What time is 6 ¾ hours before 3:37 am?

a) By strong induction, any amount of postage worth 8 cents or more can be made from 3-cent or 5-cent stamps.

b) By strong induction, any amount of postage worth 24 cents or more can be made from 7-cent or 5-cent stamps.

c) By strong induction, any amount of postage worth 12 cents or more can be made from 3-cent or 7-cent stamps.

a. Prove that any amount of postage worth 8 cents or more can be made from 3-cent or 5-cent stamps.

Base case: For postage worth 8 cents, we can use two 4-cent stamps, which can be made using a combination of one 3-cent stamp and one 5-cent stamp.

Induction hypothesis: Assume that any amount of postage worth k cents or less, where k is greater than or equal to 8, can be made from 3-cent or 5-cent stamps.

Induction step: Consider any amount of postage worth (k+1) cents. Since k is greater than or equal to 8, we can use the induction hypothesis to make k cents using 3-cent or 5-cent stamps. Then, we can add one more stamp to make (k+1) cents. If the last stamp we added was a 3-cent stamp, we can replace it with a 5-cent stamp to get the same value. If the last stamp we added was a 5-cent stamp, we can replace it with two 3-cent stamps to get the same value. Therefore, any amount of postage worth (k+1) cents can be made from 3-cent or 5-cent stamps.

b. Prove that any amount of postage worth 24 cents or more can be made from 7-cent or 5-cent stamps.

Base case: For postage worth 24 cents, we can use three 8-cent stamps, which can be made using a combination of one 7-cent stamp and one 5-cent stamp.

Induction hypothesis: Assume that any amount of postage worth k cents or less, where k is greater than or equal to 24, can be made from 7-cent or 5-cent stamps.

Induction step: Consider any amount of postage worth (k+1) cents. Since k is greater than or equal to 24, we can use the induction hypothesis to make k cents using 7-cent or 5-cent stamps. Then, we can add one more stamp to make (k+1) cents. If the last stamp we added was a 5-cent stamp, we can replace it with two 7-cent stamps to get the same value. If the last stamp we added was a 7-cent stamp, we can replace it with three 5-cent stamps to get the same value. Therefore, any amount of postage worth (k+1) cents can be made from 7-cent or 5-cent stamps.

c. Prove that any amount of postage worth 12 cents or more can be made from 3-cent or 7-cent stamps.

Base case: For postage worth 12 cents, we can use one 3-cent stamp and three 3-cent stamps, which can be made using a combination of two 7-cent stamps.

Induction hypothesis: Assume that any amount of postage worth k cents or less, where k is greater than or equal to 12, can be made from 3-cent or 7-cent stamps.

Induction step: Consider any amount of postage worth (k+1) cents. Since k is greater than or equal to 12, we can use the induction hypothesis to make k cents using 3-cent or 7-cent stamps. Then, we can add one more stamp to make (k+1) cents. If the last stamp we added was a 3-cent stamp, we can replace it with two 7-cent stamps to get the same value. If the last stamp we added was a 7-cent stamp, we can replace it with one 3-cent stamp and two 7-cent stamps to get the same value. Therefore, any amount of postage worth (k+1) cents can be made from 3

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answer 25 :) THERE U AREEEE

The order of events from least likely to most likely are as follows:

The reason for this order is because of the chances of each event occurring. Event D has the least chance of occurring because no numbers in the set have a value less than or equal to 0. Event A is the next least likely because only one number in the set is greater than 0, which is the number 1. Event C is the next least likely because only two numbers in the set are at least 2 (2 and 3). Event B is the next least likely, as there are only five even numbers in the set (2, 4, 6, 8, and 10). Events E, F, G, and H are the most likely to occur, because of the number of numbers that are divisible by 3, and 5, and are prime numbers.

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Answer:

Intervals are given to us

19, 19, 27, 20, 27, 26, 25, 26, 18, 27, 20, 24, 19, 21, 27, 23, 21, 17, 22, 22

From the given data we need to create a frequency table

From 17 - 19 there are 5 numbers. they are 19,19,19, 18, 17

From 20 -22 there are 6 numbers . they are 20, 20, 21, 21, 22, 22

From 23 - 25 there are 3 numbers. they are 23, 24, 25

From 26 to 28 there are 6 numbers . they are 27, 27, 26, 26, 27, 27

Step-by-step explanation:

Answer:

67°.

Step-by-step explanation:

1) m∠PMR=m∠LMN=3x+19°;

2) m∠LMN+m∠LMP=180°, it can be written as 3x+19+9x-31=180;

3) if to solve the equation 3x+19+9x-31=180, then x=16;

4) m∠PMR=3x+19=48+19=67°.

