function may be linear or non-linear.

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Complete question is:

Imagine X in the below is a missing value. If I were to run a median imputer on this set of data. What would the returned value be? 50 , 60 , 70 , 80 , 100 , 60 , 5000 , x (It's okay to have to look up how to do this!)

50

An error

80

70

100

The basic idea of a regression is very simple. We have some X values, we called these ______ and some Y value (this is the variable we are trying to _______.

We could have multiple Y values, but that is not but that is not re-ordered ordinals intercepts features numeric inputs.

Answer:

The area of a semicircle with diameter 28 cm is 98π cm², or 307.88 cm² to the nearest tenth.

Step-by-step explanation:

A semicircle is a two-dimensional shape that is exactly half of a circle.

The area of a circle is given by the formula:

[tex]\sf A=\pi r^2[/tex]

where A is the area of the circle, and r is the radius of the circle.

The diameter of a circle is twice its radius.

Given the diameter of the semicircle is 28 cm, the radius is:

[tex]\sf r = \dfrac{28}{2} = 14 \; cm[/tex]

Substituting this into the formula for the area of a circle, we get:

[tex]\sf A = \pi(14)^2[/tex]

[tex]\sf A = 196 \pi[/tex]

Finally, divide this by two to get the area of the semicircle:

[tex]\sf Area\;of\;semicircle = \dfrac{1}{2} \cdot 196 \pi[/tex]

[tex]\sf Area\;of\;semicircle = 98 \pi\; cm^2[/tex]

So the area of a semicircle with diameter 28 cm is 98π cm², or 307.88 cm² to the nearest tenth.

The Taylor's series for [tex]1 + 3e^x + x^2[/tex] at [tex]x=0[/tex] is :

[tex]1 + 3e^x+ x^2 = 5 + 3x + (3/2)x^2 + (1/3)x^3 + ...[/tex]

What do you mean by Taylor's series ?

The Taylor's series is a way to represent a function as a power series, which is a sum of terms involving the variable raised to increasing powers. The series is centered around a specific point, called the center of the series. The Taylor's series approximates the function within a certain interval around the center point.

The general formula for the Taylor's series of a function f(x) centered at [tex]x = a[/tex] is:

[tex]f(x) = f(a) + f'(a)(x-a) + f''(a)(x-a)^2/2! + f'''(a)(x-a)^3/3! + ...[/tex]

where [tex]f'(a), f''(a), f'''(a),[/tex] etc. are the derivatives of f(x) evaluated at [tex]x = a[/tex].

Finding the Taylor's series for [tex]1 + 3e^x + x^2[/tex] at [tex]x=0[/tex] :

We need to find the derivatives of the function at [tex]x=0[/tex]. We have:

[tex]f(x) = 1 + 3e^x + x^2[/tex]

[tex]f(0) = 1 + 3e^0 + 0^2 = 4[/tex]

[tex]f'(x) = 3e^x+ 2x[/tex]

[tex]f'(0) = 3e^0 + 2(0) = 3[/tex]

[tex]f''(x) = 3e^x + 2[/tex]

[tex]f''(0) = 3e^0 + 2 = 5[/tex]

[tex]f'''(x) = 3e^x[/tex]

[tex]f'''(0) = 3e^0 = 3[/tex]

Substituting these values into the general formula for the Taylor's series, we get:

[tex]f(x) = f(0) + f'(0)x + f''(0)x^2/2! + f'''(0)x^3/3! + ...[/tex]

[tex]f(x) = 4 + 3x + 5x^2/2 + 3x^3/6 + ...[/tex]

Simplifying, we get:

[tex]f(x) = 5 + 3x + (3/2)x^2 + (1/3)x^3 + ...[/tex]

Therefore, the Taylor's series for [tex]1 + 3e^x + x^2[/tex] at [tex]x=0[/tex] is :

[tex]1 + 3e^x+ x^2 = 5 + 3x + (3/2)x^2 + (1/3)x^3 + ...[/tex]

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Answer:

Step-by-step explanation:

First the max degree is 360

Then multiply by 5/8

360 x 5/8 = 1800/8

1800/8 = 225

Answer: 225

The cardinality of set from the given vein diagram is found as 2.

