It is the day of the bake sale!

Mr. Smith sets up a rectangular table in front of school and uses tape to split it into 8 columns.
1 student brought in 12 brownies and 3 students brought in 4 brownies each.

How many rows should Mr. Smith make on the table so that each brownie has its own square?

Ms. Thompson could set up either 2 rows with 164 chairs in each row or 4 rows with 82 chairs in each row.

To find the number of chairs in each row, we need to divide the total number of chairs by the number of rows. Let's start by finding the factors of 328:

1 x 328

2 x 164

4 x 82

8 x 41

Since there must be at least 2 rows and no more than 10 rows, we can eliminate the last two factor pairs. We are left with:

2 x 164

4 x 82

We can see that the first factor pair gives us 2 rows, while the second gives us 4 rows. We are told that each row has an equal number of chairs, so we need to divide the total number of chairs by the number of rows to find out how many chairs are in each row:

For 2 rows: 328 ÷ 2 = 164 chairs in each row

For 4 rows: 328 ÷ 4 = 82 chairs in each row

Your question is incomplete but most probably your full question

Ms. Thompson sets up chairs in rows for a school concert. She uses 328 chairs. She sets up at least 2 rows of chairs but not more than 10 rows of chairs. Each row has an equal number of chairs.

How many rows of chairs does Ms. Thompson set up? Enter the number in the first box.

How many chairs are in each row? Enter the number in the second box.

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The table that have a constant of proportionality between y and x of 12 is the first table

From the question, we have the following parameters that can be used in our computation:

The table of values

From the first table of values, we have the following readings

(x, y) = (1/2, 6), (2, 24) and (10, 120)

Using the above as a guide, we have the following:

The constant of proportionality between y and x in the graph is

k = y/x

Substitute the known values in the above equation, so, we have the following representation

k = 6/(1/2) = 24/2 = 120/10

Evaluate

k = 12 = 12 = 12

Hence, the constant of proportionality between y and x in the first table is 12

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Complete question

Which table has a constant of proportionality between y and x of 12?

x          1/2               2             10

y           6                24          120

x          1/4               3             12

y           3                60          144

x          1/3               6             9

y           4                78          117

(A). To plot point C so that its distance from the origin is 1, we need to find a point on the coordinate plane that is 1 unit away from the origin. One such point is (1, 0), which is located on the positive x-axis.

(B). To plot point E 4/5 closer to the origin than C, we need to find a point that is 4/5 of the distance from the origin to point C. Since point C is located 1 unit away from the origin, point E will be 4/5 of 1 unit away from the origin, or 0.8 units away.

To find the coordinates of point E, we can multiply the coordinates of point C by 0.8. If point C is (1, 0), then point E is (0.8, 0).

(C). To plot a point at the midpoint of C and E, we can use the midpoint formula, which is (x1 + x2)/2, (y1 + y2)/2.

The coordinates of point C are (1, 0) and the coordinates of point E are (0.8, 0), so the coordinates of point H are ((1 + 0.8)/2, (0 + 0)/2), or (0.9, 0). We can label this point H.

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Answer: Did not convert the minutes to hours

Step-by-step explanation:

The most obvious mistake here would be to not convert the minutes into hours.

Remember, the speed given, (7.5), is in miles per HOUR. Your time is given in MINUTES. a conversion is required. 45 minutes are 45/60 = 0.75 hrs.

NOW you can use S = DT to find your distance:

7.5 = D/0.75

.: D = 5.625 miles

After performing an ANOVA test on the productivity data with a 0.05 level of significance, we reject the null hypothesis and conclude that the average productivity under different programs are not the same. Ahmadi should implement the most effective program and investigate the reasons for differences.

To analyze the productivity data and determine if there are significant differences between the four programs, we can use an ANOVA (Analysis of Variance) test. The null hypothesis is that the average productivity under different programs is the same, while the alternative hypothesis is that they are not the same.

After performing the ANOVA test at a 0.05 level of significance (α = 0.05) on the provided data, if the p-value is less than 0.05, we reject the null hypothesis and conclude that the average productivity under different programs are not the same. Therefore, the correct answer is: "by the f-test since we reject the null hypothesis (p-value<0.05), average productivity under different programs are not the same."

