in the fire academy, we had a ratio of 7 to 5 passing grades to failing. how many of the 36 students failed?

Answer:

72 sec im srry if its wrong i cant see it good

Step-by-step explanation:

Answer:

Not all the means are equal.

Step-by-step explanation:

In an ANOVA table, a sum of squares for the independent variable divided by its respective degrees of freedom.

So if the omnibus null hypothesis is rejected it means that there is sufficient evidence to conclude that not all the means are equal.

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If the omnibus null hypothesis is rejected in a one-factor ANOVA, then, we can conclude that at least one of the population means is different from at least one other population mean.

ANOVA table gives us the total sum of squares ( TSS ), the residual sum of squares ( RSS ) and the estimated sum of squares ( ESS ).

It establishes the relation that:

ESS + RSS = TSS.

If we reject the omnibus null hypothesis, then it means that:

at least one of the population means is different from at least one other population mean.

Therefore, we get that, if the omnibus null hypothesis is rejected in a one-factor ANOVA, then, we can conclude that at least one of the population means is different from at least one other population mean.

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To find the subtransient current in per unit for a three-phase fault on bus 4, we need to calculate the fault current using the bus impedance matrix.

Given bus impedance matrix Zbus:

| j0.15 j0.08 j0.04 j0.07 |

| j0.08 j0.15 j0.06 j0.09 |

| j0.04 j0.06 j0.13 j0.05 |

| j0.07 j0.09 j0.05 j0.12 |

To find the fault current on bus 4, we need to find the inverse of the Zbus matrix and multiply it by the pre-fault voltage vector.

The pre-fault voltage vector V_pre-fault is given as:

| 1.0/0° |

| 1.0/0° |

| 1.0/0° |

| 1.0/0° |

Let's calculate the inverse of the Zbus matrix:

Zbus_inverse = inv(Zbus)

Now, we can calculate the fault current using the formula:

I_fault = Zbus_inverse * V_pre-fault

Calculating the fault current, we have:

I_fault = Zbus_inverse * V_pre-fault

Substituting the values and calculating the product, we get:

I_fault = Zbus_inverse * V_pre-fault

= Zbus_inverse * | 1.0/0° |

| 1.0/0° |

| 1.0/0° |

| 1.0/0° |

Please provide the values of the Zbus matrix and the pre-fault voltage vector to obtain the specific values for the fault current and the per-unit current from generator 2.

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Answer:

6

Step-by-step explanation:

id say the answer is 6 or 9 :)

Answer:

B. Polyhedron, D. Solid, C. Prism.

Step-by-step explanation:

Yw !

Answer:

yes correct

Step-by-step explanation:

Answer:

Step-by-step explanation:

b

b let me know if you are wrong

In order to show that 6x2 + 3x + 4 is O(22), we need to find suitable values for c and k in the definition of big-O.  Recall that a function f(x) is O(g(x)) if there exist positive constants c and k such that |f(x)| ≤ c|g(x)| for all x ≥ k.

Looking at the options given, we can see that the value of c must be greater than or equal to 1, since we are looking for an upper bound for the function.

Option a, where c = 100 and k = 1, is not a suitable choice because it is too large of a value for c. Similarly, option e, where c = 7 and k = 87, is not a suitable choice because it is too large of a value for k.

Option b, where c = 13 and k = 1, is a possible choice. To show this, we need to prove that there exist constants c = 13 and k = 1 such that:

\(|6x2 + 3x + 4| ≤ 13|22| for all x ≥ 1.\)

Note that |22| = 22 and |6x2 + 3x + 4| ≤ 6x2 + 3x + 4.

Thus, we need to show that:

6x2 + 3x + 4 ≤ 13(22) for all x ≥ 1.

Simplifying the inequality, we get:

6x2 + 3x + 4 ≤ 286

This inequality holds for all x ≥ 1, since the left-hand side is a quadratic function that is increasing for x ≥ 0. Therefore, option b is a suitable choice.

Option c, where c = 1 and k = 87, is not a suitable choice because it is too small of a value for c.

Option d, where c = 10 and k = 2, is not a suitable choice because it is too small of a value for k.

Option f, where c = 7 and k = 1, is also a possible choice. To show this, we need to prove that there exist constants c = 7 and k = 1 such that:

\(|6x2 + 3x + 4| ≤ 7|22| for all x ≥ 1.\)

Note that |22| = 22 and |6x2 + 3x + 4| ≤ 6x2 + 3x + 4.

Thus, we need to show that:

6x2 + 3x + 4 ≤ 154 for all x ≥ 1.

This inequality holds for all x ≥ 1, since the left-hand side is a quadratic function that is increasing for x ≥ 0. Therefore, option f is also a suitable choice.

