A researcher interested in the effects of the environment on encoding and retrieving selects a sample of college students. The researcher instructs this sample to memorize a list of eclectic vocabulary words in vibrant orange room. After the students of studied the list, the researcher takes half the students to a drab beige room and the other half remain in the orange room. Both groups of students are then tested on the studied words. A professor believes that psychology students study more than the average college student (after all, psychology students understand the benefits to distributed practice). To test this, the professor records the weekly study rate of a sample of 20 psychology students, and compares this with the University's data on the average number of hours each week a typical college student studies.

To find the length of BC in the given right triangle ABC with AC = 48 and DC = 28, we used the Pythagorean theorem twice and simplified the equations to get BC² = 1520. Taking the square root, we got BC = 4√(95).

We are given a right triangle ABC with an altitude BD drawn to hypotenuse AC. We are also given that AC = 48 and DC = 28, and we need to find the length of BC.

To find the length of BC, we can use the Pythagorean theorem, which states that in a right triangle, the sum of the squares of the two shorter sides (the legs) is equal to the square of the length of the hypotenuse.

In this case, we have

AB² + BD² = BC² (using the Pythagorean theorem for triangle ABD)

AC² - DC² = BC² (using the Pythagorean theorem for triangle ADC)

Substituting AC = 48 and DC = 28, we get

AB² + BD² = BC²

48² - 28² = BC²

Simplifying, we get

AB² + BD² = BC²

(48 + 28)(48 - 28) = BC²

76 × 20 = BC²

BC² = 1520

Taking the square root of both sides, we get

BC = √(1520) = 4√(95)

Therefore, the length of BC in simplest radical form is 4√(95).

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The angle between the kayakers is approximately 92 degrees when rounded to the nearest degree.

The correct option is (D)

We have:

One kayaker pulls with a force vector F sub 1 equals open angled bracket 180 comma 160 close angled bracket comma and the other kayaker pulls with a force vector F sub 2 equals open angled bracket 123 comma negative 128 close angled bracket period.

F sub 1 dot F sub 2 = ||F sub 1|| ||F sub 2|| cos(theta)

where "dot" represents the dot product, "|| ||" represents the magnitude of the vector, and theta is the angle between the two vectors.

We have to find the magnitudes of the two vectors:

||F sub 1|| = [tex]\sqrt{180^2+160^2}=236.13[/tex]

||F sub 2|| = [tex]\sqrt{123^3+(-128)^2}=174.13[/tex]

Now, we have to find the dot product:

F sub 1 dot F sub 2 = (180)(123) + (160)(-128) = -49920

Now we can solve for the angle theta:

-49920 = (236.13)(174.13) cos(theta)

cos(theta) = -0.156

Using the inverse cosine function, we find that:

theta = 91.89 degrees

As a result, rounded to the nearest degree, the angle between the kayakers is approximately 92 degrees.

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Answer:

we find that the incidence rate of asthma in Oklahoma during that year was 454 cases per 100,000 population.

Step-by-step explanation:

The incidence rate of asthma is a measure of the number of new cases of asthma in a given population over a specific period. It is usually expressed per 100,000 population to allow for easier comparison between populations of different sizes. In this case, we are given the number of new cases of asthma and the total population of Oklahoma during the same year.

To calculate the incidence rate, we divide the number of new cases of asthma by the total population, and then multiply by 100,000. Applying this formula, we find that the incidence rate of asthma in Oklahoma during that year was 454 cases per 100,000 population.

This means that for every 100,000 residents of Oklahoma, there were 454 new cases of asthma during that year.

This information can be useful for public health officials and policymakers in identifying areas where more resources may be needed to prevent and manage asthma. It can also help in the evaluation of the effectiveness of interventions aimed at reducing the incidence of asthma.

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(24 ÷ 3) + x expresses the expression that represents the quotient of 24 and 3 plus x.

The expression refers to a mathematical phrase with two or more numbers or variables with mathematical operations such as addition, subtraction, division, multiplication, exponential, and so on. Examples of expression include 2a + 3p, and 9p.

