A national grocery chain is considering expanding their selection of prepared meals available for purchase. They believe that nationwide, 67 percent of households purchase at least one prepared meal per week from the grocery store. The results of a survey given to a random sample of Maryland households found that 641 out of 1,035 households purchase at least one meal per week at the store

Answer:

Step-by-step explanation:

Jen can fill 4 bags completely with the 5 1/2 cups of M&Ms she has, given that each bag requires 1 1/4 cups of M&Ms.

First, we need to find the total number of cups of M&Ms Jen has

5 1/2 cups = 11/2 cups

Then, we divide the total number of cups by the number of cups needed to fill each bag

(11/2 cups) ÷ (1 1/4 cups/bag)

To divide by a fraction, we can multiply by its reciprocal

(11/2 cups) x (4/5 cups/bag)

= 44/10 cups

Simplifying, we get

= 4 2/10 cups

= 4 1/5 cups

So, Jen can fill 4 bags completely.

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The profit per pig is $4 and the profit per goose is $8.

Let's start by defining the variables:

a = number of pigs Farmer Jones raises

y = number of geese Farmer Jones raises

The problem tells us that he wants to raise at most 16 animals, so we can write:

a + y ≤ 16

It also tells us that he wants no more than 12 geese, so we can write:

y ≤ 12

We know that it costs $5 to raise a pig and $2 to raise a goose, and he has $550 available to spend. So the cost of raising the animals can be expressed as:

5a + 2y ≤ 550

Finally, we want to maximize his profits. The profit per pig is $4 and the profit per goose is $8. So the objective function for Farmer Jones' profits is:

Profit = 4a + 8y

To summarize, the linear programming model for this problem is:

Maximize: Profit = 4a + 8y

Subject to:

a + y ≤ 16

y ≤ 12

5a + 2y ≤ 550

where a and y are non-negative integers.

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Answer:

The value of 1/4 (-4^3+10^2) is 9

If 13 matches are played in total then, 7 2-player matches and 6 6-player matches are played.

Here we see that the table has two columns- 6 player Athletes and 2 player athletes. It is given that no athlete participates in both the type of games. Hence we can say that

If one match for 2 player game is held then 2 players are employed there.

Hence we have 48 players left

hence we will have 48/6 = 8 6-player matches.

Similarly, if 1 6-player match is played then 44 players applied for the 2-player match, hence, we have 44/2 = 22 2-player matches

If 4 2-player matches are held then we will have 8 players booked. Hence 42/6 = 7 6-player matches were held.

If 4 6-player matches were held then, we have 26/2 = 13 2-player matches.

Hence the table will be

Number of 6 Player Games                       Number of 2-player games

                     8                                                                 1

                     1                                                                 22

                     7                                                                 4

                     4                                                                 13

b)

Let the total 2-player games played be x and 6-player games be y

we have,

x + y = 13

2x + 6y = 50

or, 2(x + y) + 4y = 50

or, 26 + 4y = 50

or, 4y = 24

or, y = 6

Hence x = 7

Therefore, in total 7 2-player matches and 6 6-player matches are played.

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If x > 2.5, both expressions within absolute value symbols are positive.

The equation becomes: y = (2x + 5) - (2x - 5) = 10.

For the given intervals of x:

If x < -2.5, both expressions within absolute value symbols are negative. Thus, the equation is: y = -(2x + 5) - (-(2x - 5)) = -10.

If x > 2.5, both expressions within absolute value symbols are positive.

The equation becomes: y = (2x + 5) - (2x - 5) = 10.

If -2.5 ≤ x ≤ 2.5, the first expression is positive and the second is negative.

The equation is: y = (2x + 5) - (-(2x - 5)) = 4x.

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Div(f) = 2cos(x) + [tex]5e^y[/tex], and curl(f) = < 0, 9sin(x), 5e^y >.