Think about set A. The set A is said to be finite and so its cardinality is same to the amount of elements n if it includes precisely n items, where n  ≥  0. |A| stands for the cardinality of such a set A.

It turns out that there are two kinds of infinite sets that we need to determine between since one form is much "bigger" than the other. Particularly, one type is referred to as countable and the other as uncountable.

From the given figure

Set A = {8 , 8, 3, 6}

Compliment of Set B (elements not present in set B):

Set [tex]B^{c}[/tex] = {8, 8, 6(pink), 3(white)}

Thus,

(A∩ [tex]B^{c}[/tex] ) = {8, 8} (present in both set)

n (A∩ [tex]B^{c}[/tex] ) = 2 (cardinal number)

Thus, the cardinality of the set from the given vein diagram is found as 2.

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Therefore, the scale factor of the dilation of the shape is 2.

A scale factor is a ratio that describes the proportional relationship between two similar figures. It represents how much larger or smaller one figure is compared to the other, and it is calculated by dividing a corresponding measurement (such as side length, perimeter, or area) of the larger figure by the corresponding measurement of the smaller figure. Scale factor is used in mathematics, particularly in geometry and measurement, to describe the transformation of one figure into another through dilation or resizing. It is represented by a number or a ratio, such as 2:1 or 1/2, which indicates how many times larger or smaller the new figure is compared to the original.

Here,

In the given picture, it appears that the distance from the center of dilation (the origin) to the pre-image (the original figure) is 4 units, and the distance from the center of dilation to the image (the transformed figure) is 8 units. The scale factor of the dilation is equal to the ratio of the distance from the center of dilation to the image and the distance from the center of dilation to the pre-image.

So, the scale factor of the dilation is:

8 units ÷ 4 units = 2

Therefore, the scale factor of the dilation is 2.

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Answer:

23.8 cm

Step-by-step explanation:

17 * 140% = 17 * 1.4 = 23.8 cm

The image of the house would be 23.8 cm tall on the screen.

To calculate the height of the image of the house on the screen, we can use the given zoom factor of 140%.

The zoom factor of 140% means that the images on the screen are 140% as long and 140% as wide compared to when they are printed on a sheet of paper.

To calculate the height of the image on the screen, we need to multiply the printed height by the zoom factor (140% or 1.4).

Height on the screen = Printed height * Zoom factor

Height on the screen = 17 cm * 1.4

Height on the screen = 23.8 cm

Therefore, the image of the house would be 23.8 cm tall on the screen.

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The rental cost for each movie and each video game is $3.5 and $5.5 respectively.

Let

cost of each movie = x

Cost of each video game = y

5x + 3y = 34

2x + 12y = 73

Multiply (1) by 4

20x + 12y = 136

2x + 12y = 73

subtract the equations to eliminate y

18x = 63

divide both sides by 18

x = 63/18

x = 3.5

Substitute x = 3.5 into (1)

5x + 3y = 34

5(3.5) + 3y = 34

17.5 + 3y = 34

3y = 34 - 17.5

3y = 16.5

y = 16.5/3

y = 5.5

Therefore, $3.5 and $5.5 is the rental cost of each movie and video game respectively.

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Let X and Y be independent random variables, uniformly distributed in the interval [0, 1 ]. The CDF of X - Y is FZ(z) = (1/2)(1+z)^2 for -1 ≤ z ≤ 0, 1 - (1/2)(1-z)^2 for 0 ≤ z ≤ 1, 0 for z < -1 or z > 1. The PDF of X - Y is fZ(z) = z + 1 for -1 < z < 0, 1 - z for 0 < z < 1, 0 otherwise.

To find the CDF of X - Y, we first note that the range of X - Y is [0, 1]. Let Z = X - Y, then:

FZ(z) = P(Z ≤ z) = P(X - Y ≤ z)

We can write this as an integral over the joint distribution of X and Y:

FZ(z) = ∫∫[X - Y ≤ z] fXY(x, y) dx dy

Since X and Y are independent, the joint distribution is simply the product of their marginal distributions:

fXY(x, y) = fX(x) fY(y) = 1 * 1 = 1

for 0 ≤ x, y ≤ 1.