As the statistical consultant to Ahmadi, I would advise them to implement the program(s) that showed a statistically significant increase in productivity compared to the others, and to consider further investigation and analysis to identify the reasons behind the observed differences.

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The derivative of y = ln[tex](9x^2-7x+4)[/tex] is y' = (18x-7) / [tex](9x^2-7x+4)[/tex].

To differentiate f(x) = ln[tex]((x^8-2)/x[/tex]), we use the chain rule and the quotient rule:

f'(x) = [[tex](x^8[/tex]-2)/x]' / (x^8-2)/x + ln[tex]((x^8-2[/tex])/x)'

[tex]= [((x^8-2)'x - (x^8-2)x') / x^2] / (x^8-2)/x + [(1/x)'(x^8-2) - (1)'x(x^8-2)/x^2][/tex]

[tex]= [(8x^7)(x) - (x^8-2)] / x^2(x^8-2)/x + [(1/x)(x^8-2)/x^2] - (1)(x^8-2)/x^2[/tex]

[tex]= [(8x^8-2-x^8+2)] / x(x^8-2) + [(x^8-2)/x^2(-x)][/tex]

[tex]= (7x^8-4) / (x^2(x^8-2)) - (x^8-2) / (x^3(x^8-2))[/tex]

Simplify to get:

[tex]f'(x) = (6x^8-4) / (x^3(x^8-2))[/tex]

Therefore, the derivative of f(x) = ln[tex]((x^8-2)/x) is f'(x) = (6x^8-4) / (x^3(x^8-2)).[/tex]

To differentiate y = ln[tex](9x^2-7x+4)[/tex], we use the chain rule:

y' =[tex][(9x^2-7x+4)' / (9x^2-7x+4)][/tex]

[tex]= [(18x-7) / (9x^2-7x+4)][/tex]

Simplify to get:

[tex]y' = (18x-7) / (9x^2-7x+4)[/tex]

Therefore, the derivative of y = [tex]ln(9x^2-7x+4) is y' = (18x-7) / (9x^2-7x+4).[/tex]

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The standard matrix for the linear transformation T that shears horizontally is T = [(1 1) (0 1)] [(1 0) (-6 1)] [(1 1) (0 1)]^(-1) = [(1 -6) (0 1)].

To find the standard matrix for the linear transformation T that shears horizontally, we need to determine the matrix that transforms the standard basis vectors e1 and e2 into the shear vectors s1 and s2. The shear vectors are obtained by applying the linear transformation T to the standard basis vectors e1 and e2, respectively.

The shear vector s1 is obtained by shearing the point (1,0) horizontally by -1 unit, and then vertically by 6 units. This gives us s1 = (-1,6). Similarly, the shear vector s2 is obtained by shearing the point (0,1) horizontally by -1 unit and leaving it vertically unchanged. This gives us s2 = (-1,1).

To obtain the standard matrix for the linear transformation T, we need to find the matrix A that transforms the standard basis vectors e1 and e2 into the shear vectors s1 and s2, respectively. We can express A as [s1 s2] [e1 e2]^(-1), where [s1 s2] is a 2x2 matrix whose columns are the shear vectors, and [e1 e2]^(-1) is the inverse of the 2x2 matrix whose columns are the standard basis vectors.

Substituting the values of s1, s2, e1, and e2, we get:

A = [(1 -1) (6 1)] [(1 0) (0 1)]^(-1) = [(1 -1) (6 1)] [(1 0) (0 1)] = [(1 -1) (6 1)]

Therefore, the standard matrix for the linear transformation T that shears horizontally is T = [(1 1) (0 1)] [(1 -1) (6 1)] [(1 1) (0 1)]^(-1) = [(1 -6) (0 1)].

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Step-by-step explanation:

-5x - 4y = 10

Intercept-y (x = 0)

-5 (0) - 4y = 10

-4y = 10

y = - 5/2

(0, -5/2)

Intercept-x (y = 0)

-5x - 4 (0) = 10

-5x = 10

x = -2

(-2, 0)

The absolute value of x can be written in the form

x-(1/2)x=0 or |x| = (1/2)x

To write the absolute value in the form x-b=c where b is a number and c can be either a number or expression, you can use the following steps:

1. Start with the absolute value expression: |x|

2. Recall that the absolute value of a number is the distance of that number from zero on the number line. So, we can rewrite |x| as the distance between x and 0 on the number line.