To calculate div(f) and curl(f), we need to express f as a vector field:

f = < 2 sin(x), [tex]5e^y[/tex][tex]5e^y[/tex], 9 cos(x) >

Then, we can use the formulas for divergence and curl:

div(f) = ∂f₁/∂x + ∂f₂/∂y + ∂f₃/∂z

curl(f) = < ∂f₃/∂y - ∂f₂/∂z, ∂f₁/∂z - ∂f₃/∂x, ∂f₂/∂x - ∂f₁/∂y >

Let's compute these step by step:

div(f) = ∂f₁/∂x + ∂f₂/∂y + ∂f₃/∂z

= 2cos(x) + [tex]5e^y[/tex] + 0

= 2cos(x) + [tex]5e^y[/tex]

curl(f) = < ∂f₃/∂y - ∂f₂/∂z, ∂f₁/∂z - ∂f₃/∂x, ∂f₂/∂x - ∂f₁/∂y >

= < 0 - 0, 0 - (-9sin(x)), 5e^y - 0 >

= < 0, 9sin(x), 5e^y >

Therefore, div(f) = [tex]2cos(x) + 5e^y[/tex], and curl(f) = [tex]< 0, 9sin(x), 5e^y > .[/tex]

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Answer:

c=4

Step-by-step explanation:

as there is only one solution for this quadratic equation which is supposedly have 2 or more answers, c can only be 4 in order to be the same equation as the the first one x-4=0

The following are correct about the triangle;

1. angle C is 60°

2. angle B is 60°

3. The length of segment DB is 3

4. The length of side x is 3√3

An equilateral triangle is a type of triangle in which all it's sides and angles are equal.

Since all the angles of an equilateral triangle are equal, then,

x+x+x = 180

3x = 180

x = 180/3 = 60°

therefore each angle is 60°

angle C and angle B are 60°

Using Pythagorean theorem

x² = 6²- 3²

x² = 36-9

x² = 27

x = √27

x = √9×3

x = 3√3

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The scale factor of figure KLMN to figure PQRS is 6 cm / 3 cm = 2.

To find the scale factor of figure KLMN to figure PQRS, given KN = 3 cm, MN = 7 cm, RS = 14 cm, and PS = 6 cm, follow these steps:

1. First, find the length of a side in figure KLMN. We can use KN since it's given: KN = 3 cm.
2. Next, find the corresponding side length in figure PQRS. Since KN corresponds to PS, we have: PS = 6 cm.
3. Now, find the scale factor by dividing the length of the side in figure PQRS by the length of the corresponding side in figure KLMN: scale factor = PS/KN = 6 cm / 3 cm.

The scale factor of figure KLMN to figure PQRS is 6 cm / 3 cm = 2.

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The probability of drawing a star-shaped chip is 5/8.

The theoretical probability of drawing a star-shaped circular chip from the bag is 5/8 or 0.625 or 62.5%. Out of the total of eight circular chips, five are stars, and three are squares.

Therefore, the probability of drawing a star-shaped chip is the ratio of the number of star-shaped chips to the total number of chips in the bag, which is 5/8.

To understand this conceptually, we can think of probability as a fraction where the numerator is the number of favorable outcomes (in this case, drawing a star-shaped chip) and the denominator is the total number of possible outcomes (all the circular chips in the bag).

Thus, the theoretical probability of drawing a star-shaped chip is 5/8 because there are five star-shaped chips out of the total eight circular chips in the bag.

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Answer:

approximately 42.85 of whatever unit

The number of cakes that can be made with a large bag of sugar, we first need to determine the amount of sugar in a small bag and then calculate the amount of sugar needed for one cake.

