The table shows all of the possible outcomes of rolling two dice. If Jerome were to roll two dice 48 times, about how many times could he expect to roll doubles?.

Side lengths:
The length of a side of a triangle can be calculated using the distance formula, which is:

distance = √((x2 - x1)^2 + (y2 - y1)^2 + (z2 - z1)^2)

Using this formula, we can calculate the length of each side of the triangle:

- PQ: √((7 - 1)^2 + (7 - 3)^2 + (9 + 7)^2) = √(36 + 16 + 256) = √308 ≈ 17.55
- QR: √((-3 - 7)^2 + (-3 - 7)^2 + (1 - 9)^2) = √(100 + 100 + 64) = √264 ≈ 16.25
- RP: √((1 + 3)^2 + (3 + 3)^2 + (-7 - 1)^2) = √(16 + 36 + 64) = √116 ≈ 10.77

Therefore, the lengths of the sides are: 17.55, 16.25, and 10.77.

Angles:
To find the angles of a triangle, we can use the Law of Cosines, which states that:

c^2 = a^2 + b^2 - 2ab cos(C)

where c is the length of the side opposite angle C, and a and b are the lengths of the other two sides.

Using this formula, we can find each angle:

- Angle P: cos(P) = (17.55^2 + 10.77^2 - 16.25^2) / (2 * 17.55 * 10.77) = 0.3096, so P = arccos(0.3096) ≈ 1.260 radians
- Angle Q: cos(Q) = (16.25^2 + 10.77^2 - 17.55^2) / (2 * 16.25 * 10.77) = 0.0352, so Q = arccos(0.0352) ≈ 1.535 radians
- Angle R: cos(R) = (16.25^2 + 17.55^2 - 10.77^2) / (2 * 16.25 * 17.55) = 0.8342, so R = arccos(0.8342) ≈ 0.555 radians

Therefore, the angles of the triangle are: 1.260, 1.535, and 0.555 radians.


To calculate the side lengths of the triangle, we used the distance formula to find the distance between each pair of vertices. This formula calculates the distance between two points in three-dimensional space.

To calculate the angles of the triangle, we used the Law of Cosines, which relates the length of each side of a triangle to the cosine of its opposite angle. This formula allows us to find the angle opposite a given side, given the lengths of the other two sides.


The lengths of the sides of the triangle are 17.55, 16.25, and 10.77, and the angles are 1.260, 1.535, and 0.555 radians.

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All correct proportions include the following:

A. \(\frac{AC}{CE} =\frac{BD}{DF}\)

D. \(\frac{CE}{DF} =\frac{AE}{BF}\)

In Mathematics and Geometry, two geometric figures are said to be similar when the ratio of their corresponding side lengths are equal and their corresponding angles are congruent.

Hence, the lengths of the pairs of corresponding sides or corresponding side lengths are proportional to one another when two (2) geometric figures are similar.

Since line segment AB is parallel to line segment CD and parallel to line segment EF, we can logically deduce that they are congruent because they can undergo rigid motions. Therefore, we have the following proportional side lengths;

\(\frac{AC}{CE} =\frac{BD}{DF}\)

\(\frac{CE}{DF} =\frac{AE}{BF}\)

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Answer:

x² + (y + 3)² = 25

Step-by-step explanation:

the centre (C) of the circle is at the midpoint of the diameter.

using the midpoint formula

midpoint = ( \(\frac{x_{1}+x_{2} }{2}\) , \(\frac{y_{1}+y_{2} }{2}\) )

with (x₁, y₁ ) = P (3, 1 ) and (x₂, y₂ ) = Q (- 3, - 7 )

C = ( \(\frac{3-3}{2}\) , \(\frac{1-7}{2}\) ) = ( \(\frac{0}{2}\) , \(\frac{-6}{2}\) ) = (0, - 3 )

the radius r is the distance from the centre to either P or Q

using the distance formula

r = \(\sqrt{(x_{2}-x_{1})^2+(y_{2}-y_{1})^2 }\)

with (x₁, y₁ ) = C (0, - 3 ) and (x₂, y₂ ) = P (3, 1 )

r = \(\sqrt{(3-0)^2+(1-(-3)^2}\)  

 = \(\sqrt{3^2+(1+3)^2}\)

 = \(\sqrt{3^2+4^2}\)

 = \(\sqrt{9+16}\)

 = \(\sqrt{25}\)