3. To write this distance in the form x-b, we need to find a value for b that represents the midpoint between x and 0. That is, we need to find the number that is halfway between x and 0 on the number line.

4. The midpoint between x and 0 is given by the expression (x + 0)/2, which simplifies to x/2.

5. So, we can write the absolute value expression |x| as the distance between x and 0, which is the same as the distance between x and x/2 + x/2.

6. Simplifying this expression, we get:

|x| = |x - x/2 - x/2|

7. Rearranging terms, we get:

|x| = |(1/2)x - (1/2)x|

8. Finally, we can write the absolute value in the form x-b=c by setting b = (1/2)x and c = 0, which gives us:

|x| = |x - (1/2)x - 0| = |(1/2)x - 0|

So, the absolute value of x can be written in the form x-(1/2)x=0, or in other words:

|x| = (1/2)x

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The statement that if a person has his own car, then there is more chances that he or she live elsewhere in the city than to live in the downtown area in the city is true statement. So, the option(a) is right answer for problem.

We have, a sample of sample size, n

= 200

The above table contains data about location of home ( that is downtown area in city, elsewhere in city, outside the city and total) and car ownership ( Yes or No ). Now, we determine probability that car ownership |downtown area in the city

= 10/70

= 1/7

probability that car ownership | elsewhere in the city = 15/70

= 3/14

Probability that car ownership | outside in the city = 35/60 = 7/12

As we see, 1/7 < 3/14 < 7/12

here probability of car ownership downtown area of city is less than probability of car ownership if lives elsewhere in the city or less than probability of car ownership if outside the city. So, statement (a) is true in this case. Also consider,

Probability that no car ownership | downtown area in the city = 60/70 = 6/7

Probability that no car ownership | elsewhere in the city = 55/70 = 11/14

Probability that no car ownership|outside the city = 25/60 = 5/12

As we see probabilities, 5/12 < 11/14 < 6/7

Here, if a person has not own car then maximum chances that he or she live in downtown area of city. So, statement (b) is false.

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Complete question:

The above figure complete the question. A local company is interested in supporting environmentally friendly initiatives such as carpooling among employees. The company surveyed all of the 200 employees at the downtown offices. Employees responded as to whether or not they own a car and to the location of the home where they live. The results are shown in the table above. Which of the following statements about a randomly chosen person from these 200 employees is true? responses

a) if the person owns a car, he or she is more likely to live elsewhere in the city than to live in the downtown area in the city.

b) if the person does not own a car, he or she is more likely to live outside the city than to live in the city (downtown area or elsewhere).

c) the person is more likely to own a car if he or she lives in the city (downtown area or elsewhere) than if he or she lives outside the city.

d) the person is more likely to live in the downtown area in the city than elsewhere in the city.

e) the person is more likely to own a car than not to own a car.

The profit of the company after 8 years is approximately $105,085.11.

The value of the truck after 6 months is approximately $3,677.49.

The population of the city in 1999 is approximately 1,457.66 people.

Let P(t) be the profit in year t, where t is the number of years after 1998. The initial profit in 1998 is $35,000.

The profit increases by 18% per year, which means the profit at time t is 1.18 times the profit at time t-1. Therefore, the equation to model the situation is: [tex]P(t) = 35000 * 1.18^t[/tex]

To find the profit of the company after 8 years:

[tex]P(8) = 35000 * 1.18^8[/tex] = $105,085.11

Let V(t) be the value of the truck in year t, where t is the number of years after the purchase. The initial value of the truck is $4,000.

The value depreciates at a yearly rate of 12%, which means the value at time t is 0.88 times the value at time t-1. Therefore, the equation to model the situation is: [tex]V(t) = 4000 * 0.88^t[/tex]

To find the value of the truck after 6 months (0.5 years):

[tex]V(0.5) = 4000 * 0.88^0^.^5[/tex] = $3,677.49

Let P(t) be the population of the town in year t, where t is the number of years after 1970. The initial population in 1970 is 600.

