The length of a diagonal of a square is 37√2 millimeters. Find the perimeter of the square

(a)There's solid prove to demonstrate that the standard deviation of gap breadth surpasses 0.01 millimeters. (b) Employing a control calculator or computer program, ready to decide that a test measure of roughly 44 is required to realize a control of at slightest 0.8 to detect a 50% increment in standard deviation at a centrality level of 0.1. (c) Assuming the same noteworthiness level of 0.1, ready to utilize a control calculator or program to discover that a test measure of around 22 is required to attain a control of at slightest 0.8.

(a) To test in case the standard deviation of gap distance across surpasses 0.01 millimeters, we are able utilize a one-tailed t-test with a noteworthiness level of 0.1. The invalid speculation is that the standard deviation is less than or rise to to 0.01 millimeters, and the elective theory is that the standard deviation is more noteworthy than 0.01 millimeters. We expect that the basic dispersion of the gap breadths is around ordinary.

Utilizing the equation for the t-test, we get:

[tex]t = (s / \sqrt{} (n-1)) / (0.01)[/tex]

[tex]t = (0.008 / \sqrt{} (14)) / (0.01)[/tex]

t = 2.26

The degrees of opportunity for this test is n-1 = 14. From a t-distribution table, we discover that the p-value for a one-tailed test with 14 degrees of opportunity and t=2.26 is roughly 0.021. Since the p-value is less than the noteworthiness level of 0.1, we dismiss the invalid speculation.

(b) To discover the likelihood that the test in part (a) will identify a 50% increment in standard deviation, we have to be calculate the control of the test. The control of a test is the likelihood of dismissing the invalid theory when the elective theory is genuine.

The control of the test depends on a few components, counting the test measure, the noteworthiness level, and the impact measure. In this case, the effect size is the contrast between the actual standard deviation and the hypothesized esteem, communicated in standard deviation units.

(c) If the real standard deviation is 0.0125 millimeters and we need to distinguish this with a control of at least 0.8, we ought to decide the test measure required for the test. Assuming the same noteworthiness level of 0.1, ready to utilize a control calculator or program to discover that a test measure of around 22 is required to attain a control of at slightest 0.8.

We have utilized a one-tailed t-test to decide that there's solid prove to show that the standard deviation of gap breadth surpasses 0.01 millimeters. We have too calculated the control of the test to distinguish a 50% increment in standard deviation and the test measure required to distinguish a standard deviation of 0.0125 millimeters with a control of at slightest 0.8.

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The correct statement regarding the end behavior of the graph is given as follows:

C. As x approaches positive infinity, D(x) approaches negative infinity.

The end behavior of a function is given by the limit of the function is the input x goes to either negative infinity or positive infinity.

For this problem, the function is a quadratic function with negative leading coefficient, meaning that it will approach negative infinity when x approaches negative infinity and when x approaches positive infinity.

This means that the correct option is given by option C.

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The arc length of a semicircle with a diameter of 9 is approximately 14.13 units.

The arc length of a semicircle is half of the circumference of a full circle with the same radius. Therefore, to find the arc length of a semicircle with a diameter of 9, we first need to find the radius. The radius is half the diameter, so it is 4.5.

The circumference of a full circle with a radius of 4.5 is 2πr, where r is the radius. Substituting r=4.5, we get:

C = 2π(4.5) = 9π

Therefore, the arc length of a semicircle with a diameter of 9 is half of 9π, or 4.5π. To find the numerical value of this arc length, we can use the approximation π ≈ 3.14:

arc length = 4.5π ≈ 4.5(3.14) = 14.13

So the arc length of a semicircle with a diameter of 9 is approximately 14.13 units.

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Point W has the polar coordinates negative five comma two pi over 3.

We have to given that;

To find the coordinate for the point (- 5, 2π/3).

Now, We can formulate;

Coordinates of W = (- 5, 2π/3).

Thus, The correct point which shows the polar coordinates negative five comma two pi over 3 is,

⇒ Point W

Therefore, Point W has the polar coordinates negative five comma two pi over 3.

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The equation that can be used to find x, the number of yards that Aj walks and runs each week is 2x = 400 + 90.