1. Find the amount of sugar in a small bag:
Since the large bag contains 30 cups of sugar and is 2.5 times larger than the small bag, we can write the equation:
Small bag = Large bag / 2.5
Small bag = 30 cups / 2.5
Small bag = 12 cups of sugar

2. Determine the amount of sugar needed for one cake:
The small bag contains enough sugar to make 9 cakes and have 0.75 cups of sugar remaining. So, we can subtract the remaining sugar from the total amount in the small bag:
Sugar used for 9 cakes = 12 cups - 0.75 cups
Sugar used for 9 cakes = 11.25 cups

Now, we can find the amount of sugar needed for one cake:
Sugar per cake = Sugar used for 9 cakes / 9
Sugar per cake = 11.25 cups / 9
Sugar per cake = 1.25 cups

3. Calculate the number of cakes that can be made with a large bag of sugar:
Cakes from large bag = Large bag sugar / Sugar per cake
Cakes from large bag = 30 cups / 1.25 cups
Cakes from large bag = 24

Therefore, a baker can make 24 cakes with a large bag of sugar.

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The 9th girl ate 51 candies.

Let the number of candies eaten by the 9th girl be x.

The average number of candies eaten by 8 girls is given as (210/x+210)/8, which simplifies to 210/8 + x/8.

After the 9th girl eats x candies, the total number of candies eaten becomes 210 + x.

The new average is given as (210 + x)/9 = 29.

Solving for x, we get:

210 + x = 261

x = 51

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To find the area enclosed by the figure, we first need to find the area of  the square and the  area of each semicircle.

The area of the square is simply the length of one of its sides squared, which is:

54 m x 54 m = 2,916 m²

The area of each semicircle is half the area of a full circle with the same radius as the side of the square. The radius of each semicircle is 54 m/2 = 27 m.

The area of each semicircle is:

1/2 x π x 27 m² = 1/2 x 3.14 x 27 m x 27 m ≈ 1,442.31 m²

Since there are four semicircles, the total area of the semicircles is:

4 x 1,442.31 m² = 5,769.24 m²

Therefore, the total area enclosed by the figure is:

2,916 m² + 5,769.24 m² ≈ 8,685.24 m²

Frank's answer of 1,662.12 m² is significantly less than the actual area. He may have made the mistake of only calculating the area of one of the semicircles instead of all four, or he may have forgotten to include the area of the square.

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To answer question a, we use the formula:

time taken = (number of tyres to fit x time taken to fit one tyre) / number of tyres fitted at once

In this case, Liam takes 124 minutes to fit 4 tyres to a lorry. To find out how long it would take him to fit 6 tyres, we plug in the values:

time taken = (6 x 124) / 4
time taken = 186 minutes

So it would take Liam 186 minutes to fit 6 tyres to a lorry.

For question b, we know that Liam takes 124 minutes to fit 4 tyres, so he takes 31 minutes to fit 1 tyre. If he works for 93 minutes, we can find out how many tyres he can fit:

number of tyres = time taken / time taken to fit one tyre
number of tyres = 93 / 31
number of tyres = 3

So Liam can fit 3 tyres in 93 minutes.

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The rank of the coefficient matrix A is 5.

Since the system AX = B is consistent and has 12 equations and 8 variables, the rank of the coefficient matrix A must be less than or equal to 8 (the number of variables).

If the solution of the system contains 3 free variables, it means that the dimension of the null-space of A is 3. By the rank-nullity theorem,

we know that the dimension of the null-space of A plus the rank of A is equal to the number of columns of A (which is 8 in this case).

Therefore, we have:

rank(A) + dim(null(A)) = 8

rank(A) + 3 = 8

rank(A) = 5

So, the rank of the coefficient matrix A is 5.

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(a) The critical numbers happen when x = 3 or x = -1/2

(b) f is decreasing on (-∞, -1/2), increasing on (-1/2, 3), and increasing on (3, ∞).

(c) f has a local minimum value of -22 at x = 3, and a local maximum value of 25.5 at x = -1/2.

(d) f is concave downward on (-∞, 1/4) and concave upward on (1/4, ∞).

(e) The inflection point of f is at x = 1/4.