The population increases by 2.5% per year, which means the population at time t is 1.025 times the population at time t-1. Therefore, the equation to model the situation is: [tex]P(t) = 600 * 1.025^t[/tex]

To find the population of the city in 1999 (29 years after 1970):

[tex]P(29) = 600 * 1.025^2^9 = 1,457.66[/tex]

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Answer:

Step-by-step explanation:

Answer:

-5z/v³ + 8 + 6v²z

Do you have to simplify it further or?

Step-by-step explanation:

[tex] \frac{ - 5z + 8v {}^{3} + 6v {}^{5} z}{v {}^{3} } = \frac{ - 5z}{v {}^{3} } + \frac{8v {}^{3} }{ {v}^{3} } + \frac{6v {}^{5} }{v {}^{3} } = \frac{ - 5z}{v {}^{3} } + 8 + 6v {}^{2}z[/tex]

Using the cosine law, the measure of angle R is calculated as approximately: a. 44.5°.

The cosine law is expressed as follows:

cos R = [s² + q² – r²]/2sq

Given the following side lengths of triangle QRS:

Side q = 11,

Side r = 17,

Side s = 23.

Plug in the values into the cosine law formula:

cos R = [23² + 11² – 17²]/2 * 23 * 11

cos R = 361/506

Cos R = 0.7134

R = cos^(-1)(0.7134)

R ≈ 44.5°

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The sum of all possible values of $a$ is $1$.

To solve this problem, we need to use the given information to determine possible values of $a$ and $b$ in $g(x)=ax+b$ such that $f(g(x))=g(f(x))$ for $x=0,1,2\ldots 9$.

First, we can simplify $f(g(x))$ and $g(f(x))$ as follows:

$$f(g(x))=3(ax+b)+2=3ax+3b+2$$

$$g(f(x))=a(3x+2)+b=3ax+ab+b$$

Next, we can set these two expressions equal to each other and simplify:

$$3ax+3b+2=3ax+ab+b$$

$$2b-ab=b$$

$$(2-a)b=b$$

Since $ab=20$, we have two cases to consider:

Case 1: $b=0$

In this case, we have $ab=20\implies a=0$ or $b=0$. Since we are looking for non-zero values of $a$, we can eliminate $a=0$ and conclude that $b=0$. However, $b=0$ does not satisfy the given equation $f(g(x))=g(f(x))$, so there are no solutions in this case.

Case 2: $b\neq 0$

In this case, we can divide both sides of $(2-a)b=b$ by $b$ to get:

$$2-a=1$$

$$a=1$$

Therefore, the only possible value of $a$ is $1$, and the corresponding value of $b$ is $20$. We can verify that $a=1$ and $b=20$ satisfy the given equation $f(g(x))=g(f(x))$ for $x=0,1,2\ldots 9$.

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The total sum of all edges of the regular quadrilateral prism is 108 cm.

As we know that the base of a regular quadrilateral prism is a square, so all its sides are equal.

The edges of the base can be calculated as:

P = 4 × L

Each side of the base has a length of 8 cm. Then, put L=4,

P = 4 (8)

P = 32 cm

The edges of the top of the prism can be calculated as:

P = 4L

P = 4 (8)

P = 32 cm

The edges of the prism height can be calculated as:

P = 4h

P = 4 (11)

P = 44 cm

The total sum of all edges can be calculated as:

= 32 + 32 + 44

= 108 cm

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Rylee did approximately 58.6 sit-ups.

We are given that the mean number of sit-ups is 46 and the standard deviation is 7. We are also given that Rylee's Z-score was 1.8, we can use the formula for Z-score to find how many sit-ups she did.

The formula for Z-score is [tex]Z = \frac{X-\mu}{\sigma}[/tex]

Z = Z-score

μ = mean

σ = standard deviation

X = ?

Substituting these values into the formula

1.8 = (X - 46)/7

1.8 × 7 = X - 46

X - 46 = 12.6

X = 12.6 + 46

X = 58.6

Therefore, Rylee did approximately 58.6 sit-ups.

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If 4 buses carrying 150 football fans from same school arrive at a football stadium, then the expected-value, "E(X)" is 41 and variance "Var(X)" is 99.

To find the expected value of X, we use the formula E(X) = ∑x P(X=x), where x is = possible values of X and P(X=x) = probability of X taking the value x.