Let's assume that Aj runs x yards each week. Then, the number of yards that Aj walks each week would be (x + 90) yards, since he walks 90 more yards than he runs.

The total distance that Aj covers each week would be the sum of the distance that he runs and the distance that he walks, which is given as 400 yards.

So, we can write an equation as:

Distance covered by Aj = Distance that he runs + Distance that he walks

or,

Simplifying this equation gives:

Therefore, the equation that can be used to find x, the number of yards that Aj walks and runs each week is 2x = 400 + 90.

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The population mean is not equal to 72 at a 10% significance level.

a) The standard error of the mean is given by the formula:

SE = σ/√n

where σ is the population standard deviation, n is the sample size. Since the population standard deviation is not known, we use the sample standard deviation as an estimate. Therefore,

SE = s/√n = 1.3/√76 ≈ 0.149

b) We need to find the critical value of t* with 75 degrees of freedom (df = n-1) and a 90% confidence level. Using a t-table or calculator, we find that the critical value is approximately t* = ±1.663.

c) To construct the 90% confidence interval, we use the formula:

CI = X ± t*(SE)

where X is the sample mean, t* is the critical value, and SE is the standard error of the mean. Substituting the values, we get:

CI = 77 ± 1.663(0.149) = (76.739, 77.261)

Therefore, we are 90% confident that the true population mean μ lies within the interval (76.739, 77.261).

d) To test the hypothesis at α=0.100, we compare the confidence interval with the null hypothesis. If the null hypothesis falls outside the confidence interval, we reject it at the given level of significance.

Since 72 is not within the confidence interval of (76.739, 77.261), we can reject the null hypothesis at α=0.100. This means we have sufficient evidence to conclude that the population mean is not equal to 72 at a 10% significance level.

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The change in temperature of the charcoal from 180°F to 60°F is equal to a decrease of approximately 2.2 degrees Celsius.

To convert Fahrenheit to Celsius, we can use the formula:

Celsius = (Fahrenheit - 32) × 5/9

We know that the initial temperature of the charcoal was 180°F, and it experienced a temperature drop of 120°F. To find the final temperature in Fahrenheit, we can subtract 120°F from 180°F:

Final temperature in Fahrenheit = 180°F - 120°F = 60°F

Now, we can convert the final temperature from Fahrenheit to Celsius using the formula above:

Celsius = (60°F - 32) × 5/9

Celsius = (28°F) × 5/9

Celsius = -2.2222...

Rounding the result to one decimal place, we get:

Celsius = -2.2 degrees Celsius (approx.)

It's worth noting that the Celsius scale is based on the metric system, which is the standard measurement system used in most countries worldwide. In contrast, the Fahrenheit scale is primarily used in the United States and a few other countries, making it less universal. Understanding how to convert between the two scales is crucial in various scientific, engineering, and technical fields.

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Complete question:

A piece of charcoal that was initially at 180°F experiences a temperature drop of 120°F. Express this change of temperature in Celsius degrees.

The evaluation gives 8.

Quotient is division of two given integers; which is expressed as a fraction. It can be expressed in the form of either proper fraction or improper fraction.

Considering the given question, we have;

quotient of 25 and 5 = 25/ 5

Then increased by 3, we have;

25/5 + 3

find the LCM of the expression

25/5 + 3 = (25 + 15)/5

              = 40/5

              = 8

Therefore on evaluation, the final answer is 8.

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The  ratio of the volume of the cone to the volume of the cylinder is approximately 0.33 : 1 (rounded to the nearest hundredth).

The volume of the cone can be calculated using the formula:

V = (1/3)πr^2h

where r is the radius of the cone and h is the height of the cone. Substituting the given values, we get:

V = (1/3)π(3)^2(11) = 103.67 cubic feet

Therefore, the volume of the cone is 103.67 cubic feet (rounded to the nearest hundredth).

To find the radius of the cylinder, we can use the formula for the volume of a cylinder:

V = πr^2h

where r is the radius of the cylinder and h is the height of the cylinder. We are given that the volume of the cylinder is 310.86 cubic feet and that the height is 11 feet, so we can solve for r:

310.86 = πr^2(11)
r^2 = 310.86 / (11π)
r ≈ 2.3 feet

Therefore, the radius of the cylinder is approximately 2.3 feet (rounded to the nearest hundredth).