 = 5

the equation of a circle in standard form is

(x - h)² + (y - k)² = r²

where (h, k ) are the coordinates of the centre and r is the radius

here (h, k ) = (0, - 3 ) and r = 5 , then

(x - 0 )² + (y - (- 3) )² = 5² , that is

x² + (y + 3)² = 25

hi

(y-7)  (y+6) =   y²+6y -7y -42

                  =    y² -y -42

The area of the shaded region of the figure is 100.53 square cm

The area of the shaded part is solved by finding the area of the two parts and subtracting

This can be done using the formula below

area of the shaded part = π(R² - r²)

Where 'R = 4 cm + 2 cm = 6 cm

r = 2 cm

plugging in the values we have

area of the shaded part = π(R² - r²)

area of the shaded part = π(6² - 2²)

area of the shaded part = π(36 - 4)

area of the shaded part = 32π

area of the shaded part = 100.53 square cm

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To find the change in population of insects between day 0 and day 3, we need to calculate the population at both times and subtract them.

Using the given rate equation, we can calculate the population at day 0 and day 3 as follows:
At day 0 (t=0):
Population = 200(10^0) + 13(0^2) = 200
At day 3 (t=3):
Population = 200(10^3) + 13(3^2) = 20,130
Therefore, the change in population between day 0 and day 3 is:
20,130 - 200 = 19,930
So the population of insects increased by 19,930 between day 0 and day 3.

To find the change in the population of insects between day 0 and day 3 with the given rate of 200 + 10t + 13t^2, follow these steps:
1. Plug in t = 0 (day 0) into the rate equation: 200 + 10(0) + 13(0)^2 = 200
2. Plug in t = 3 (day 3) into the rate equation: 200 + 10(3) + 13(3)^2 = 200 + 30 + 117 = 347
3. Subtract the day 0 population from the day 3 population: 347 - 200 = 147
The change in the population of insects between day 0 and day 3 is 147.

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To calculate the coverage Cameron can get for the same premium rate, we need to consider the remaining term and his current age.

Cameron had a 30-year term policy, but he let it expire. Since he is now 40 years old, the remaining term would be 30 - 40 = 10 years. Next, we calculate the total premium paid for the original policy. Cameron paid $18.20 per month, so the total premium paid over 30 years is 18.20 * 12 * 30 = $6,552.

To find the coverage he can get for the same premium rate, we divide the total premium paid by the remaining term:

Coverage = Total premium paid / Remaining term

Coverage = $6,552 / 10 = $655.20 per year

To calculate the coverage percentage loss, we compare the new coverage amount with the original coverage:

Coverage loss percentage = ((Original coverage - New coverage) / Original coverage) * 100

Coverage loss percentage = (($325,000 - $655.20) / $325,000) * 100

The coverage loss percentage will depend on the exact calculations above, but it will provide the percentage by which the coverage has decreased compared to the original policy.

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Answer:

65

Step-by-step explanation:

sin y =12/13 but can't find cos x as sin x can't exceed from 1.

It can be solved by using identities of trigonometry.

What are the  trigonometric identities?

These are the equivalent ways to define trigonometric functions.

Basic identities are:

sin²x + cos²x = 1

1 + tan²x = sec²x

1 + cot²x = cosec²x

For cos x:
cos²x = 1 - sin²x

          = 1 - (5/3)²

          = 1- (25/9)

          = -16/9

i.e. not possible

For sin y:

sin²y = 1 - cos²y

         = 1 - (5/13)²

         = 1 - 25/169

         = 144/169

So, sin y = ±(12/13)

but as mentioned x and y are in 1st quadrant ,

so, sin y = 12/13.

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In order to pull a 2-cm nail with a resistance of 1500 N out of a piece of wood, a carpenter exerted a force of 250 N using a claw hammer. The amount of work done by the carpenter to pull out the nail can be calculated using the formula W = F x d.

The work done by the carpenter to pull out the nail can be calculated using the formula W = F x d, where W is work, F is force, and d is distance. In this case, the carpenter exerted a force of 250 N to pull the nail out of the wood, and the nail traveled a distance of 2 cm. To use this formula, we need to convert the distance into meters, which gives us 0.02 m. Plugging in the values, we get W = 250 N x 0.02 m = 5 J.

This calculation tells us that the carpenter did 5 Joules of work to remove the nail. This work was necessary to overcome the resistance of the nail, which was 1500 N. Without the application of sufficient force, the nail would not have budged from its position in the wood. The carpenter's claw hammer provided the necessary force to extract the nail, and the amount of work done was directly proportional to the amount of force exerted and the distance the nail traveled. In this case, the carpenter was able to apply sufficient force to pull the nail out of the wood and complete the task at hand.

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1. The given statement is False.