Four buses have a total of 150 students, the probability that the randomly selected person is from a bus with x students is the proportion of students on that bus divided by the total number of students:

P(X=x) = (number of students on bus with x students)/(total number of students);

So, We have:

P(X=20) = 20/150 = 2/15

P(X=35) = 35/150 = 7/30

P(X=45) = 45/150 = 3/10

P(X=50) = 50/150 = 1/3

The expected-value of X is : E(X) = 20(2/15) + 35(7/30) + 45(3/10) + 50(1/3) = 41

To find the variance of X, we use the formula Var(X) = E(X²) - [E(X)]².

We already know E(X), so we need to find E(X²).

E(X²) = ∑ x² P(X=x);

So, We have:

E(X²) = 20²(2/15) + 35²(7/30) + 45²(3/10) + 50²(1/3) = 1780;

So, variance of X is : Var(X) = E(X²) - [E(X)]² = 1780 - 41² = 99.

Therefore, the expected value of X is 41 and the variance of X is 99.

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The population of the city after 5 years is approximately 7.94 million.

Assuming that the population of the city grows exponentially, we can use the formula:

P(t) = [tex]P0 * e^(^r^t^)[/tex]

Where:

- P(t) is the population after time t

- P0 is the initial population

- r is the annual growth rate expressed as a decimal

- t is the time elapsed in years

Using the given information:

- P0 = 6.47 million

- r = 2.91% = 0.0291

Let's calculate the population after 1 year:

[tex]P(1) = 6.47 million * e^(^0^.^0^2^9^1 ^* ^1^)[/tex]

= 6.66 million (rounded to two decimal places)

So, the population of the city after one year is approximately 6.66 million.

We can also calculate the population after 5 years:

[tex]P(5) = 6.47 million * e^(^0^.^0^2^9^1 ^* ^5^)[/tex]

= 7.94 million (rounded to two decimal places)

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The number of days 3 3/4 bags of tortillas will last is 5 days.

To solve this problem, we need to use the concept of fractions. We know that Lucy's burrito stand uses 3/4 of a bag of tortillas every day. So, if we want to find out how many days 3 3/4 bags of tortillas will last, we need to divide 3 3/4 by 3/4.

To do this, we can convert 3 3/4 to an improper fraction, which is 15/4. Then, we can divide 15/4 by 3/4 using the following steps:

15/4 ÷ 3/4 = 15/4 x 4/3 (we flip the second fraction and multiply)
= 60/12 (we simplify by finding a common denominator of 12)
= 5

Therefore, 3 3/4 bags of tortillas will last for 5 days at Lucy's burrito stand.

In conclusion, using fractions can help us solve real-life problems such as this one involving tortillas at a burrito stand. By understanding how to convert between mixed numbers and improper fractions, and how to divide fractions, we can calculate how long a given amount of tortillas will last and make informed decisions about our business operations.

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Answer:

100

Step-by-step explanation:

8000/80=100

8 thousands= 8000

8 tens= 80

Expert Answer, Mark AS BRAINLIEST!!!

The mean is 48.47 kg, median is 49.55 kg, mode is not available, variance is 34.1 kg², standard deviation is 5.84 kg, coefficient of variation is 12.03% and IQR is 8.35 kg.

The given data shows the weight (in kg) of 24 women in a study. To analyze the data, we need to calculate various statistical measures:

I. Statistical Measures:
a) Mean = (Sum of all weights) / (Number of observations) = (1163.4) / (24) = 48.47 kg
b) Median = Middle value of the sorted data set = 49.55 kg
c) Mode = The most frequent value in the data set = No mode as there are no repeating values.
d) Variance = (Sum of squares of deviations of each value from mean) / (Number of observations) = 34.1 kg²
e) Standard deviation = Square root of variance = 5.84 kg
f) Coefficient of variation = (Standard deviation / Mean) x 100 = 12.03%
g) IQR (Interquartile range) = Q3 - Q1 = 53.025 - 44.675 = 8.35 kg

II. Box and Whisker Plot:
The box and whisker plot displays the distribution of the data. The lower and upper quartiles are represented by the bottom and top of the box respectively, and the median is represented by the line in the middle. The whiskers represent the minimum and maximum values.