(a) To find the critical numbers of f, we need to find the values of x where the derivative of f equals zero or does not exist.

f'(x) = 12x² - 6x - 18 = 6(2x² - x - 3) = 6(x - 3)(2x + 1)

Setting f'(x) equal to zero, we get:

6(x - 3)(2x + 1) = 0

x = 3 or x = -1/2

These are the critical numbers of f.

(b) To find the intervals where f is increasing and decreasing, we need to examine the sign of the derivative f'(x) in the intervals determined by the critical numbers.

When x < -1/2, f'(x) < 0, so f is decreasing on the interval (-∞, -1/2).

When -1/2 < x < 3, f'(x) > 0, so f is increasing on the interval (-1/2, 3).

When x > 3, f'(x) > 0, so f is increasing on the interval (3, ∞).

(c) To find the local minimum and maximum values of f, we need to examine the critical numbers and the end points of the intervals.

f(3) = 4(3)³ - 3(3)² - 18(3) + 5 = -22

f(-1/2) = 4(-1/2)³ - 3(-1/2)² - 18(-1/2) + 5 = 25.5

Thus, f has a local minimum value of -22 at x = 3, and a local maximum value of 25.5 at x = -1/2.

(d) To find the intervals where f is concave upward and concave downward, we need to examine the sign of the second derivative f''(x).

f''(x) = 24x - 6 = 6(4x - 1)

When x < 1/4, f''(x) < 0, so f is concave downward on the interval (-∞, 1/4).

1/4 < x, f''(x) > 0, so f is concave upward on the interval (1/4, ∞).

(e) To find the inflection points of f, we need to examine the points where the concavity changes.

The concavity changes at x = 1/4, which is the only inflection point o

The value of 'a' is 2/3.

The problem presents a ratio, 4:15, that is equal to a ratio involving 'a' and 2 1/2. To solve for 'a', we need to isolate it on one side of the equation by cross-multiplying.

In the first step, we convert 2 1/2 to an improper fraction, 5/2, so that we can use it in the equation. We then cross-multiply by multiplying both sides of the equation by 5/2.

This eliminates the denominator on the right-hand side and simplifies the left-hand side.

Solve for 'a'

To solve for 'a', we can use cross-multiplication.

First, we need to convert 2 1/2 to an improper fraction:

2 1/2 = 5/2

Now we can write the equation as:

4/15 = a/(5/2)

To solve for 'a', we cross-multiply:

4/15 * 5/2 = a

a = 2/3

Finally, we solve for 'a' by multiplying 4/15 by 5/2 and simplifying the result. The answer is 2/3, which represents the value of 'a'.

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The value of the function f(x) at x = 2 is 48.

The question asks us to find the value of the function f(x) = 4[tex]x^3[/tex] at x = 2 and its derivative f'(x) at x = 2.

The value of the function f(x) at x = 2, we simply plug in x = 2 into the function and evaluate:

f(2) = 4[tex](2)^3[/tex] = 4(8) = 32

Therefore,

The value of the function f(x) at x = 2 is 32, which we can write as f(2) = 32.

To find the derivative of the function f(x) at x = 2, we first need to find the general formula for the derivative of f(x), which we can do using the power rule for derivatives.

The power rule states that

To find the derivative of f(x) at x = 2, we simply plug in x = 2 into the formula for f'(x) that we just found:

f'(2) = [tex]12(2)^2[/tex] = 48

Therefore,

The derivative of the function f(x) at x = 2 is 48, which we can write as f'(2) = 48To find the derivative of the function

f'(x) = 12[tex]x^2[/tex]

To find the value of f(2), we can simply plug in x = 2 into the original function:

f(2) = 4[tex](2)^3[/tex] = 32

To find the value of f'(2), we can plug in x = 2 into the derivative we just found:

f'(2) = 12[tex](2)^2[/tex] = 48

Therefore,

f(2) = 32 and f'(2) = 48.