2. The given statement is False.

3. The given statement is True.

False. The CLT states that the sample mean approaches a normal distribution as the sample size increases, regardless of the population distribution being sampled from.

False. An increasing trend in a time plot suggests that there is a relationship between the observations, but it does not necessarily imply that they are not iid. Additional tests or analyses would be needed to confirm or refute independence.

Correct. μ is defined as the population means salary for statisticians who have limited work experience, which is the parameter being tested in the hypotheses.

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If the average service rate is 6 customers per hour, and assuming the negative exponential distribution is used to describe the randomness of the service time distribution. The probability that the service time will be less than or equal to 6 minutes is 0.549 [B]

The average service rate of 6 customers per hour corresponds to a mean service time of 1/6 = 0.1 hour = 6 minutes. To find the probability that the service time will be less than or equal to 6 minutes, we need to find the cumulative distribution function (CDF) of the exponential distribution evaluated at 6 minutes. The CDF of the exponential distribution is given by:

F(x) = 1 - exp(-λx)

where,

λ = 1/mean = 1/0.1 = 10. Plugging in x = 6, we get:

F(6) = 1 - exp(-10 * 6) = 1 - exp(-60) ≈ 0.549

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In mathematical terms, if X is the sum of successes in n different Bernoulli trials, and p is the likelihood of success in each trial, then X may be represented as a normal distribution with mean increasing as n rises.

μ = np

variance σ^2 = np(1-p)

The binomial distribution is a discrete probability distribution that produces just two outcomes in an experiment: success or failure. In a binomial distribution, the chance of success must remain constant during the trials under consideration. For example, because there are only two possible outcomes when tossing a coin, the chance of flipping a coin is 1/2 or 0.5 for each experiment we do.

Here,

When the sample size (n) is high, the distribution of the total of successes (k) in n separate Bernoulli trials approaches a normal distribution, according to the normal approximation for the binomial distribution.

In mathematical terms, if X is the total of successes in n separate Bernoulli trials, with p being the chance of success in each trial, then X may be represented by a normal distribution with mean as n increases.

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Answer:

above sea level

Step-by-step explanation:

The frequency for the number of computer science students enrolled in Stat 155 is 127, and the relative frequency is approximately 0.552 (to 3 decimal places).

In Stat 155, there are 230 students enrolled, and 127 of them are majoring in computer science. The frequency for the number of computer science students enrolled in Stat 155 is 127. To find the relative frequency, divide the frequency by the total number of students:

Relative frequency = (Number of computer science students) / (Total number of students)
Relative frequency = 127 / 230 ≈ 0.552

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The probability of observing a sample mean between 84 and 86 when n = 9 is approximately 0.5878.

To calculate p(84 ≤ x ≤ 86) when n = 9, we first need to determine the distribution of the sample mean. Since the sample size is n = 9, we can use the central limit theorem to assume that the distribution of the sample mean is approximately normal with mean μ = 85 and standard deviation σ = σ/√n = σ/3, where σ is the population standard deviation.

Next, we need to standardize the values of 84 and 86 using the formula z = (x - μ) / (σ / √n). Plugging in the values, we get:

z(84) = (84 - 85) / (σ/3) = -1 / (σ/3)
z(86) = (86 - 85) / (σ/3) = 1 / (σ/3)

To calculate the probability between these two z-scores, we can use a standard normal table or a calculator with a normal distribution function. The probability can be expressed as:

P(-1/σ ≤ Z ≤ 1/σ) = Φ(1/σ) - Φ(-1/σ)

where Φ is the cumulative distribution function of the standard normal distribution.

Therefore, to calculate p(84 ≤ x ≤ 86) when n = 9, we need to determine the value of σ and use the formula above. If σ is known, we can plug in the value and calculate the probability. If σ is unknown, we need to estimate it using the sample standard deviation and replace it in the formula.

For example, if the sample standard deviation is s = 2, then σ = s * √n = 2 * √9 = 6. Plugging in this value in the formula, we get:

P(-1/6 ≤ Z ≤ 1/6) = Φ(1/6) - Φ(-1/6) = 0.2061 - 0.7939 = 0.5878

Therefore, the probability of observing a sample mean between 84 and 86 when n = 9 is approximately 0.5878.

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Answer:

Step-by-step explanation:

The probability of observing a sample mean between 84 and 86 when n = 9 is approximately 0.5878.

To calculate p(84 ≤ x ≤ 86) when n = 9, we first need to determine the distribution of the sample mean. Since the sample size is n = 9, we can use the central limit theorem to assume that the distribution of the sample mean is approximately normal with mean μ = 85 and standard deviation σ = σ/√n = σ/3, where σ is the population standard deviation.