The ratio of the volume of the cone to the volume of the cylinder is the volume of the cone divided by the volume of the cylinder. Using the values we calculated, we get:

V(cone) / V(cylinder) = 103.67 / 310.86 ≈ 0.33 : 1

Therefore, the ratio of the volume of the cone to the volume of the cylinder is approximately 0.33 : 1 (rounded to the nearest hundredth).

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Answer:

The count of the basic operation in the algorithm for an instance of input size n=10² is: C(10²) = (10²)² log10 ((10²)³) = (10,000) log10 (1,000,000) ≈ 40,000

The total running time of the algorithm can be estimated using: T(n) = 0.9 * C(n) * C where C is the time taken by the basic operation.

In this case, C = 2ns or 2 x 10⁻⁹s. Substituting the values, we get: T(10²) = 0.9 * 40,000 * 2 x 10⁻⁹ = 7.2 x 10⁻⁶ s

Converting this to minutes, we get: 7.2 x 10⁻⁶ s * 1 min/60 s ≈ 1.2 x 10⁻⁷ min

Therefore, the estimated total running time of the algorithm to solve a problem instance with input size n=10² is approximately 1.2 x 10⁻⁷ minutes.

Step-by-step explanation:

The estimated total running time of the algorithm to solve a problem instance with input size n=10² is approximately 1.998 minutes.

To estimate the total running time of the algorithm, we need to calculate the number of times the basic operation is executed and multiply it by the time it takes to execute it.

First, let's calculate the number of times the basic operation is executed for an input size of n=10². We can do this by plugging n=100 into the equation for C(n):

C(100) = 100² log10 (100³)

C(100) = 10000 log10 (1000000)

C(100) = 10000 * 6

C(100) = 60000

So the basic operation is executed 60,000 times for an input size of n=10².

Next, let's calculate the time it takes to execute the basic operation:

C = 2ns

C = 2 * 10^-9 s

C = 2 * 10^-9 / 60 (converting to minutes)

C = 3.33 * 10^-11 min

Finally, we can estimate the total running time of the algorithm:

Total running time = basic operation time * number of times basic operation is executed

Total running time = 3.33 * 10^-11 min * 60,000

Total running time = 1.998 min

Therefore, the estimated total running time of the algorithm to solve a problem instance with input size n=10² is approximately 1.998 minutes.

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The requried, system of equations is y = 2x + 10 and y = 4.50x.

Let's use the variables x and y to represent the number of movies rented and the total cost, respectively. Then, the two plans can be represented by the following equations:

For Netflix members:

y = 2x + 10

For non-members:

y = 4.50x

In the first equation, the $10 represents the flat fee that is charged regardless of how many movies are rented, and the $2x represents the additional cost based on the number of movies rented.

In the second equation, the $4.50x represents the cost per movie for non-members.

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Approximately 35% of students prefer swimming as their preferred type of exercise. So, correct option is B.

To find the marginal relative frequency for students who prefer swimming as their preferred type of exercise, we need to add up the percentage of boys and girls who prefer swimming.

The percentage of boys who prefer swimming is 16% and the percentage of girls who prefer swimming is 19%.

So, the total percentage of students who prefer swimming is:

16% + 19% = 35%

Therefore, the marginal relative frequency for students who prefer swimming as their preferred type of exercise is 35%.

So, correct option is B.

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The total price Donte paid for the computer was $155.52.

The regular price of the computer was $180.

Donte got a 20% discount, which means he paid 100% - 20% = 80% of the regular price.

So, Donte paid 80% of $180, which is

(80/100) x $180 = $144.

Next, an 8% sales tax was added to the cost of the computer.

The amount of tax is

(8/100) x $144 = $11.52

Therefore, the total price Donte paid for the computer was

$144 + $11.52 = $155.52.

sales tax is a consumption tax imposed by the government on the sale of goods and services. A conventional sales tax is levied at the point of sale, collected by the retailer, and passed on to the government.

Sales tax is always a percentage of a product's value which is charged at the point of exchange or buy and is indirect.