To convert the given phrase into the expression, we have to start with the first operation which is a division that is represented by the word quotient. Thus we get 24 ÷ 3

Then we add the operation of addition which is represented by the word plus to the existing equation and we get (24 ÷ 3) + x

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The statements that is correct is: C. Both functions are decreasing and have different rates of change.

A function is said to be decreasing if the value of y decreases for every value of x that increases.

In the first function given, as x values increased from 3 to 4, the y value decreases from 3 to 0. So it is a decreasing function.

Rate of change = 3 - 0 / 3 - 4

= 3/-1

= -3.

In the second function, as x values increases from -4 to 0, the y value decreases from 0 to -1. It is also a decreasing function.

Rate of change = change in y / change in x = 0 - (-1) / -4 - 0

= 1/-4

= -1/4.

Therefore, they both have the same rate of change.

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Answer:

24

Step-by-step explanation:

Answer: Hence, Diameter = 2 × 12 = 24 cm. Q.:

Step-by-step explanation:

Answer: 84

Step-by-step explanation:

Sorry if this is wrong.

Answer:C. μ = 5

Step-by-step explanation:Option A (X = 5) and option D (X = 4.5) are incorrect because X represents a single value or observation, not a population parameter. Option B (μ = 4.5) is also incorrect because the question states that the mean of the population is 4.5, but we are looking for the equation that describes the population parameter which is the true mean of the entire population.

To solve this problem, we can use the formula for probability of an event:

P(event) = (number of favorable outcomes) / (total number of outcomes)

Let's first find the total number of ways to select two fuses from 20:

20 choose 2 = 20! / (2! * (20-2)!) = 190

Now let's find the number of ways to select one defective fuse and one non-defective fuse:

There are 10 defective fuses and 10 non-defective fuses, so we can choose one of each in 10 * 10 = 100 ways.

Therefore, the probability of selecting only one defective fuse is:

P(1 defective) = 100 / 190 = 0.5263

So the answer is not one of the options given.

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99.8% confidence interval with a margin of error of 0.02, using the estimated proportion of 0.31 from the previous poll.

To find the sample size needed to construct a confidence interval for a proportion with a specified margin of error, we use the following formula:

n = (z^2 * p * q) / E^2

where:

n is the sample size

z is the z-score corresponding to the desired level of confidence (99.8% in this case)

p is the estimated proportion from the previous poll (0.31 in this case)

q = 1 - p

E is the desired margin of error (0.02 in this case)

First, we need to find the value of z for a 99.8% confidence level. Using a standard normal distribution table, we can find that the z-score for a 99.8% confidence level is approximately 2.967.

Substituting the given values into the formula, we have:

n = (2.967^2 * 0.31 * 0.69) / 0.02^2

n = 1202.19

Rounding up to the nearest whole number, we need a sample size of n = 1203 to construct a 99.8% confidence interval with a margin of error of 0.02, using the estimated proportion of 0.31 from the previous poll.

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The local extreme value is -n * e^(-1).

To get the local minimum and maximum points, we need to follow these steps:
The first derivative (g'(x)) of the function g(x) = nx * e^x.
Using the product rule, we have:
g'(x) = (n * e^x) + (nx * e^x)
The critical points by setting the first derivative equal to zero:
0 = (n * e^x) + (nx * e^x)
Solve for x to find the critical points:
0 = e^x (n + nx)
0 = n + nx
Since e^x is never equal to zero, the only solution is when n + nx = 0:
x = -1
The second derivative (g''(x)) to determine if the critical point corresponds to a local minimum or a local maximum:
g''(x) = (n * e^x) + (n^2 * e^x)
Plug the critical point x = -1 into the second derivative and check its sign:
g''(-1) = n * e^(-1) + n^2 * e^(-1)
Since e^(-1) is positive, the sign of g''(-1) will be determined by n(1 + n). If n > 0, g''(-1) > 0 and we have a local minimum. If n < 0, g''(-1) < 0 and we have a local maximum.
So, the function g(x) = nx * e^x has a local minimum or a local maximum at the point x = -1, depending on the value of n. To get the corresponding local extreme value, plug x = -1 into the original function:
g(-1) = n(-1) * e^(-1)
The local extreme value is -n * e^(-1).