Next, we need to standardize the values of 84 and 86 using the formula z = (x - μ) / (σ / √n). Plugging in the values, we get:

z(84) = (84 - 85) / (σ/3) = -1 / (σ/3)

z(86) = (86 - 85) / (σ/3) = 1 / (σ/3)

To calculate the probability between these two z-scores, we can use a standard normal table or a calculator with a normal distribution function. The probability can be expressed as:

P(-1/σ ≤ Z ≤ 1/σ) = Φ(1/σ) - Φ(-1/σ)

where Φ is the cumulative distribution function of the standard normal distribution.

Therefore, to calculate p(84 ≤ x ≤ 86) when n = 9, we need to determine the value of σ and use the formula above. If σ is known, we can plug in the value and calculate the probability. If σ is unknown, we need to estimate it using the sample standard deviation and replace it in the formula.

For example, if the sample standard deviation is s = 2, then σ = s * √n = 2 * √9 = 6. Plugging in this value in the formula, we get:

P(-1/6 ≤ Z ≤ 1/6) = Φ(1/6) - Φ(-1/6) = 0.2061 - 0.7939 = 0.5878

Therefore, the probability of observing a sample mean between 84 and 86 when n = 9 is approximately 0.5878.

We do not have enough evidence to conclude that the average height of 7th graders has increased.

Null Hypothesis (H0): The average height of 7th graders has not increased (μ <= 145 cm)

Alternative Hypothesis (Ha): The average height of 7th graders has increased (μ > 145 cm)

We will use a one-tailed test since we are testing for an increase in height.

We need to compute the test statistic to make a decision.

z = (X - μ) / (σ / √n)

where X is the sample mean, μ is the population mean, σ is the population standard deviation, and n is the sample size.

Here, X = 147 cm, μ = 145 cm, σ = 20 cm, and n = 200.

z = (147 - 145) / (20 / √200) = 1.41

Using a one-tailed table with a significance level of 0.01, the critical value is 2.33.

Since the calculated test statistic (1.41) is less than the critical value (2.33), we fail to reject the null hypothesis.

Therefore, we do not have enough evidence to conclude that the average height of 7th graders has increased.

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It's 38*23 =874 dollars

Answer:

21 is your answer

Step-by-step explanation:

The combination formula is used to calculate the number of ways to choose r items from a set of n items without regard to order. The formula is:

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

where n is the total number of items and r is the number of items to be chosen.

Using this formula, we can solve the problem when n = 7 and r = 5 as follows:

C(7,5) = 7! / (5! * (7-5)!)

= (7 x 6 x 5 x 4 x 3 x 2 x 1) / [(5 x 4 x 3 x 2 x 1) x (2 x 1)]

= (7 x 6) / (2 x 1)

= 21

Therefore, there are 21 ways to choose 5 items from a set of 7 items without regard to order.

Answer:

D. a=1 and b=4

Step-by-step explanation:

x^2-2x+(-2/2)^2 =3+(-2/2)^2

x^2-2x+(-1)^2=3+(-1)^2

x^2-2x +1 =4

(x-1)^2=4

Therefore a=1 and b=4

The equilibrium constant Kp for this reaction is equal to 0.105 atm. The balanced chemical equation for the given reaction is: CaCO3(s) ⇌ CaO(s) + CO2(g)The equilibrium pressure

P = 0.105 atmThe temperature, T = 350°C To calculate the equilibrium constant Kp for the reaction, we need to use the partial pressure of the gases involved at equilibrium. In this case, we have only one gas, which is carbon dioxide (CO2).

The balanced equation for the reaction is:

CaCO3 (s) ⇌ CaO (s) + CO2 (g)

Given: Pressure at equilibrium (P) = 0.105 atm

Since there is only one gas in the reaction, the equilibrium constant Kp can be calculated as follows:

Kp = P(CO2)

Therefore, Kp = 0.105 atm.

The equilibrium constant Kp for this reaction is equal to 0.105 atm.