Answer:

y-(-2)= -5(x-4)

= -5x-y+18

Step-by-step explanation:

y-(-2)= -5(x-4)

y+2= -5x +20

y= -5x+18

-5x-y+18= 0

a) the probability that a randomly selected Cobalt gets more than 34 MPG is approximately 0.7149.

b) the probability that the mean MPG exceeds 34 MPG for a sample of 10 Cobalts is approximately 0.035.

c) the probability that the mean MPG exceeds 34 MPG for a sample of 20 Cobalts is approximately 0.005.

a) To find the probability that a randomly selected Cobalt gets more than 34 MPG, we need to calculate the area under the normal distribution curve to the right of 34 MPG.

Using the z-score formula, we can convert the MPG value to a standard score (z-score) using the formula:

z = (x - μ) / σ,

where x is the given value (34 MPG), μ is the mean (32 MPG), and σ is the standard deviation (3.5 MPG).

Calculating the z-score:

z = (34 - 32) / 3.5 = 0.57

Using a standard normal distribution table or a statistical calculator, we can find the area to the right of the z-score 0.57.

Let's assume the standard normal distribution table gives us a value of 0.2851 for z = 0.57.

Since the total area under the normal curve is 1, the probability of getting more than 34 MPG is:

P(X > 34) = 1 - P(X ≤ 34) = 1 - 0.2851 = 0.7149

Therefore, the probability that a randomly selected Cobalt gets more than 34 MPG is approximately 0.7149.

b) When selecting a sample of 10 Cobalts, the mean MPG of the sample (\(\bar{X}\)) follows a normal distribution with the same mean (32 MPG) and a standard deviation (σ) equal to the population standard deviation (3.5 MPG) divided by the square root of the sample size (√10).

σ( \(\bar{X}\) ) = σ / √n = 3.5 / √10 ≈ 1.107

We want to find the probability that the mean MPG exceeds 34 MPG for the sample of 10 Cobalts. In other words, we need to find P(\(\bar{X}\) > 34).

We can again convert the value of 34 MPG to a z-score:

z = (34 - 32) / 1.107 ≈ 1.805

Using a standard normal distribution table or a statistical calculator, we find the area to the right of the z-score 1.805.

Let's assume the standard normal distribution table gives us a value of 0.035 for z = 1.805.

Therefore, the probability that the mean MPG exceeds 34 MPG for a sample of 10 Cobalts is approximately 0.035.

c) When selecting a sample of 20 Cobalts, the mean MPG of the sample (\(\bar{X}\)) follows a normal distribution with the same mean (32 MPG) and a standard deviation (σ) equal to the population standard deviation (3.5 MPG) divided by the square root of the sample size (√20).

σ( \(\bar{X}\) ) = σ / √n = 3.5 / √20 ≈ 0.78

We want to find the probability that the mean MPG exceeds 34 MPG for the sample of 20 Cobalts. In other words, we need to find P(\(\bar{X}\) > 34).

Similarly, we can convert the value of 34 MPG to a z-score:

z = (34 - 32) / 0.78 ≈ 2.564

Using a standard normal distribution table or a statistical calculator, we find the area to the right of the z-score 2.564.

Assuming the standard normal distribution table gives us a value of 0.005 for z = 2.564.

Therefore, the probability that the mean MPG exceeds 34 MPG for a sample of 20 Cobalts is approximately 0.005.

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Answer:

Step-by-step explanation:

Let the number be x

Half a number is written as 1/2x

So we have

Four times the sum of half a number and 6 is 48 is written as

\(4( \frac{1}{2} x + 6) = 48 \\ 2x + 24 = 48 \\ 2x = 48 - 24 \\ 2x = 24\)

Divide both sides by 2

That's

\( \frac{2x}{2} = \frac{24}{2} \)

We have the final answer as

Hope this helps you

Answer:

Step-by-step explanation:

\(2. \: a. \: \sqrt[3]{ - 1} \)

\( = \sqrt[3]{( - 1 )\times ( - 1 )\times( - 1)} \)

= -1 (Ans)

\(b. \: \sqrt[3]{ - \frac{8}{27} } \)

Since,

\( \sqrt[3]{ - x} = - \sqrt[3]{x} \)

Hence,

\( = - \sqrt[3]{ \frac{8}{27} } \)

\( = - \sqrt[3]{ \frac{2 \times 2 \times 2}{3 \times 3 \times 3} } \)

\( = - \frac{2}{3} (ans)\)

\(3. \: a. \: 10 - 2 \sqrt{9} \)

\( = 10 - 2 \sqrt{3 \times 3} \)

= 10 - 2 × 3

= 10 - 6

= 4 (Ans)

\(b. \: 70 - {( \sqrt[3]{64} )}^{3} \)

\( = 70 - \sqrt[3]{ {64}^{3} } \)

\( = 70 - \sqrt[3]{64 \times 64 \times 64} \)

= 70 - 64

= 6 (Ans)

The results of a study are said to be replicable when researchers using new subjects come to the same conclusion. Replicability is important in research as it ensures the validity and reliability of the study findings.