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The instantaneous time rate of change of V after 15 minutes of draining is -200 L/min

To find the instantaneous time rate of change of V after 15 minutes of draining, we need to differentiate the given equation V = 6000(1 - t/30) with respect to time t and then evaluate the derivative at t=15.

1. Differentiate the equation with respect to t:
dV/dt = -6000(1/30)
dV/dt = -200 L/min

2. Evaluate the derivative at t=15:
dV/dt at t=15 is -200 L/min.

The instantaneous time rate of change of V after 15 minutes of draining is -200 L/min.

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An average wind velocity of approximately 5.8 m/s is required to produce 3.8 MW of electrical power with the given wind turbine specifications.

Solar PV farms: These are large-scale installations of solar panels that use photovoltaic technology to generate electricity from sunlight.

Concentrated solar thermal plants: These plants use mirrors or lenses to concentrate sunlight onto a receiver, which heats a fluid to produce steam that drives a turbine to generate electricity.

Wind energy farms: These are large-scale installations of wind turbines that convert the kinetic energy of wind into electrical energy.

Biogas plants: These plants use organic matter such as agricultural waste, food waste, or sewage to produce biogas, which can be burned to generate electricity or used as a fuel for transportation.

A 1.5 kW solar panel, working at full capacity for 5 hours, will produce 1.5 kW x 5 hours = 7.5 kWh (kilowatt-hours) of electrical energy.

1 watt-hour (Wh) = 3600 joules (J)

7500 Wh = 7500 x 3600 J = 27,000,000 J (27 million joules)

The power output of a wind turbine is given by:

P = (1/2) x (air density) x (blade area) x (wind velocity)^3 x (power coefficient)

where blade area = π x (blade length)^2, and power coefficient is a dimensionless efficiency factor that depends on the design of the turbine.

To produce 3.8 MW of electrical power, we have:

3.8 MW = 3,800 kW = 3,800,000 W

Assuming the electrical, gear, and generator efficiencies are all independent and multiply together, the total efficiency is 0.93 x 0.91 x 0.95 = 0.797

So, the mechanical power output of the turbine must be:

P_mech = P_elec / efficiency = 3,800,000 W / 0.797 = 4,769,064 W

Plugging in the given values and solving for wind velocity:

4,769,064 W = (1/2) x 1.17 kg/m³ x π x (49 m)^2 x (wind velocity)^3 x 0.46

wind velocity = (4,769,064 W / (0.5 x 1.17 kg/m³ x π x (49 m)^2 x 0.46))^(1/3) ≈ 5.8 m/s

Therefore, an average wind velocity of approximately 5.8 m/s is required to produce 3.8 MW of electrical power with the given wind turbine specifications.

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We can interpret this confidence interval as: With 95% confidence, we can say that the true difference between

a) Hypothesis test for comparing variances between two data sets:

Null hypothesis: The variance of data set A is equal to the variance of data set B.

Alternative hypothesis: The variance of data set A is not equal to the variance of data set B.

We can use the F-test to compare the variances between the two data sets. The test statistic is calculated as:

[tex]F = s1^2 / s2^2[/tex]

where [tex]s1^2[/tex] is the sample variance of data set A and [tex]s2^2[/tex] is the sample variance of data set B.

Using the given information, we can calculate the test statistic as:

F = 0.45 / 0.32 = 1.41

Using an alpha level of 0.05 and degrees of freedom of 28 and 21 (n1-1 and n2-1), we can find the critical values for F as 0.46 and 2.33.

Since the calculated F value of 1.41 falls between the critical values, we fail to reject the null hypothesis and conclude that there is not enough evidence to suggest that the variance of data set A is different from the variance of data set B.

b) Hypothesis test for comparing means between two data sets:

Null hypothesis: The mean weight of newborns whose parents smoke cigarettes is equal to the mean weight of newborns whose parents do not smoke cigarettes.

Alternative hypothesis: The mean weight of newborns whose parents smoke cigarettes is not equal to the mean weight of newborns whose parents do not smoke cigarettes.

Since the variances of the two data sets are not significantly different from each other, we can use a two-sample t-test assuming equal variances to compare the means between the two data sets.