Thus, we have:

FZ(z) = ∫∫[X - Y ≤ z] dx dy

= ∫∫[Y ≤ X - z] dx dy

= ∫0^1 ∫0^(x-z) 1 dy dx + ∫0^1 ∫(x-z)^1 1 dy dx

= ∫0^(1+z) (1-z) dx

= (1/2)(1+z)^2 for -1 ≤ z ≤ 0

= 1 - (1/2)(1-z)^2 for 0 ≤ z ≤ 1

Therefore, the CDF of X - Y is:

FZ(z) =

(1/2)(1+z)^2 for -1 ≤ z ≤ 0

1 - (1/2)(1-z)^2 for 0 ≤ z ≤ 1

0 for z < -1 or z > 1

To find the PDF of X - Y, we differentiate the CDF:

fZ(z) = dFZ(z)/dz =

z + 1 for -1 < z < 0

1 - z for 0 < z < 1

0 otherwise

Therefore, the PDF of X - Y is:

fZ(z) =

z + 1 for -1 < z < 0

1 - z for 0 < z < 1

0 otherwise

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Answer:

Step-by-step explanation:

Sure! Here's a number line showing the distance of 0.75 kilometers:

0 -------------|-------------|------------- 0.75 km

The "0" on the left represents the starting point (such as home), and the "|---|" in the middle represents the distance of 0.75 kilometers to the destination (such as school).

Answer:

N and E is 90 degrees

N and S is 180 degrees

N and W is 90 degrees

The discount rate that should be applied to a new project's cash flows is the Weighted Average Cost of Capital (WACC). To calculate WACC, you need to first calculate the cost of debt. This is done by taking the face value of the bonds ($1000) multiplied by the coupon rate (8%) multiplied by (1 - the tax rate (26%)), which equals 5.92%. The cost of debt is then calculated by taking the market value of the debt (6,380 x $1,000 x 99.1%) and dividing this by the total market value of the debt plus the market value of the equity (6,380 x $1,000 x 99.1% + 76,650 x $50 + 14,250 x $80), which equals 5.22%.

Next, you need to calculate the cost of equity using the Capital Asset Pricing Model (CAPM). This is done by taking the risk-free rate (3.21%) plus the market risk premium (9.79%) multiplied by the firm's beta (1.47), which equals 17.18%.

The WACC is then calculated by taking the cost of equity multiplied by the proportion of equity (76,650 x $50 + 14,250 x $80 divided by the total market value of the debt plus the market value of the equity) plus the cost of debt multiplied by the proportion of debt (6,380 x $1,000

Answer:

The unit circle has many different angles that each have a corresponding point on the circle. The coordinates of each point give us a way to find the tangent of each angle. The tangent of an angle is equal to the y-coordinate divided by the x-coordinate.

Answer:

56

90

56

Step-by-step explanation:

easy easy lol.

aaaaa

The inequality is X > 0 and a word sentence represent the graph is  X  the graph of a number line with an open circle on zero and an arrow pointing to the right.

The inequality X > 0 represents the graph of a number line with an open circle on zero to left and an arrow pointing to the right. This means that any value of X that is greater than zero is a valid solution for the inequality.

In other words, X can be any positive number, such as 1, 2, 3, and so on. However, X cannot be zero or any negative number, as those values do not satisfy the inequality. Therefore, the word sentence that represents this inequality is "X is greater than zero."

This means that X must be a positive number, and it can be any value that is greater than zero.

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A vector x orthogonal to the row space of A, and a vector y orthogonal to column space, and a vector z orthogonal to the null space. The orthogonal vector is :

A = [tex]\left[\begin{array}{ccc}1&2&1\\2&1&0\\1&-2&2\end{array}\right][/tex]

The orthogonal complement of the subspace V contains any vector perpendicular to V. This orthogonal subspace is denoted V⊥. (pronounced "V perp").

By this definition, null space is the orthogonal complement of row space. Every x perpendicular to the line satisfies Ax = 0 and lies in null space.

vice versa. If v is orthogonal to null space, it must be in row space. Otherwise, we can add this v as an extra row of the matrix without changing its null space. The rice space will become larger, breaking the rule of r+(n−r) = n.