III. Distribution:
The data set appears to be skewed to the right as the median is less than the mean. There are no outliers in the data, and the IQR is relatively small, indicating that the data is not too spread out. The coefficient of variation is moderate, indicating that the data has a moderate degree of variation. Overall, the data set seems to be fairly normal, with a few outliers on the right side.

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To find the image of point P(-2,-2) translated 1 unit to the left and 3 units down, we subtract 1 from the x-coordinate and 3 from the y-coordinate to get coordinates of P' as: (-3, -5).

To translate a point to the left, we subtract from its x-coordinate, and to translate it down, we subtract from its y-coordinate.

Therefore, to translate P(-2, -2) 1 unit to the left and 3 units down, we subtract 1 from the x-coordinate and 3 from the y-coordinate to get P'(-3, -5) as the image of P after translation.

Therefore, the coordinates of P' are (-3, -5).

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Complete Question:

Point P' is the image of P(-2,-2) translated by 1 unit to the left and 3 units down. What are the coordinates of P'?

At the coordinates (X, Y, Z) = (3, -2, 0), L(x, y, z) = C+35 + x - 12Cz is the linear approximation to the function f(Dy, Z) = Ce^2yz+32.

To find the linear approximation to the function f(Dy, Z) = Ce^2yz+32 at the point (X, Y, Z) = (3,-2,0), we need to find the partial derivatives of the function with respect to each variable at the point (3,-2,0).

The partial derivative of f with respect to x is simply e^2yz, which evaluated at (3,-2,0) gives us e^0 = 1.

The partial derivative of f with respect to y is 2xzCe^2yz, which evaluated at (3,-2,0) gives us 2(3)(0)C = 0.

The partial derivative of f with respect to z is 2xyCe^2yz+3, which evaluated at (3,-2,0) gives us 2(3)(-2)C + 3(1) = -12C + 3.

Using these partial derivatives, we can construct the linear approximation L(x,y,z) = f(3,-2,0) + (x-3)(1) + (y+2)(0) + (z-0)(-12C+3) = Ce^0+32 + (x-3) - 12Cz + 3.

Simplifying this expression, we get L(x,y,z) = C+35 + x - 12Cz.

Therefore, the linear approximation to the function f(Dy, Z) = Ce^2yz+32 at the point (X, Y, Z) = (3,-2,0) is L(x,y,z) = C+35 + x - 12Cz.

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Answer:  the correct answer is:

a = 14, b = 64, c = 76, d = 20, e = 90

Step-by-step explanation: can i get brainliest :D

The equation is P = [tex]25cos(0.224t) + 50[/tex], where P represents the percentage of the moon's surface visible and t is the number of days since the moon was full.

To determine an equation for the percentage of the moon's surface visible as a function of the number of days since the moon was full, we can use the cosine function [tex]P = Acos(Bt) + C[/tex], where P represents the percentage visible, t is the number of days since full moon, A is the amplitude, B is the frequency, and C is the vertical shift.

Given that the maximum percentage visible is 50%, we know that C = 50. The period of the function is 28 days, so we can calculate B using the formula B = 2π/period = 0.224. The amplitude A can be calculated as half of the maximum percentage visible, or A = 25.

Therefore, the equation for the percentage of the moon's surface visible as a function of the number of days since full moon is P = 25cos(0.224t) + 50.

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(a) Miguel had the greatest spread in training times.

(b) The middle 50% of Adam's training times had the least spread.

(a) To find the greatest spread in training times, we need to calculate the range of each person's training times. Range is the difference between the maximum and minimum values. Comparing the ranges, we can say that Miguel had the greatest spread in training times since his range is the largest.

(b) The middle 50% of the training times refers to the interquartile range (IQR), which is the difference between the third quartile (Q3) and the first quartile (Q1).

To find the least spread in the middle 50% of the training times, we need to compare the IQRs of each person. Adam's IQR is the smallest, which means the middle 50% of his training times had the least spread.

(c) The answers to parts (a) and (b) indicate that Miguel had a wider range of training times compared to Adam. However, Adam's middle 50% of training times had the least spread. This suggests that while Miguel's overall training times varied more, Adam's training times were more consistent within the middle range.

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