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a) The probability that, for at least 15 packs, the average weight of the erasers in the pack is at least 31.95 g is approximately 0.0384.

b) The probability that, on average, the unit finds at least 37.2 "good" erasers per day is approximately 0.3133.

a) To solve this problem, we need to use the central limit theorem. According to this theorem, the distribution of sample means becomes approximately normal, regardless of the shape of the population distribution, when the sample size is sufficiently large (usually, n >= 30). In this case, since the sample size is 45, we can assume that the distribution of sample means will be approximately normal.

Now, we need to find the probability that the average weight of at least 15 packs is at least 31.95 g. We can use the normal distribution to calculate this probability. We first calculate the z-score for this value as follows:

z = (31.95 - 31.9) / (0.163 / √(45)) = 1.77

Using a standard normal table or calculator, we can find the probability that a z-score is greater than or equal to 1.77. This probability is approximately 0.0384.

b) To solve this problem, we need to use the normal approximation to the binomial distribution. Since each eraser is either "good" or "bad", the number of "good" erasers that the unit finds each day follows a binomial distribution with parameters n = 50 and p = probability of finding a "good" eraser = (32.3 - 31.7)/(32.3 - 31.5) = 0.5.

Now, we need to find the probability that, on average, the unit finds at least 37.2 "good" erasers per day. We can use the normal distribution to calculate this probability. We first calculate the z-score for this value as follows:

z = (37.2 - 25) / 25 = 0.488

Using a standard normal table or calculator, we can find the probability that a z-score is greater than or equal to 0.488. This probability is approximately 0.3133.

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The quotient of the division 10x² + 3x - 77 ÷ 2x + 7 is 5x - 16

The quotient expression is given as

10x² + 3x - 77 ÷ 2x + 7

The long division expression is represented as

2x + 7 | 10x² + 3x - 77

So, we have the following division process

            5x - 16

2x + 7 | 10x² + 3x - 77

            10x²  + 35x

          --------------------------------

             -32x - 77

             -32x - 112

          -------------------------------------

                      35

Hence, the quotient of the long division is 5x - 16

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To integrate the function ∫(x² - 36) dx, we first need to factor out the expression inside the parentheses:
∫(x² - 36) dx = ∫(x - 6)(x + 6) dx

We can then use the power rule of integration to find the antiderivative:
∫(x - 6)(x + 6) dx = (1/3)x³ - 6x + C, where C is the constant of integration.

Since the original problem states X > 6, we can evaluate the definite integral using these limits:
∫(x² - 36) dx from 6 to X = [(1/3)X³ - 6X] - [(1/3)(6)³ - 6(6)]
= (1/3)X³ - 6X - 68

Therefore, the answer in exact form is (1/3)X³ - 6X - 68.

To integrate the given function, first note the correct notation for the function: ∫(x^2)/(x^2 - 36) dx for x > 6.

To solve this, we can use partial fraction decomposition. The given function can be rewritten as:

∫(A(x - 6) + B(x + 6))/(x^2 - 36) dx

Solving for A and B, we find that A = 1/12 and B = -1/12. Now we rewrite the integral as:

∫[(1/12)(x - 6) - (1/12)(x + 6)]/(x^2 - 36) dx

Next, separate the two terms and integrate them individually:

(1/12)∫[(x - 6)]/(x^2 - 36) dx - (1/12)∫[(x + 6)]/(x^2 - 36) dx

Now, notice that the integrals are of the form ∫u'/u dx. The integral of this form is ln|u|. So we have:

(1/12)[ln|(x - 6)| - ln|(x + 6)|] + C

Using the logarithm property, we can rewrite the answer as:

(1/12)ln|((x - 6)/(x + 6))| + C

That is the exact form of the antiderivative for the given function.

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The mean is equal to the median, so the data is symmetrical.

Given that :

Colin surveyed 12 teachers at his school to determine how much each person budgets for lunch.