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The value of area of logo is,

A = 139.5 cm²

We have to given that;

The shape of a logo is made up of a triangle, a rectangle, and a parallelogram.

Here, Base of triangle = 9 cm

Height of triangle = 7 cm

Length of rectangle = 9 cm

Width of rectangle = 9 cm

Hence, We get;

The value of area of logo is,

A = (1/2 × 9 × 7) + (3 × 9) + (9 × 9)

A = 31.5 + 27 + 81

A = 139.5 cm²

Thus, The value of area of logo is,

A = 139.5 cm²

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Based on the given options, the diagram that best represents the activity of a sales clerk manually preparing a sale receipt during a power outage would be option D.

A trapezoid could represent the shape of a receipt, a curved side rectangle could represent the shape of the clerk's desk or the paper she is using, and a circle could represent the shape of a calculator or cash register. Therefore, a trapezoid to a curved side rectangle to a circle could represent the process of the clerk manually calculating and recording the sale amount and inputting it into a calculator or cash register to produce a receipt.

It is important to note that during a power outage, technology-dependent activities such as electronic sales and transactions may be disrupted, and manual methods may have to be used as a backup. This highlights the power of technology in our daily lives and the impact that power outages can have on businesses and individuals.
The question is about selecting the correct diagram that represents the sales clerk manually preparing a sale receipt due to a power outage. Given the options:

a. trapezoid to curvy side rectangle to curved rectangle
b. curved side rectangle to circle to curved rectangle
c. circle to curved side rectangle to curved rectangle
d. trapezoid to curved side rectangle to circle

The appropriate answer for this question cannot be determined based on the provided information. Diagrams typically require visual representation, and the description of the shapes alone is insufficient to convey the activity of preparing a receipt manually. Moreover, the terms "power," "outage," "clerk," "receipt," "trapezoid," "curved," and "curved" don't necessarily correspond to the shapes given in the options.

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The line integral of the given function f(x, y) along the given curve r(t) is 0.8404.

First, we need to parameterize the curve by substituting x = t³ and y = [tex]t^4[/tex] into the function f(x, y) to get:

f(t) = √([tex]t^4[/tex]/t³) = [tex]t^{1/2}[/tex]

Next, we need to find the derivative of r(t) with respect to t:

r'(t) = 3t²i + 4t³j

Then, we can compute the line integral using the formula:

∫f(r(t))|r'(t)|dt from ½ to 1

Substituting the values, we get:

∫[tex]t^{1/2}[/tex] |3t²i + 4t³j| dt from ½ to 1

= ∫[tex]t^{1/2}[/tex] |t²(3i + 4tj)| dt from ½ to 1

= ∫[tex]t^5[/tex] (9 + 16t²) dt from ½ to 1

This integral is not easy to solve analytically, so we can use numerical methods to find an approximate value. Using a numerical integration method such as Simpson's rule, we get:

≈ 0.8404

Therefore, the line integral of f(x, y) = √y/x along the curve r(t) = t³i + [tex]t^4[/tex]j, ½ ≤ t ≤ 1 is approximately 0.8404.

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Answer: The answer is 2 and 3 or B and C which ever way you want it.

The reason its 2and3 is because you can see its 60 degree angle.

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The reason its 3 also is because they are all congruent and its the only other right answer that fits.

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Hence, The answer is 2 and 3 or B and C.

Step-by-step explanation: Please give Brainliest.

Hope this helps!!!!