Using the given information, we can calculate the test statistic as:

t = (x1bar - x2bar) / (sqrt[([tex]s^2[/tex] / n1) + ([tex]s^2[/tex] / n2)])

where x1bar and x2bar are the sample means,[tex]s^2[/tex] is the pooled sample variance, n1 and n2 are the sample sizes.

Using an alpha level of 0.05 and degrees of freedom of 48 (n1 + n2 - 2), we can find the critical values for t as ±2.01.

Using the given information, we can calculate the test statistic as:

t = (7.25 - 7.68) / (sqrt[(0.[tex]385^2[/tex] / 30) + ([tex]0.28^2[/tex] / 23)]) = -1.2

Since the calculated t value of -1.23 falls between the critical values, we fail to reject the null hypothesis and conclude that there is not enough evidence to suggest that the mean weight of newborns whose parents smoke cigarettes is different from the mean weight of newborns whose parents do not smoke cigarettes.

c) Confidence interval for the difference between means:

Using the given information, we can calculate the 95% confidence interval for the difference between means as:

(x1bar - x2bar) ± tα/2,df * (sqrt[([tex]s^2 / n1[/tex]) + (s^2 / n2)])

where tα/2,df is the t-value for the given alpha level and degrees of freedom.

Using the calculated values from part b), we can find the 95% confidence interval as:

(7.25 - 7.68) ± 2.01 * (sqrt[(0.385^2 / 30) + ([tex]0.28^2[/tex] / 23)]) = (-0.779, 0.179)

We can interpret this confidence interval as: With 95% confidence, we can say that the true difference between

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a)We have x2 = 21.44 as our test statistic.

b) the critical value for a chi-squared distribution with 4 degrees of freedom and a 10% significance level is 7.78.

c) the distribution of appointments is not the same for all weekdays.

a. To calculate the test statistic for the chi-squared goodness-of-fit test, we need to find the expected frequency for each day of the week if the appointments were distributed evenly. Since there are five days of the week, we would expect 500/5 = 100 appointments for each day of the week.

Day Appointment Observed Frequency Expected Frequency (O - E)^2 / E Monday 98 100 (98-100)^2/100 = 0.04 Tuesday 116 100 (116-100)^2/100 = 2.56 Wednesday 122 100 (122-100)^2/100 = 4.84 Thursday 98 100 (98-100)^2/100 = 0.04 Friday 66 100 (66-100)^2/100 = 13.96 Total:

The test statistic is the sum of the squared differences between the observed and expected frequencies, divided by the expected frequencies for all categories:

x2 = Σ(Observed - Expected)² / Expected

x2 = (0.04 + 2.56 + 4.84 + 0.04 + 13.96)

x2 = 21.44

We have x2 = 21.44 as our test statistic.

b. To determine the critical value(s) for the hypothesis test, we need to use the chi-squared distribution table with (5 - 1) = 4 degrees of freedom and a significance level of 10%. Looking at the table, the critical value for a chi-squared distribution with 4 degrees of freedom and a 10% significance level is 7.78.

c. To conclude whether to reject the null hypothesis or not based on the test statistic, we compare it with the critical value. Since our test statistic (21.44) is greater than the critical value (7.78), we reject the null hypothesis that the appointments are distributed evenly from Monday to Friday each week. This means that the distribution of appointments is not the same for all weekdays.

d. We reject the null hypothesis.

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The initial number of bacteria in the colony was approximately 13.5.

The value of n, the number of bacteria at the start of the experiment, can be calculated using the formula n =

[tex](b/2)^(2/t),[/tex]

where b is the number of bacteria at any given time and t is the time in days.

Plugging in the given values, we get: n =

[tex](10,000/2)^(2/6)[/tex]

n =

[tex]2,500^(1/3)[/tex]

n ≈ 13.5. This formula is derived from the fact that the growth of bacteria is often modeled by an exponential function, where the rate of growth is proportional to the current population size.

In this case, the number of bacteria is doubling every 2 days (since [tex]2^(1/2) = 2^(2/4) = 2^(4/8) = ...),[/tex] so we can rewrite the original equation as n • [tex]2^(t/2)[/tex]. Using the given information that there are 10,000 bacteria 6 days after the experiment starts, we can plug in these values and solve for n using the derived formula.