In summary, options b and f are both suitable choices for c and k in the definition of big-O to show that 6x2 + 3x + 4 is O(22).

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Answer:

brand b is the better buy, brand a unit price is 0.0427, and brand b is 0.0346

Step-by-step explanation:

hope it helped :)

Answer:

To round 4.3741 to the nearest hundredth, we need to look at the digit in the second decimal place, which is 7. Since 7 is greater than or equal to 5, we need to round up the previous digit, which is 4. Therefore, rounding 4.3741 to the nearest hundredth gives us:

4.37

So the answer is 4.37.

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Answer:

the answer to this problem is 4.3700

Step-by-step explanation: i hope this helps

whats the question....?

Answer:

x = 8

Step-by-step explanation:

3x + 2 + 7x + 8 = 90

10x + 10 = 90

10x = 80

x = 8

Check:  3x + 2 = 3(8) + 2 = 26°

7x + 8 = 7(8) + 8 = 56 + 8 = 64°

sin 26 = cos 64

The given sample's confidence interval is 4.73 ± 0.1199, or from 4.61 to 4.85

The z-score is a numerical assessment of a value's connection to the mean of a set of values, expressed in terms of standards from the mean, that is used in statistics.

Given data;

Mean  = 4.73

Standard deviation  = 0.865

Sample size  = 200

interval  = 95%

Confidence interval=?

95% of samples contain the population mean (μ) within the confidence interval of 4.73 ± 0.1199.

Hence, the sample's confidence interval is 4.73 ± 0.1199, from 4.61 to 4.85

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The answer is 10.5, but don't put 10.5 if it's just asking for the number and not the unit only put (10)!!!

Answer:

u = 24

Step-by-step explanation:

-7/8u = -21

Solve for u

Multiply each side by -8/7

-8/7 *-7/8u = -21 *-8/7

u =3*8

u = 24

Answer:

u = 1/24

Write the equation:

–7/8u = –21

Get rid of fraction:

–7/2^3 • u = –21

–7 = ( – (8u) ) 21

Solve:

–7 = –8u + 21

–7 + 8u = –8u + 8u + 21

–7 + 8u = 21

–7 + 7 + 8u = 21 + 7

8u = 21 + 7

8u = 28

u = 1/24

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Answer:

27 yd²

Step-by-step explanation:

Refer to attachment.

Hope it helps :)

Answer:

(-6,2)

Step-by-step explanation:

for the reflection just go on the opposite side of the y axis and then it would be negative but then going left 2 more would add to 4.

The number of minutes it will take for them to complete a lap at the same time is given as follows:

12 minutes.

Considering periodic events, they will complete a lap at the same time for the least common factor of the periods of each event.

The periods for each person are given as follows:

Then the least common factor is obtained as follows:

4 - 6 - 3|2

2 - 3 - 3|2

1 - 3 - 3|3

1 - 1 - 1

Hence:

lcf(4, 6, 3) = 2² x 3 = 12 minutes.

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The numbers from least to greatest number is as follows;

4.06, 4.59, 4.6, 4.72

The numbers from least to greatest is as follows;

Ordering from least to the greatest simply means we are ordering the number from the smallest to the largest number.

Therefore,

4.06, 4.6, 4.72, 4.59

The numbers from least to greatest number is as follows;

4.06, 4.59, 4.6, 4.72

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The original cost of the car was $12,935.74 (rounded to the nearest $100).

(a) To find the depreciation rate for a car that originally cost $20,000 and is worth $12,000 after 4 years, we can use the formula:

\(C = (1 - r)^n * C_0\)

where C is the value of the car after n years, C_0 is the original cost of the car, and r is the yearly depreciation rate.

Plugging in the values given, we get:

\(12,000 = (1 - r)^4 * 20,000\)

Solving for r, we get:

\(1 - r = (12,000/20,000)^(1/4) = 0.8185\)

r = 1 - 0.8185 = 0.1815

The depreciation rate for the car is 18.2% (rounded to the nearest tenth of a percent).

(b) To find the original cost of a car that has a yearly depreciation rate of 11% and is worth $10,000 after 2 years, we can use the same formula:

\(C = (1 - r)^n * C_0\)

Plugging in the values given, we get:

\(10,000 = (1 - 0.11)^2 * C_0\)

Solving for C_0, we get:

\(C_0 = 10,000 / (0.89)^2 = $12,935.74\)

The original cost of the car was $12,935.74 (rounded to the nearest $100).

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Answer:

456 drinks

Step-by-step explanation:

just multiply 11.40×40 and you get 456. that's how many drinks you can afford. hope i helped.

If the expression be 23 x² + 3x + 8 then the constant exists 8.