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The answer is:

Work/explanation:

First, use the distributive property and distribute 3 through the parentheses:

\(\sf{3(2a+6)}\)

\(\sf{6a+18}\)

Now we can plug in 4 for a:

\(\sf{6(4)+18}\)

\(\sf{24+18}\)

\(\bf{42}\)

Here's how you can use the Trapezoidal Rule to approximate the integral of the function \(f(x) = \frac{1}{{(25+x^2)}^{\frac{3}{2}}}\) from -2 to 2 in MATLAB:

```matlab

a = -2;    % Lower limit

b = 2;     % Upper limit

n = 1000;  % Number of subintervals (increase for higher accuracy)

h = (b - a) / n;   % Step size

x = a:h:b;         % Generate evenly spaced x values

y = 1 ./ (25 + x.^2).^1.5;  % Evaluate the function at x

approximation = h * (sum(y) - (y(1) + y(end)) / 2);  % Trapezoidal Rule approximation

fprintf('Approximation: %.6f\n', approximation);

```

1. We define the lower limit `a` as -2, the upper limit `b` as 2, and the number of subintervals `n` as 1000 (you can adjust `n` for higher accuracy).

2. We calculate the step size `h` by dividing the range (`b - a`) by the number of subintervals (`n`).

3. We generate an array `x` of evenly spaced values from `a` to `b` using the step size `h`.

4. We evaluate the function `f(x)` at each point in `x` and store the results in the array `y`.

5. Finally, we use the Trapezoidal Rule formula to approximate the integral by summing the values in `y` and adjusting for the endpoints, multiplying by the step size `h`.

The Trapezoidal Rule approximation for the integral of the function \(f(x) = \frac{1}{{(25+x^2)}^{\frac{3}{2}}}\) from -2 to 2 is the value calculated using the MATLAB code above.

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Answer:

After expansion and simplifying the expression gives;

Explanation:

Given the expression;

Expanding using Distributive Property;

simplifying;

Therefore after expansion and simplifying the expression gives;

The point on the graph that shows the price of 2 ounces of medicine, given the formula, is 4 currency units.

The graph shows the relationship between the ounces of medicine and the price of those ounces in the formula y = 2 x. The value of x in this instance, is the number of ounces of medicine to be sold.

The value of y would be the price of those two ounces to be sold. The formula, y = 2 x means that the price of a single ounce of medicine, is 2 currency units.

For 2 ounces therefore, the price is:

y = 2 x

y = 2 x 2

y = $ 4

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The line that is tangent to both parabolas at this point will have the slope 2x = 2 * 0 = 0 and y-intercept equal to 0, giving us the equation y = 0.

To graph the two parabolas y = x^2 and y = -x^2 + 2x - 5 in the same coordinate plane, we can use a tool such as a graphing calculator or plot the points of the parabolas by hand. The first parabola, y = x^2, will be a downward-facing parabola centered at the origin with a vertex at (0,0), while the second parabola, y = -x^2 + 2x - 5, will be an upward-facing parabola that is shifted to the right by 1 unit and downward by 5 units.

Next, we want to find the lines that are simultaneously tangent to both parabolas. To do this, we find the derivative of each parabolic function, which gives us the slope of the tangent line at any point on the graph.

For the first parabola, y = x^2, the derivative is 2x, so the slope of the tangent line at any point (x, x^2) is 2x.

For the second parabola, y = -x^2 + 2x - 5, the derivative is 2x - 2, so the slope of the tangent line at any point (x, -x^2 + 2x - 5) is 2x - 2.

To find the equations of the two lines simultaneously tangent to both parabolas, we set the slopes equal to each other and solve for x. This gives us the x-coordinate of the points of tangency on both parabolas, and from there we can use the original functions to find the y-coordinate.

For example, setting the slopes equal to each other, we get:

2x = 2x - 2

0 = -2

So, x = 0, and this point (0, 0) lies on both parabolas. Therefore, the line that is tangent to both parabolas at this point will have the slope 2x = 2 * 0 = 0 and y-intercept equal to 0, giving us the equation y = 0.

We can repeat this process to find the equation of the second tangent line.

Note: There may be other points of tangency that are not shown on the graph, and these can be found using the same process.

Therefore, the line that is tangent to both parabolas at this point will have the slope 2x = 2 * 0 = 0 and y-intercept equal to 0, giving us the equation y = 0.

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The distance between the given coordinates  (−5, −19) and (−5, 32) is given by 51 units.

Let us consider the coordinates of two given points be,

(x₁ , y₁ ) = ( -5 , -19 )

(x₂ , y₂ ) = (-5 , 32 )

Distance formula between two points is equals to,

Distance = √ ( y₂ - y₁)² + ( x₂ - x₁ )²

Substitute the values of the coordinates we have,

⇒ Distance = √ ( 32 - (-19))² + ( -5- (-5) )²

⇒ Distance = √ (32 +19)² + (-5 + 5)²

⇒ Distance = √ 51² + 0²

⇒ Distance =  51 units.

Therefore, the distance between the two points is equal to 51 units.

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