I can answer more questions if you want.

a. With 99% confidence, the difference in proportions is between -0.023 and 0.223. b. 99% of the confidence intervals will contain the true population proportion, and about 1% will not contain the true population difference in proportions.

a. To construct a confidence interval for the difference in proportions, we can use the following formula:

CI = (p1 - p2) ± zα/2 * √((p1*(1-p1)/n1) + (p2*(1-p2)/n2))

where:

p1, p2 = proportion of MCC, FSU students who believe they can achieve the American dream

n1, n2 = sample size of MCC, FSU students

zα/2 = critical value from the standard normal distribution for a 99% confidence level, which is 2.576

So,

CI = (0.75 - 0.65) ± 2.576 * √((0.75*(1-0.75)/100) + (0.65*(1-0.65)/100))

CI = 0.10 ± 0.123

CI = (−0.023, 0.223)

Therefore, with 99% confidence, the difference in proportions of MCC and FSU students who believe they can achieve the American dream is between -0.023 and 0.223.

b. Approximately 99% of these confidence intervals will contain the true population proportion of the difference in proportions of MCC students, and FSU students who believe they can achieve the American dream.

And about 1% will not contain the true population difference in proportions. This is because we constructed a 99% confidence interval.

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The area of the equilateral triangle that has an apothem of 14cm and a side length of 48.5 cm is 1019.25 square centimeters.

An equilateral triangle is a triangle in which all sides are equal and all angles are 60 degrees. The apothem of an equilateral triangle is the perpendicular distance from the center of the triangle to one of its sides.

To find the area of the equilateral triangle, we can use the formula:

Area = (1/2) x apothem x perimeter

where perimeter is the sum of the lengths of all three sides of the triangle.

In this case, the apothem is given as 14 cm and the side length is given as 48.5 cm. Since the triangle is equilateral, all three sides are equal to 48.5 cm.

Therefore, the perimeter of the triangle is:

Perimeter = 3 x 48.5 cm = 145.5 cm

Now we can substitute the values of the apothem and perimeter into the formula for the area:

Area = (1/2) x 14 cm x 145.5 cm = 1019.25 cm²

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Answer:

6 is the middle point of AD

There is no significant difference between the mean tensile strength of the silicone rubber at different curing temperatures. we should be cautious in interpreting the results of the ANOVA test and consider other factors such as the practical significance of the differences in tensile strength.

(a) To test the hypothesis that curing temperatures affect the tensile strength of the silicone rubber, we can perform an analysis of variance (ANOVA). ANOVA is used to compare the means of two or more groups and determine if there is a significant difference between them.

We can use the following null and alternative hypotheses:

Null Hypothesis (H0): The mean tensile strength of the silicone rubber is the same at all curing temperatures.

Alternative Hypothesis (HA): The mean tensile strength of the silicone rubber is different at least for one curing temperature.

We can use the ANOVA test to determine if there is a significant difference between the means of the groups. The ANOVA table is shown below:

Source | SS | df | MS | F

Between | 0.014 | 2 | 0.007 | 3.06

Within | 0.118 | 18 | 0.007

Total | 0.132 | 20

The F-statistic for the ANOVA is 3.06 and the p-value is 0.068. Since the p-value is greater than 0.05, we fail to reject the null hypothesis. Therefore, we can conclude that there is no significant difference between the mean tensile strength of the silicone rubber at different curing temperatures.

(b) The box plots of the data are shown below:

       +-----+-----+-----+

       |  25 |  40 |  55 |

       +-----+-----+-----+

       |     |     | 2.03|  -

       |     | 2.22|     |  -

       |     |     | 2.07|  -

       | 2.09| 2.14|     |  -

       | 2.10| 2.18|     |  -

       |     |     | 2.03|  -

       | 2.01| 2.11| 2.15|  -

       |     | 2.05|     |  -

       | 2.02| 2.18|     |  -

       +-----+-----+-----+

The box plots do not support the conclusion of the ANOVA test. The box plots show that the median and interquartile range for the tensile strength at 55°C are lower than those at 25°C and 40°C. This suggests that the mean tensile strength at 55°C may be lower than at 25°C and 40°C. However, the ANOVA test failed to reject the null hypothesis of equal means. This discrepancy may be due to the small sample size and the variability of the data. Therefore, we should be cautious in interpreting the results of the ANOVA test and consider other factors such as the practical significance of the differences in tensile strength.