The column space extent of A. These two vectors are the basis of col(A) , but they are not normalized.

In this case, the columns of A are already orthonormal, so you don't need to use the Gram-Schmidt procedure. To normalize a vector and then divide it by its norm:

                   [tex]\left[\begin{array}{ccc}1&2&1\\2&4&3\\3&6&4\end{array}\right][/tex]

and the vector after orthogonal process is:

                  [tex]\left[\begin{array}{ccc}1&2&1\\2&1&0\\1&-2&2\end{array}\right][/tex]

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there is an 85% chance that the two cards drawn at random from an ordinary deck of 52 cards will not be sixes.

The probability of drawing a card from an ordinary deck without replacement can be determined using the concept of conditional probability. Conditional probability is the probability of an event occurring, assuming that another event has already occurred.

In order to calculate the probability that the two cards drawn are not sixes, we can use the formula:

P(A and B) = P(A) x P(B|A)

Where A and B represent two independent events, P(A) is the probability of event A occurring, and P(B|A) is the conditional probability of event B occurring given that event A has already occurred.

The probability of drawing the first card that is not a six is:

P(A) = 48/52 = 0.9231

The probability of drawing the second card that is not a six, given that the first card drawn was not a six, is:

P(B|A) = 47/51 = 0.9216

Therefore, the probability of drawing two cards at random from an ordinary deck of 52 cards and having neither of them be a six is:

P(A and B) = P(A) x P(B|A) = 0.9231 x 0.9216 = 0.8503 or approximately 85%.

This means that there is an 85% chance that the two cards drawn at random from an ordinary deck of 52 cards will not be sixes.

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The area of the enlarged logo will be 9 times larger than the original logo.

When an object is enlarged or scaled up how does it area change ?

When an object is enlarged or scaled up by a factor of [tex]k[/tex], both its length and width are multiplied by [tex]k[/tex]. Therefore, the new length is [tex]k[/tex] times the original length, and the new width is [tex]k[/tex] times the original width.

The area of the new object is the product of the new length and width, which is ([tex]k[/tex] times the original length) multiplied by ([tex]k[/tex] times the original width), or [tex]k^2[/tex] times the original area.

Therefore, the area of an object increases by a factor of [tex]k^2[/tex] when the object is enlarged or scaled up by a factor of [tex]k[/tex].

Calculating how many times larger the area of the enlarged logo will be :

The original logo has dimensions of 3 inches by 5 inches, so its area is 3 x 5 = 15 square inches.

Mr. James wants to enlarge the logo so that the dimensions are 3 times larger than the original. This means the new dimensions will be 9 inches by 15 inches.

To determine how many times larger the area of the enlarged logo will be, we need to compare the areas of the original logo and the enlarged logo. The area of the enlarged logo is 9 x 15 = 135 square inches.

To find out how many times larger the area of the enlarged logo is compared to the original logo, we divide the area of the enlarged logo by the area of the original logo:

135 square inches ÷ 15 square inches = 9

Therefore, the area of the enlarged logo will be 9 times larger than the area of the original logo.

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(a) The country will first experience shortages of food in approximately 26.6 years

(b) If the country doubled its initial food supply and maintained a constant rate of increase in the supply, shortages would still occur in approximately 38 years.

(c) If the country doubled the rate at which its food supply increases, in addition to doubling its initial food supply, shortages would still occur in approximately 55.4 years.

a. Let P(t) be the population of the country at time t (in years), and F(t) be the food supply of the country at time t.

We know that P(0) = 4 million, and P'(t) = 0.05P(t), which means that the population is increasing by 5% per year.

We also know that F(0) = 8 million, and F'(t) = 0.25 million, which means that the food supply is increasing by 0.25 million people per year.