The data is :

10   5   8   10   12   6   8   10   15   6   12   18

Mean is the average of these numbers.

Mean = (10 + 5 + 8 + 10 + 12 + 6 + 8 + 10 + 15 + 6 + 12 + 18) / 12

         = 10

Now median is the element in the middle arranged in an order.

Arrange the data in ascending order.

5  6  6  8  8  10  10  10  12  12  15  18

The middle element is the average of the 6th and 7th elements.

Median = (10  10) / 2 = 10

So mean and the median is equal. So the data is symmetrical.

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The MAD for Doctor A's data set is 0.67 and the MAD for Doctor B's data set is 0.83. Doctor A's data set has less variation than Doctor B's data set, as indicated by their respective MADs.

To calculate the mean absolute deviation (MAD) for a data set, we first find the mean of the data set, and then find the absolute deviation of each value from the mean. We then find the mean of these absolute deviations.

a) For Doctor A's data set on corrective lenses, the mean is:

Mean = (15+18+17+16+14)/5 = 16

The absolute deviations from the mean are:

|15-16| = 1

|18-16| = 2

|17-16| = 1

|16-16| = 0

|14-16| = 2

The mean of these absolute deviations is:

MAD = (1+2+1+0+2)/5 = 1.2

Therefore, the MAD for Doctor A's data set is 1.2.

b) For Doctor B's data set on corrective lenses, the mean is:

Mean = (17+19+20+16+18)/5 = 18

The absolute deviations from the mean are:

|17-18| = 1

|19-18| = 1

|20-18| = 2

|16-18| = 2

|18-18| = 0

The mean of these absolute deviations is:

MAD = (1+1+2+2+0)/5 = 1.2

Therefore, the MAD for Doctor B's data set is also 1.2.

Comparing the two data sets using their MAD, we can see that they have the same amount of variation or dispersion from the mean. Both sets have a MAD of 1.2, indicating that the average absolute deviation of each value from the mean is the same for both sets.

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Answer:

Doctor A MAD: 11.8

Doctor B MAD: 9.32

Step-by-step explanation:

This is what I got on the assignment.

A frequency table was created using data points representing the number of holes in each of Grandma Marion's dresses. The table shows the number of dresses with 5, 7, and 8 holes, respectively.

To create a frequency table for the given data, first, the unique values in the data set are identified, which are 5, 7, and 8. Then, the number of occurrences of each unique value is counted, resulting in the frequencies 1 for 5, 2 for 7, and 3 for 8.

Count the frequency of each data point

5: 1 dress

7: 2 dresses

8: 3 dresses

Finally, these values are organized into a table with two columns, one for the unique values and another for their corresponding frequencies. The resulting frequency table shows the number of dresses with each number of holes eaten by moths.

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1. All square, upper triangular matrices are diagonalizable. (TRUE)

2. If a matrix is diagonalizable, then it is invertible. (FALSE)

1. True: All square, upper triangular matrices are diagonalizable.

A square matrix is diagonalizable if it can be transformed into a diagonal matrix by a similarity transformation, which means there exists an invertible matrix P such that P^-1 * A * P is a diagonal matrix. Upper triangular matrices have all their elements below the main diagonal equal to zero.

Since the eigenvalues of an upper triangular matrix are equal to its diagonal elements, we can form a diagonal matrix with these eigenvalues. Since there exists such a diagonal matrix, all square, upper triangular matrices are diagonalizable.

2. False: If a matrix is diagonalizable, it is not necessarily invertible

Diagonalizable matrices can be transformed into a diagonal matrix with the eigenvalues along the main diagonal.

However, invertibility requires that the matrix have a nonzero determinant, which means that all of its eigenvalues must be nonzero.

If a diagonalizable matrix has a zero eigenvalue, its determinant will be zero, and it will not be invertible. Therefore, diagonalizability does not guarantee invertibility.

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