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The angle measurement of the triangle would be 9. 59 degrees

The different trigonometric identities are given as;

The ratios of the trigonometric identities are represented as;

sin θ = opposite/hypotenuse

cos θ = adjacent/hypotenuse

tan θ = opposite/adjacent

From the information given, we have;

Opposite side = 1

Hypotenuse side = 6

Using the sine identity, we have;

sin θ = 1/6

Divide the values

sin θ = 0. 1666

Find the inverse of the sine

θ = 9. 59 degrees

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The area of the arrow of the painting is A = 70 units²

Given data ,

Mr. Barth is painting an arrow on the school parking lot.

The coordinates are (-2, 2), (5, 2), (5, 6), (12, 0), (5, -6), (5, -2), (-2, -2)

The area of the arrow would be:

Area of Arrow = Area of Triangle + Area of Rectangle

Let the base of the triangle be = 12 units

Let the height of the triangle is = 7 units

So , area of triangle = 42 units²

Area of rectangle = 7 x 4 = 28 units

Hence , the area of arrow A = 70 units²

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An inequality that represents the situation is s > 140.

Let's use "s" to represent the number of milligrams of sodium in a serving.

Since a serving of food does not qualify as low sodium if it contains more than 140 milligrams of sodium, we can write the inequality:

s > 140

This inequality reads "s is greater than 140", indicating that any value of "s" that is greater than 140 milligrams of sodium per serving does not qualify as low sodium.

To graph this inequality, we can represent "s" on the vertical axis and mark the value of 140 with a dashed line.

Since the inequality is greater than 140, we shade the area above the line to represent all the possible values of "s" that do not qualify as low sodium.

The resulting graph would look like given in the attached image.

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The complete question:

Write and graph an inequality that represents the number of sodium 's' in a serving that does not qualify as low sodium.

For a food to be labeled low sodium, there must be no more than 140 milligrams of sodium per serving.

If f(x)= - 3(x - m)2 + p Parabola vertical point T(2,5), then m + p is equal to 27.

Since the given parabola is vertical and has a vertex at T(2,5), we know that the equation is of the form f(x) = a(x-2)^2 + 5, where a is a constant.

We also know that f(x) = -3(x-m)^2 + p, which is in the same form as the first equation.

So, we can equate the two equations and get:

a(x-2)^2 + 5 = -3(x-m)^2 + p

Expanding the squares, we get:

a(x^2 - 4x + 4) + 5 = -3(x^2 - 2mx + m^2) + p

Simplifying and collecting like terms, we get:

ax^2 + (-4a + 6m)x + (4a - 3m^2 + p - 5) = 0

Since this equation must hold for all values of x, the coefficients of x^2 and x must be equal to zero.

Therefore, we have:

a = -3    (from the given equation f(x) = -3(x-m)^2 + p)
-4a + 6m = 0    (from the equation above)
-4(-3) + 6m = 0
12 + 6m = 0
m = -2

Substituting m = -2 and a = -3 into the equation above, we get:

4a - 3m^2 + p - 5 = 0

4(-3) - 3(-2)^2 + p - 5 = 0

-12 - 12 + p - 5 = 0

p = 29

Therefore, m + p = -2 + 29 = 27.

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The ratio of the area of a regular hexagon to the area of an equilateral triangle is 3/2.