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To answer your question involving independent events, algebra, and Kolmogorov's zero-one law. That to prove that if B ∈ T, then P(B) is either 0 or 1,

Follow these steps:

1. Define An as a sequence of independent events and T as the tail σ-algebra generated by the events An, An+1, ...

2. Introduce the concept of a tail event: A tail event is an event B such that B belongs to the tail σ-algebra T.

3. Apply Kolmogorov's zero-one law: This law states that for any tail event B belonging to T, the probability of B is either 0 or 1.

Proof:

Step 1: Given An as a sequence of independent events, let T be the tail σ-algebra generated by the events An, An+1, ...

Step 2: Let B be a tail event such that B ∈ T.

Step 3: By Kolmogorov's zero-one law, for any tail event B ∈ T, the probability of B is either 0 or 1.

Therefore, if B ∈ T, then P(B) is either 0 or 1.

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Answer:

8384742

Step-by-step explanation:

Answer:

there is no cow there because they have been barn

The probability of a sample of 144 producing a mean of 50 or larger if the population has a mean of 49 and a standard deviation of 5 is approximately 0.0082 or 0.82%.

To solve this problem, we can use the central limit theorem, which states that the distribution of sample means approaches a normal distribution as the sample size increases, regardless of the shape of the population distribution.

First, we need to calculate the standard error of the mean (SEM) using the formula:

a) the probability that a household views television between 5 and 12 hours a day is approximately 0.7357.

b)a household must view approximately 13.70 hours of television per day to be in the top 3% of all television viewing households.

c) the probability that a household views television more than 4 hours a day is approximately 0.9599.

(a) We need to find the probability that a household views television between 5 and 12 hours a day. Let X be the random variable representing daily television viewing per household. Then, we need to find P(5 < X < 12). Using the standard normal distribution table or a calculator with normal distribution functions, we can compute:

z1 = (5 - 8.35) / 2.5 = -1.34

z2 = (12 - 8.35) / 2.5 = 1.46

P(-1.34 < Z < 1.46) ≈ 0.7357

Therefore, the probability that a household views television between 5 and 12 hours a day is approximately 0.7357.

(b) We need to find the value of X such that the probability of a household viewing more than X hours of television per day is 0.03. Using a standard normal distribution table or a calculator with inverse normal distribution functions, we can compute:

z = InvNorm(0.97) ≈ 1.88

z = (X - 8.35) / 2.5

X = 2.5z + 8.35 ≈ 13.70

Therefore, a household must view approximately 13.70 hours of television per day to be in the top 3% of all television viewing households.

(c) We need to find the probability that a household views television more than 4 hours a day. Using the standard normal distribution table or a calculator with normal distribution functions, we can compute:

z = (4 - 8.35) / 2.5 = -1.74

P(Z > -1.74) ≈ 0.9599

Therefore, the probability that a household views television more than 4 hours a day is approximately 0.9599.

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The total volume of ice cream in term of π is 42π³

The volume of an object is the amount of space occupied by the object or shape, which is in three-dimensional space.

The total volume of the ice cream = volume of cone + volume of half sphere

volume of a cone = 1/3 πr²h

= 1/3 × π × 3² × 8

= π×3 ×8

= 24π in³

volume of the half sphere = 4/6πr³

= 4/6 ×π × 3³

= 108π/6

= 18π in³

therefore the total volume of the ice cream

= 24π + 18π

= 42π in³

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The simplified expression for the cost of renting the apartment for n months is 900n + 1350.

We have,

To simplify the expression 900(n + 1) + 450, we can start by using the distributive property of multiplication over addition, which states that:

a(b + c) = ab + ac.

So, we have:

900(n + 1) + 450

= 900n + 900(1) + 450 (applying the distributive property)

= 900n + 900 + 450

= 900n + 1350

Therefore,

The simplified expression for the cost of renting the apartment for n months is 900n + 1350.

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The  (q/p)^Sn, n>=1 is a martingale.

To show that (q/p)^Sn, n>=1 is a martingale, we need to show that it satisfies the three conditions of a martingale:

The expected value of (q/p)^Sn is finite for all n.

