help me plz ASAP please help me ​

a)Surface relaxation is the phenomenon in which the surface atoms of a crystal shift from their ideal positions to a more stable configuration, which has a lower energy state.

b) Surface reconstruction is a process in which the surface structure of a crystal is altered to create a new surface structure that has a lower surface energy. This process occurs when the original surface structure of the material is unstable and can be changed by breaking and forming new bonds between the surface atoms.

a) This process is caused by the surface energy of the material, which forces the atoms to move and rearrange themselves in a way that minimizes the surface energy. Surface relaxation typically results in a change in the atomic spacing and arrangement near the surface of the material

b) Surface reconstruction can occur under various conditions, such as changes in temperature, pressure, or chemical environment. The result of surface reconstruction is a change in the surface structure, which can affect the chemical and physical properties of the material.

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The price of 1 kg of fries is $6.60.

To find out the price of 1 kg of fries, we need to use proportion. Since we know that 420 grams of fries cost $2.77, we can set up a proportion with the price and the weight:

Price / Weight = Cost per unit

We can solve for the price of 1 kg of fries by setting the weight to 1 kg (1000 grams) and solving for the price:

Price / 1000g = $2.77 / 420g

Price = ($2.77 / 420g) * 1000g

Price = $6.60

Therefore, the price of the given 1 kg of fries is $6.60.

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Answer:

This is the simplest radical form of 128.

8√2 .

Step-by-step explanation:

√128=√16×√8→4×√4×√2=8√2 .

a.  The probability is approximately \(0.1847\).

b. The probability of exactly 2 times falling between 1 and 5 seconds is approximately \(0.3038\).

(a) To find the probability that between 1 and 5 seconds elapse between emits, we can use the Poisson distribution. Given a mean rate of 3 emits per second, the parameter \($\lambda$\) for the Poisson distribution is also 3.

Let \($X$\) be the random variable representing the waiting time between emits. We want to find \($P(1 \leq X \leq 5)$\).

Using the Poisson distribution formula, we can calculate this probability:

\($P(1 \leq X \leq 5) = P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5)$\)

\(P(X = k) = \frac{e^{-\lambda} \lambda^k}{k!}\)

Substituting \($\lambda = 3$ and $k = 1, 2, 3, 4, 5$\), we have:

\($P(1 \leq X \leq 5) = \frac{e^{-3} 3^1}{1!} + \frac{e^{-3} 3^2}{2!} + \frac{e^{-3} 3^3}{3!} + \frac{e^{-3} 3^4}{4!} + \frac{e^{-3} 3^5}{5!}$\)

Calculating this expression, we find that the probability is approximately \(0.1847\).

(b) When selecting 10 times between emissions randomly, the number of times falling between 1 and 5 seconds follows a binomial distribution. The probability of exactly 2 times falling in this range can be calculated using the binomial distribution formula:

\($P(X = 2) = \binom{10}{2} \cdot (P(1 \leq X \leq 5))^2 \cdot (1 - P(1 \leq X \leq 5))^{(10 - 2)}$\)

Substituting the probability from part (a), we have:

\($P(X = 2) = \binom{10}{2} \cdot (0.1847)^2 \cdot (1 - 0.1847)^{(10 - 2)}$\)

Calculating this expression, we find that the probability of exactly 2 times falling between 1 and 5 seconds is approximately 0.3038.

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As per the given statement a.) The probability that between 1 and 5 seconds elapse between emits is 5.26%. b.) The probability that exactly 2 of the emissions are between 1 and 5 seconds is 27.05%.

a. To find the probability that between 1 and 5 seconds elapse between emits in a Poisson process with a mean rate of 3 per second, we can use the exponential distribution.

The exponential distribution is characterized by the parameter lambda \((\(\lambda\))\), which is equal to the mean rate of the process. In this case, \(\(\lambda = 3\).\) The probability density function (PDF) of the exponential distribution is given by:

\(\[ f(t) = \lambda e^{-\lambda t} \]\)

To find the probability that between 1 and 5 seconds elapse between emits, we need to calculate the integral of the PDF from 1 to 5 seconds:

\(\[ P(1 \leq t \leq 5) = \int_{1}^{5} \lambda e^{-\lambda t} dt \]\)

Integrating the PDF, we have:

\(\[ P(1 \leq t \leq 5) = \left[ -e^{-\lambda t} \right]_{1}^{5} \]\)

\(\[ P(1 \leq t \leq 5) = -e^{-3 \cdot 5} - (-e^{-3 \cdot 1}) \]\)

\(\[ P(1 \leq t \leq 5) = -e^{-15} + e^{-3} \]\)

\(\[ P(1 \leq t \leq 5) \approx 0.0526 \]\)

Therefore, the probability that between 1 and 5 seconds elapse between emits is approximately 0.0526, or 5.26%.

b. If we randomly select 10 times between emissions, and we assume they are independent, we can model the situation as a binomial distribution.