The addition, subtraction, multiplication, and division arithmetic operators are used to write a group of numbers together to form a numerical statement in mathematics. The expression of a number can take on various forms, including verbal form and numerical form.

A mathematical expression is a finite combination of symbols that is well-formed in accordance with context-specific norms.

A mathematical expression is a phrase that includes at least two numbers or variables, at least one arithmetic operation, and the expression itself. Any one of the following mathematical operations can be used. A sentence has the following structure: Number/variable, Math Operator, Number/Variable is an expression.

During the course of a program's execution, a constant's value cannot change. As a result, the value is constant, as implied by its name. During the course of a program's execution, a variable's value can change. As a result, the value might change, as implied by its name.

Let the expression be 23 x² + 3x + 8

then the constant exists 8.

Therefore, the correct answer is option C. 8.

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Answer:

5:3

Step-by-step explanation:

Therefore, based on the given information, the estimated age of the tomb is approximately 3,970 years.

To estimate the age of the tomb based on the radioactive carbon-14 (C-14) content in the cypress beam, we can use the decay equation:

A = A₀ * e*(kt)

Where:

A = Final amount of radioactive substance (55% of the original C-14 content)

A₀ = Initial amount of radioactive substance (100% of the original C-14 content)

k = Decay constant (ln(2) / half-life of C-14)

t = Time (age of the tomb)

Given that the half-life of C-14 is approximately 5600 years, we can calculate the decay constant:

k = ln(2) / 5600 years

Now we can plug in the values:

0.55A₀ = A₀ * e*((ln(2) / 5600 years) * t)

Dividing both sides by A₀:

0.55 = e*((ln(2) / 5600 years) * t)

Taking the natural logarithm of both sides:

ln(0.55) = (ln(2) / 5600 years) * t

Now we can solve for t, the age of the tomb:

t = (ln(0.55) * 5600 years) / ln(2)

Evaluating the expression:

t ≈ 3,970 years

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Answer:

Graph it as ○--->

The open circle because the answer is greater than 8 and we are going to the left because it is greater than.

Answer:

JL = 7.5

Step-by-step explanation:

since the triangles are similar then the ratios of corresponding sides are in proportion , that is

\(\frac{JL}{AC}\) = \(\frac{JK}{AB}\) ( substitute values )

\(\frac{JL}{2.5}\) = \(\frac{10.5}{3.5}\) ( cross- multiply )

3.5 × JL = 2.5 × 10.5 = 26.25 ( divide both sides by 3.5 )

JL = 7.5

Answer:

2² * 13

Step-by-step explanation:

Prime Factorization refers to the representation of the number as the product of prime numbers.

Dividing it by 2 until it reaches a point that it cannot be divided by 2 (since 2 is a prime number):

52/2 = 26/2 = 13

Since 13 is a prime number:

13/13 = 1

Hence,

52 = 2 * 2 * 13

= 2² * 13

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The process that generates more values that are more than 2 standard deviations from the mean is called a "fat-tailed" distribution.

In a fat-tailed distribution, the probability of extreme values occurring is higher compared to a normal distribution.

To understand this concept, let's consider two processes: Process A and Process B.

Process A generates a dataset with a normal distribution, while Process B generates a dataset with a fat-tailed distribution.

In a normal distribution, the majority of the data falls close to the mean, with fewer values farther away.

The probability of values more than 2 standard deviations from the mean is relatively low.

On the other hand, in a fat-tailed distribution, there is a higher likelihood of extreme values occurring.

This means that the probability of values more than 2 standard deviations from the mean is higher.

For example, let's say we have two datasets: Dataset A generated by Process A and Dataset B generated by Process B.

We calculate the mean and standard deviation for both datasets.

In Dataset A, if the mean is 50 and the standard deviation is 5, then values more than 2 standard deviations away from the mean would be greater than 60 or less than 40.

In Dataset B, if the mean is 50 and the standard deviation is also 5, then values more than 2 standard deviations away from the mean would have a wider range.

These values could be significantly higher than 60 or lower than 40, depending on the specific distribution of Dataset B.

Therefore, if Process B generates a fat-tailed distribution, it is more likely to produce values that are more than 2 standard deviations from the mean compared to Process A and its normal distribution.

In summary, the process that generates more values that are more than 2 standard deviations from the mean is a fat-tailed distribution, which has a higher likelihood of extreme values occurring.

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Answer:

1:3 red to blue

Step-by-step explanation:

you have x shirts. Let's call the number of red shirts r and the number of blue shirts b.

r = x * 1/4

b = x - r = x - (x * 1/4) = x * 3/4

r:b = (x * 1/4):(x * 3/4) = (1/4):(3/4) = 1:3