For all n, E[(q/p)^Sn+1 | Fn] = (q/p)^Sn, where Fn is the sigma-algebra generated by the first n transitions.

(q/p)^Sn is adapted to the filtration Fn.

First, we note that the expected value of (q/p)^Sn is finite for all n since q/p < 1, and thus (q/p)^n approaches zero as n approaches infinity.

Next, we consider the second condition. Let F_n be the sigma-algebra generated by the first n transitions, and let X_n = (q/p)^Sn. We need to show that E[X_n+1 | F_n] = X_n.

We can write (q/p)^(n+1) = (q/p)^n * (q/p), so we have:

E[X_n+1 | F_n] = E[(q/p)^(n+1) | F_n]

= E[(q/p)^n * (q/p) | F_n]

= (q/p)^n * E[(q/p) | F_n]

= (q/p)^n * [(q/p) * P(up) + (p/q) * P(down)]

= (q/p)^n * [(q/p) * p + (p/q) * q]

= (q/p)^n * (p + q)

= (q/p)^n * 1

= X_n

Thus, the second condition is satisfied.

Finally, we need to show that X_n is adapted to the filtration F_n. This is true since X_n only depends on the first n transitions, which are included in F_n.

Therefore, we have shown that (q/p)^Sn, n>=1 is a martingale.

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(a) the probability that the daily revenue exceeds $5,000 is approximately 0.1587.

(b) E(P) = E(N(10 - C)) = E(10N) - E(NC) = 4000 - E(N)E(C) + Cov(N, C)

= 4000 - 400*1000 + 12000 = -120000

(c) The expected daily profit is -$120,000.

(d) the variance of the daily profit is $56,250,000,000.

What is probability?

Probability is a measure of the likelihood of an event occurring. It is expressed as a number between 0 and 1, where 0 indicates that the event is impossible and 1 indicates that the event is certain.

(a) Let X be the daily revenue. Then X = 10N, and we have:

E(X) = E(10N) = 10E(N) = 10(400) = 4000

[tex]Var(X) = Var(10N) = 10^2Var(N) = 10^2(50^2) = 25000[/tex]

Using the standardization formula, we have:

[tex]P(X > 5000) = P(Z > (5000-4000)/\sqrt(25000)) = P(Z > 1)[/tex]

Using a standard normal table or calculator, we find P(Z > 1) = 0.1587.

Therefore, the probability that the daily revenue exceeds $5,000 is approximately 0.1587.

(b) The total daily profit is given by:

P = N(10 - C)

Using the formula for the covariance between N and C, we have:

Cov(N, C) = rhosigma(N)sigma(C) = 0.850300 = 12000

Then we have:

E(P) = E(N(10 - C)) = E(10N) - E(NC) = 4000 - E(N)E(C) + Cov(N, C)

= 4000 - 400*1000 + 12000 = -120000

(c) The expected daily profit is -$120,000.

(d) To compute the variance of P, we use the formula:

Var(P) = Var(N(10 - C)) = 100Var(N)Var(10 - C) + 210Cov(N, 10 - C) + Var(10 - C)Var(N)

We have already computed Var(N) and Cov(N, 10 - C) in part (a) and (b). Also, we have:

Var(10 - C) = Var(10) + Var(C) - 2Cov(10, C) = 0 + 300^2 - 2(0) = 90000

Plugging in the values, we get:

Var(P) = 100(25000)(90000) + 2(10)(12000) + 90000(25000)

= 56250000000

Therefore, the variance of the daily profit is $56,250,000,000.

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Answer:

∠ PQR = 18°

Step-by-step explanation:

the exterior angle of a triangle is equal to the sum of the 2 opposite interior angles.

∠ PQR is an exterior angle of the triangle , then

∠ PQR = ∠OPQ + ∠ QOP , that is

4x - 10 = x + 9 + x - 5

4x - 10 = 2x + 4 ( subtract 2x from both sides )

2x - 10 = 4 ( add 10 to both sides )

2x = 14 ( divide both sides by 2 )

x = 7

Then

∠ PQR = 4x - 10 = 4(7) - 10 = 28 - 10 = 18°