The area of a regular hexagon is given by:

A[tex]_{H}[/tex] = (3√3)/2 · a²

where a is the length of the side of the hexagon.

a = 1 unit:

A[tex]_{H}[/tex] = (3√3)/2 · 1²

A[tex]_{H}[/tex] = (3√3)/2 unit²

The area of an equilateral triangle is given by:

A = (√3)/4 · b²

where b is the length of the side of the triangle.

b = 2 units:

A = (√3)/4 · 2²

A = (√3)/4 · 4

A = √3 unit²

ratio = A[tex]_{H}[/tex]/A

ratio = [(3√3)/2] / √3

ratio = 3/2  

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Answer:

Step-by-step explanation:

1/2*b*h

1/2*4*5

4/2*5

2*5

10

The value of c2 in the notation of the text is 4mk. The units of c2 are Ns/m.
To find the position of the mass after t seconds, we need to solve the differential equation:
m d^2x/dt^2 + c dx/dt + kx = 0
where m is the mass of the spring, c is the damping constant, k is the spring constant, and x is the position of the mass.
We can write the solution to this equation in the general form:
x(t) = G1eat cos( Bt) + ce" sin(St)
where a and B are constants that depend on the initial conditions of the system, and G1 and c are constants determined by those initial conditions.
To find the constants G1 and c, we need to use the initial conditions given in the problem: the spring is stretched 0.5 meters beyond its natural length and then released with zero velocity. This means that x(0) = 0.5 and dx/dt(0) = 0.

Substituting these initial conditions into the general solution, we get:
x(t) = G1e^(-ct/2m) cos( ωt) + c e^(-ct/2m) sin( ωt)
where ω = sqrt(k/m - c^2/4m^2) is the angular frequency of the motion.
To find the constants G1 and c, we differentiate x(t) with respect to time and use the initial condition dx/dt(0) = 0:
dx/dt = -G1c/2m e^(-ct/2m) sin( ωt) + c e^(-ct/2m) cos( ωt)
dx/dt(0) = 0 = c
Therefore, c = 0.
To find G1, we use the initial condition x(0) = 0.5:
x(0) = G1 cos(0) + 0 = G1 = 0.5
Therefore, the position of the mass after t seconds is:
x(t) = 0.5e^(-ct/2m) cos( ωt)
where c = 0 and ω = sqrt(k/m).
Plugging in the given values, we get:
x(t) = 0.5e^(-0t/2*7) cos( sqrt(40/7)t) = 0.5cos(2.226t)
So the position of the mass after t seconds is 0.5cos(2.226t) meters.

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The interquartile range (IQR) is a more appropriate measure of dispersion than the range for the chess club's tournament play data, as it considers the middle 50% of the data and gives a better representation of the overall variability in the distribution.

When we are interested in describing the variability or spread of data distribution, we typically use a measure of dispersion or spread. The most commonly used measures of dispersion are the range, the interquartile range (IQR), variance, and standard deviation.

In the case of the chess club's tournament play, we have a list of the number of games won by each member. To calculate the range, we simply subtract the minimum value from the maximum value. However, the range is a very crude measure of dispersion because it only considers the two extreme values and ignores the rest of the data.

A more appropriate measure of dispersion, in this case, would be the interquartile range (IQR), which is defined as the difference between the 75th percentile and the 25th percentile of the data. The IQR gives us a better sense of the spread of the middle 50% of the data, which is more representative of the overall variability in the data distribution.

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The number of possible outcomes is 10 times with five heads and no two heads adjacent to each other is 6.

To determine the number of possible outcomes of tossing a coin 10 times that contain five heads with no two heads adjacent to each other, we can follow these steps:

1. Since there are 10 tosses, we need to place 5 heads (H) and 5 tails (T) in a sequence.
2. To ensure no two heads are adjacent, we must place each head in between tails, which creates 6 possible positions for the heads (i.e., _T_T_T_T_T_).
3. Now, we need to distribute 5 heads into these 6 positions. This is a combination problem, so we'll use the formula for combinations:

C(n, k) = n! / (k! * (n-k)!),

where n = the number of available positions, k = the number of heads, and ! denotes factorial.
4. Applying the formula:

C(6, 5) = 6! / (5! * (6-5)!)

= 6! / (5! * 1!)

= 720 / (120 * 1)

= 6.

So, the number of possible outcomes of tossing a coin 10 times with five heads and no two heads adjacent to each other is 6.

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