When the food supply is just enough to feed the population, we have P(t) = F(t), so we can solve for t as follows:

4 million x (1 + 0.05)^t = 8 million + 0.25 million x t

[tex]4(1 + 0.05)^t = 8 + 0.25t\\\\t \approx 26.6 \ years[/tex]

b. If the country doubled its initial food supply, then F(0) = 16 million. We can use the same equation as before and solve for t:

4 million x  (1 + 0.05)^t = 16 million + 0.25 million x t

[tex]4(1 + 0.05)^t = 16 + 0.25t\\\\t \approx 38 \ years[/tex]

c. If the country doubled the rate at which its food supply increases and doubled its initial food supply, then we have F(0) = 16 million and F'(t) = 0.5 million. Using the same equation as before, we get:

4 million x  (1 + 0.05)^t = 32 million + 0.5 million x t

[tex]4(1 + 0.05)^t = 32 + 0.5t\\\\t \approx 55.4 \ years[/tex]

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A parabola with a vertex at (1,3) and an opening downhill is depicted by the equation.

A parabola is an equation of a curve with a spot on it that is equally spaced from a fixed point and a fixed line.

In mathematics, a parabola is a roughly U-shaped, mirror-symmetrical plane circle. The same curves can be defined by a number of apparently unrelated mathematical descriptions, which all correspond to it. A point and a line can be used to depict a parabola.

Equation given: 3x² - 6x + 4y - 9 = 0. When the given equation's graph is plotted, it is discovered that the parabola that is created is opened downward and has a vertex at the spot. ( 1,3). The graph and the following response are attached.

The equation that depicts a parabola with a vertex at (1,3) opening downward is option C, making it the right choice.

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Answer:

Parabola with a vertex at (1, 3) opening downward.

Step-by-step explanation:

Answer: The poster should have width 17.50 CM

Step-by-step explanation:

Given that boasted I have a total area 245 cm square area of a poster is 200 and 45 20 m square. And it is in the rectangle format. So we know that the poster is always in the rectangle format and the area of rectangular area is equal to L N T W. So 245 will be equal to Ln tW. From this. We need to find L. So L is equal to 245 divided by W. Now consider the diplomatic representation of the poster. So it has mentioned that there is a margin around the edges of six cm at the top. This is 6cm and four cm at the sites. All the four sides 3 sides are four cm. So from this we need to find land and the doctor posted area that is printed area. The first wine printed with www. Z. Quilter. Now let this be total birth will be W. And this will be four and this will be four. Therefore posted with will be we need to find this part alone. So W -4 -4 will give the this part with. So W -4 -4. So post printed with PW will be equal to W -8. Similarly printed lunch will be equal to The total length is already we have found 245 by W. And we need to find this part length. So we have to subtract six and four from the total length so that that will give them this part length, So -6 -4. So printed length will be equal there 245 Divided by W -10. And we know that formula for area of a rectangle. Urz D is equal to L W. Now substitute the printed with and printed length in the area formula. We have to find the printed area. So Printed area Zeke Walter W -8 into 245 Divided by W -10. To simplify this, we get 325 minus 10. W -100 and 30,960 W. to the power of -1. Now fine D A by D. T. Not D D. This is D. W. So this is equal to differentiation of constant alma zero, this is minus 10 plus 900 and 60 W. to the power of -2 and equate this to be equal to zero. We out to find the maximum width so D A by D W is equal to zero, therefore minus 10 1960 W. To the power of -2 will be equal to zero. To simplify this, we get them maximum with value the W is equal term I wrote off 196. Therefore the value of W. Z quilter plus or minus 14. We will neglect the negative values since we cannot be negative. So one we assume the positive values. So what will be equal to 14 and length will be equal to 245 divided by 14, So which will be equal to 17.50 centimeters. And they have concluded that At W is equal to 14 cm and lunch will be equal to 17.50 cm needed to print the largest area. I hope you found the answer to school. Thank you.

Answer:

Step-by-step explanation:

D

(a) The interval (-∞, ∞).

(b) The interval (-∞, ∞).

(c) The interval (-∞, ∞).

(d) The interval (-π/2, π/2) \ {0}.

(a) The longest interval in which the given initial value problem is certain to have a unique twice-differentiable solution is the interval where the coefficient function, 3t, is continuous and bounded. Since 3t is a continuous and bounded function for all t in the interval (-∞, ∞), the given initial value problem is certain to have a unique twice-differentiable solution for all t in (-∞, ∞).