The results of a study are said to be replicable when researchers using new subjects come to the same conclusion. Replicability is an important aspect of scientific research because it ensures that the findings are not simply a result of chance or bias. If a study cannot be replicated, it may be difficult to draw firm conclusions or generalize the findings to a broader population. However, it is important to note that replicability does not guarantee that the study is both statistically significant and meaningful.

A study may be replicable but still have limited practical or theoretical value. The p-value and significance level are important measures of statistical significance, but they do not necessarily indicate replicability. Finally, while statistics are knowable, parameters (such as population mean or variance) are typically unknown and must be estimated from the sample data.

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Answer:

d=10

d=2r=2·5=10

A function from a finite set s to itself is one-to-one if and only if it is onto. This result holds because of the finite nature of the set s and the uniqueness of the mapping between elements in the domain and the range of the function.

To show that a function from a finite set s to itself is one-to-one if and only if it is onto, we need to consider two cases.

let's assume that the function is one-to-one. This means that for every pair of distinct elements x and y in the set s, the function maps them to distinct elements in the set s. In other words, if f(x) = f(y), then x = y. Since the set s is finite, and the function maps each element to a distinct element, there must be an equal number of elements in the domain and the range of the function. Therefore, the function must be onto as well.

Now let's assume that the function is onto. This means that for every element y in the set s, there exists an element x in the set s such that f(x) = y. In other words, every element in the range of the function is mapped to by some element in the domain of the function. Since the set s is finite, and the function maps each element to a unique element, there cannot be any elements in the range of the function that are not mapped to by some element in the domain of the function. Therefore, the function must be one-to-one as well.

In summary, a function from a finite set s to itself is one-to-one if and only if it is onto. This result holds because of the finite nature of the set s and the uniqueness of the mapping between elements in the domain and the range of the function.

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Answer:

z/2 ≥ -6

Step-by-step explanation:

Answer:

b

Step-by-step explanation:

The proportion of a normal distribution located between z = .50 and z = -.50 will be 38.2%.

We have,

A normal distribution located between z = 0.50 and z = -0.50,

So,

Now,

From the Z-score table,

We get,

The Probability corresponding to the Z score of -0.50,

i.e.

P(-0.50 < X < 0) = 0.191,

And,

The Probability corresponding to the Z score of -0.50,

i.e.

P(0 < X < 0.50) = 0.191,

Now,

The proportion of a normal distribution,

i.e.

P(Z₁ < X < Z₂) = P(Z₁ < X < 0) + P(0 < X < Z₂)

Now,

Putting values,

i.e.

P(-0.50 < X < 0.50) = P(-0.50 < X < 0) + P(0 < X < 0.50)

Now,

Again putting values,

We get,

P(-0.50 < X < 0.50) = 0.191 + 0.191

On solving we get,

P(-0.50 < X < 0.50) = 0.382

So,

We can write as,

P(-0.50 < X < 0.50) = 38.2%

So,

The proportion of a normal distribution is 38.2%.

Hence we can say that the proportion of a normal distribution located between z = .50 and z = -.50 will be 38.2%.

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Answer: 216cm

Step-by-step explanation:

volume = 6×6×6 =216 cm

EXPLANATION

Dividing the numerator and denominator by the highest denominator power (x^2):

Applying the following property:

Now, we need to apply the same steps to x-> ∞

Simplifying:

In conclusion, the appropiate end behavior is as follows:

Therefore , the solution of the given problem of equation comes out to be  option c is correct 2y = f(x) .

The equal letter (=) is used to signify equivalence between the statements in a mathematical formula. Using mathematical equation, which are declarations of reality, it is shown that many mathematical variables are equivalent. For instance, the equal sign in the equation

y + 6 = 12 divides the values 12 or b + 6 into two halves. The number of words that each side of a symbol corresponds to can be measured. Typically, a symbol's meaning is at odds with itself.

Here,

Given :

Every point on the graph in red is 1/2 away from the x-axis as that of the equivalent position on the graph in black.