(b) The longest interval in which the given initial value problem is certain to have a unique twice-differentiable solution is the interval where the coefficient functions, t(t - 4), 3t, and 4, are continuous and bounded. Since t(t - 4), 3t, and 4 are continuous and bounded functions for all t in the interval (-∞, ∞), the given initial value problem is certain to have a unique twice-differentiable solution for all t in (-∞, ∞).

(c) The longest interval in which the given initial value problem is certain to have a unique twice-differentiable solution is the interval where the coefficient functions, cost and In|t|, are continuous and bounded. Since cost and In|t| are continuous and bounded functions for all t in the interval (-∞, ∞), the given initial value problem is certain to have a unique twice-differentiable solution for all t in (-∞, ∞).

(d) The longest interval in which the given initial value problem is certain to have a unique twice-differentiable solution is the interval where the coefficient functions, x - 2, 1, and (x - 2)tanx, are continuous and bounded. Since x - 2, 1, and (x - 2)tanx are continuous and bounded functions for all x in the interval (-π/2, π/2) \ {0} , the given initial value problem is certain to have a unique twice-differentiable solution for all x in (-π/2, π/2) \ {0}.

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The given question is incomplete, the complete question is:

determine the longest interval in which the given initial value problem is certain to have a unique twice- differentiable solution. Do not attempt to find the solution. (a) ty" + 3y = 1, y(1) = 1, y'(1) = 2   (b)   t(t – 4)y" + 3ty' + 4y = 2, y(3) = 0, y'(3) = -1 (c)   y" + (cost)y' + 3( In |t|) y = 0, y(2) = 3, y'(2) = 1 (d) (x - 2)y"+y' +(x - 2)(tan x) y = 0, y(3) = 1, y'(3) = 2

Answer:

1 - [tex]\frac{2\sqrt{2} }{3}[/tex]

Step-by-step explanation:

[tex]\frac{\sqrt{3}-\sqrt{6} }{\sqrt{3}+\sqrt{6} }[/tex]

rationalise the denominator by multiplying the numerator and denominator by the conjugate of the denominator.

the conjugate of [tex]\sqrt{3}[/tex] + [tex]\sqrt{6}[/tex] is [tex]\sqrt{3}[/tex] - [tex]\sqrt{6}[/tex]

= [tex]\frac{(\sqrt{3}-\sqrt{6})(\sqrt{3}-\sqrt{6}) }{(\sqrt{3}+\sqrt{6})(\sqrt{3}-\sqrt{6}) }[/tex] ← expand numerator/ denominator using FOIL

= [tex]\frac{3-\sqrt{18}-\sqrt{18}+6 }{3-\sqrt{18}+\sqrt{18}+6 }[/tex]

= [tex]\frac{9-2\sqrt{18} }{3+6}[/tex]

= [tex]\frac{9-2(3\sqrt{2}) }{9}[/tex]

= [tex]\frac{9-6\sqrt{2} }{9}[/tex]

= [tex]\frac{9}{9}[/tex] - [tex]\frac{6\sqrt{2} }{9}[/tex]

= 1 - [tex]\frac{2\sqrt{2} }{3}[/tex]

Answer: $27.98

Step-by-step explanation:

22.00 × .2= 4.40

22 + 4.40 = 26.40

26.40 × .06 = 1.584

26.40 + 1.584 = 27.984

Round to the nearest hundred so the total paid by the customer would be 27.98

Answer:

Step-by-step explanation:

We'Re looking at a normal distribution here- let's start by drawing it out to the mean me- is 315 grams standard. Deviation is 16 point. We want to know the weight corresponding to each of these events and we can use either the appendix which i assume is a z, school table or excel so the first 1. We want the highest 20 percent up here somewhere. This is what we would call the 80 percent. It separates the bottom 80 percent from the highest 20 percent. So how do we work the well? We need to start by getting the z score for it. So how do we get that, while in excel you're going to use the norm inverse function which looks like this? So it's calls norm and then in here you just put in x, where x is your percentage and that will spit out the z score and i'm using this rather than a table, because it will give me a more exact value. So we're going to do that, and so here the percent is the 80, so it's not .8 and that spits out the z score of nort .8416 to 4 decimal places. But i'm always going to this exact score, because we now have to turn it from a z score to a piece of real data. Z score is a measure of how many standard deviations away from the mean a value is so we're. Looking for the value not .84 standard deviations, above the mean or if we write it like this x, is equal to z, sigma plus mu. So here we take our exact zedscore, because we can still just use excels, multiply it by 16 and add it on to the mean and we'll get our value of 328.447328 .47 grams to 2 decimal places for part b. We want to be middle 60 percent. Now we need to cut off points and within here where, in this interval we have 60 percent of values, which means we have 40 percent of values, not in here. So we can label our 3 sections. These 2 add up to 40 percent, so they have to be 20 percent. Each are called nor .2 and not .2, and then this middle bit here is nor .6 for the total of 100 percent point. So we need these cut off points when you, the z, scores first, and because these are equal distances away from a mine they're going to have the same z score. Just 1 is going to be positive or negative, so z is going to be equal to plus and minus. Let'S look at the lower 1. This is the 20 percent here so into this excel command. We put 20 percent nor .2 and out of it we get the z score of minus, not .8416. So it's actually very similar to the top question, because the top question asked you for the 80 percent be 80 percent. Is the upper cut off point here? 20? Is below cutoff point, so we already have the up 1, let's just calculate below 1, so we've got to be minus, nor .8416 multiplied by 16, but the standard deviation and on to the mean- and we get 301.53 and the upper cut off point is from part A so that's the middle 60 percent makes more space a part c. We want to be highest 80 percent. So now what we want is the cut off between the lowest 20 and the highest 80, which we've just got from part c part b. It'S this lower 1, here, 301.53 grams. That'S an easy! 1! Now we want the lowest 15 percent, so the lowest 15 percent is the 15 percent. So we go to our exylgamant put in the 15 for percent, so that would be no .15 and it's a z score of minus 1.036 keeping the exact value put into this formula. We multiply our z by 16, as on 315, to get 298.42 grams.

The weight that corresponds to this event are approximately 344.03 grams and 375.97 grams.

To find the weight that corresponds to each event, we need to use the standard normal distribution, which has a mean of 0 and a standard deviation of 1. We can convert the given mean and standard deviation to z-scores using the formula:

z = (x - μ) / σ

where x is the weight we want to find, μ is the mean (360 grams), and σ is the standard deviation (9 grams).

Then, we can use a standard normal distribution table or calculator to find the probability of each event, and convert it back to a weight using the inverse of the z-score formula:

x = μ + z * σ

where z is the z-score that corresponds to the desired probability.

Event 1: The weight is less than 345 grams.

z = (345 - 360) / 9 = -1.67

Using a standard normal distribution table or calculator, we find that the probability of a z-score less than -1.67 is approximately 0.0475.

x = 360 + (-1.67) * 9 = 344.03 grams

Therefore, the weight that corresponds to this event is approximately 344.03 grams.

Event 2: The weight is between 355 and 365 grams.

First, we need to find the z-scores that correspond to the two boundaries:

z1 = (355 - 360) / 9 = -0.56

z2 = (365 - 360) / 9 = 0.56

Using a standard normal distribution table or calculator, we find that the probability of a z-score less than -0.56 is approximately 0.2123, and the probability of a z-score less than 0.56 is approximately 0.7123. Therefore, the probability of a z-score between -0.56 and 0.56 is:

0.7123 - 0.2123 = 0.5

x1 = 360 + (-0.56) * 9 = 355.16 grams

x2 = 360 + (0.56) * 9 = 364.84 grams

Therefore, the weight that corresponds to this event is any weight between 355.16 and 364.84 grams.

Event 3: The weight is greater than 375 grams.

z = (375 - 360) / 9 = 1.67

Using a standard normal distribution table or calculator, we find that the probability of a z-score greater than 1.67 is approximately 0.0475.

x = 360 + (1.67) * 9 = 375.97 grams

Therefore, the weight that corresponds to this event is approximately 375.97 grams.

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