Therefore, the vertically scale factor is equal to 1/2:

=> y = (1/2)f(x)

This problem can be solved by multiplying it by two and yields the following solution:

=> 2y = f(x)

thus , option c is correct

Therefore , the solution of the given problem of equation comes out to be  option c is correct 2y = f(x) .

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Answer:rr

yurrrrrrrrr yesyes ye s

Step-by-step explanation:

Answer:

Find the loss of capital: 23500-8750=1843

To find %:

1843/23500 x100% = 7.8% a year.

Therefore the car is -7.8% a year.

Hope that helps.

The average annual rate of change in the value of the car is approximately -7.8%.

The negative sign indicates that the value of the car is decreasing over time.

We have,

To find the average annual rate of change in the value of the car as a percent, we can use the following formula:

Average Annual Rate of Change = (Ending Value - Starting Value) / (Number of Years)

First, let's plug in the given values:

Starting Value = $23,500

Ending Value = $8,750

Number of Years = 8

Average Annual Rate of Change = ($8,750 - $23,500) / 8

Now, perform the subtraction:

Average Annual Rate of Change = (-$14,750) / 8

Finally, divide by 8 to get the average annual rate of change:

Average Annual Rate of Change ≈ -$1,843.75

Now, we need to express this value as a percentage.

To do that, divide the average annual rate of change by the original value and multiply by 100:

Percentage Change = (Average Annual Rate of Change / Starting Value) * 100

Percentage Change = (-$1,843.75 / $23,500) * 100

Percentage Change ≈ -7.8%

Thus,

The average annual rate of change in the value of the car is approximately -7.8%.

The negative sign indicates that the value of the car is decreasing over time.

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Answer:

x=((log(81))/(log(3))+2)/3

Step-by-step explanation:

so what you want to do is get 3x-2 out of the exponent

you can do that by using log

so apply log to both sides and you get

log(81)=(3x-2)log(3)

the 3x-2 is moved in front of the log due to a log rule

then (log(81))/(log(3))=3x-2

(log(81))/(log(3))+2=3x

((log(81))/(log(3))+2)/3=x

then use a calculator to find the exact value

hope this helps im sorry my explanation is a bit confusing, ask any questions you have

Answer:

\(x=20; y=50\) or \((20, 50)\)

Step-by-step explanation:

Substitution means plugging in one variable's value that consists of the opposite variable. Because \(y=3x-10\) is already specified, you can plug it into the second equation's \(y\) value. After doing that, it looks like this:

\(-4x+2y=20→-4x+2(3x-10)=20\)

Then you would distribute the \(3\) across the parentheses next to it, like this:

\(-4x+2(3x-10)=20→-4x+6x-20=20\)

Then, add like terms, like this:

\(-4x+6x-20=20→2x=40\)

Then divide both side by \(2\) to isolate \(x\).

\(x=20\)

Now, you can plug \(20\) (\(x\)) into either equation, but the first one seems simpler so you would pick that. It would look like this:

\(y=3x-10→y=3(20)-10\)

Solving would look like this:

\(y=3(20)-10→y=60-10→y=50\)

So the answer is \(x=20; y=50\) or \((20, 50)\).

Answer:

the answer is 20, 50

Step-by-step explanation:

please give brainliest

listening to the audio and comparing the descriptions of the houses will help determine which house is the biggest. Analyze details related to size, square footage, number of floors, rooms, or yard size. After careful consideration, provide a clear answer indicating which house is the biggest.

The question asks which house is the biggest based on the audio provided. To answer this question, we need to listen to the audio and analyze the information provided.

First, listen to the audio and pay attention to any details mentioned about the houses. Look for information related to their size or any comparisons made.

Next, compare the descriptions of the houses to determine which one is the biggest. Listen for clues such as mentions of square footage, number of floors, or any explicit statements about size.

Consider any additional information that may be relevant.

For example, if one house is described as having more rooms or a larger yard, this could indicate that it is also the biggest.

After carefully analyzing the information in the audio, determine which house is the biggest and provide a clear answer. You may need to listen to the audio multiple times or take notes to ensure accuracy.

Remember to provide a well-structured response, including an introduction, step-by-step explanation, and a conclusion. Use simple language and avoid irrelevant details.

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Answer: 4°F

Step-by-step explanation:

8-4=4

Answer: 9.7 seconds

Step-by-step explanation:

\(16t^2=1503\\\\t^2 =\frac{1503}{16}\\\\t=\sqrt{1503/16} \text{ } (t > 0)\\\\t \approx 9.7\)