The probability of having exactly 2 times between emissions between 1 and 5 seconds can be calculated using the binomial probability formula:

\(\[ P(X = k) = \binom{n}{k} p^k (1-p)^{n-k} \]\)

where \(\( n \)\) is the number of trials (10), \(\( k \)\) is the number of successful trials (2), and \(\( p \)\) is the probability of success (probability that between 1 and 5 seconds elapse between emits).

From part a, we know that \(\( P(1 \leq t \leq 5) \approx 0.0526 \).\)

Plugging in these values into the formula:

\(\[ P(X = 2) = \binom{10}{2} (0.0526)^2 (1 - 0.0526)^{10 - 2} \]\)

\(\[ P(X = 2) = 45 \cdot (0.0526)^2 \cdot (0.9474)^8 \]\)

\(\[ P(X = 2) \approx 0.2705 \]\)

Therefore, the probability that exactly 2 of the times between emissions are between 1 and 5 seconds is approximately 0.2705, or 27.05%.

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2x - 7

x = 5

The value of the expression.

2x - 7

By putting the value of x, we get

2(5) - 7 = 10 - 7 = 3

The value of the expression is 3.

Answer:

B

Step-by-step explanation:

When compared to the number of persons who were there in 2008, the number of people who were present in 2010 was roughly 1.26 times higher.

In 2008, the professional football team's home games averaged an attendance of around 4.32 times 10-5 people. The number of people who attended from their homes reached around 5.44 times 10-5 in the year 2010. We can determine how many times larger the attendance was in 2010 in comparison to 2008 by dividing the number of people who attended in 2010 by the number of people who attended in 2008.

The approximate value that is arrived at after taking 5.44 x 10-5 and dividing it by 4.32 x 10-5 is 1.26. As a direct consequence of this, the total number of individuals who participated in the event in 2010 was roughly 1.26 times more than the total number of people who participated in the event in 2008.

Between the years 2008 and 2010, there was an increase in attendance that was approximately equivalent to a 26 percent increase. Attendance at the professional football team's games has increased, which is a direct reflection of the growing interest in, and support for, the team over the course of the past two years.

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Answer: 3 and 7

Step-by-step explanation: 3 and 7 because they don't go into anything less than 21, so multiply them together, and you'll get 21

Answer:

qwertzuiopsadfghjkly

Answer:

3

Step-by-step explanation:

y+3=-y+9

Move -y to the left and move +3 to the right and change the + into a -

y+y=-3+9

Calculate

2y=6

Divide both members by 2

y=3

Answer:

1. x= -1 & x= -13

2. x= -3 & x= -4

3. x= -3 & x= -5

4. x= -6 & x= -2

5. x= -5 & x= -2

8. x= -7 & x= -3

9. x= -8 & x= -2

10. x= -3 & x= -3

11. x= -5 & x= -4

12. x= -2 & x= -2

13. x= -6 & x= -7

14. x= -6 & x= -3

15. x= -10 & x= -7

\(\longrightarrow\)Let us write all expression in factorised form .

\(\leadsto\) Common factors are 2,x,y,z and their product will give us highest common factor

\(\implies \sf 2\times x\times y \times z\)

\(\implies \sf 2xyz\)

Answer:

Similar to the question: They figure that they will drive 792 miles round trip. If their car gets 25 miles per gallon and the current cost of gasoline is $2.35, what will the gas for their trip cost? $74.45 George is taking the same trip to San Antonio (792 miles)

Step-by-step explanation:

Hope it helps :)

From $1786 Bill Dupp’s Lumber Yard delivered 620 cubic feet of sand.

A numerical expression is a mathematical statement written in the form of numbers and unknown variables. We can form numerical expressions from statements.

Given, Bill Dupp’s Lumber Yard charges $2.80 for each cubic foot of sand delivered, plus $50 to deliver the sand.

Let, 'x' be the number of cubic foot of sand delivered.

Therefore,

2.80x = 50 + 1786

2.80x = 1786 - 50.

2.80x = 1736.

x = 1736/2.80.

x = 620 cubic foot of sand.

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Answer:

Step-by-step explanation:

IT IS 12

Answer hi guys wanted to say if you think about it 12 is the perfect spot so it 12

Step-by-step explanation:

Answer:

Which plave value is it? if it was 10's, it would be 11.

\(\huge\text{\bf Hey there!}\)

\(\large\text{GUIDE}\downarrow\\\large\text{Tenths place has \bf 1 decimal place}\\\large\text{Hundredths has \bf 2 decimal places (also known as cents)}\\\large\text{Thousandths has \bf 3 decimal places}\\\large\text{Ten thousandths has \bf 4 decimal places}\\\large\text{Hundred thousanths has \bf 5 decimal places}\\\large\text{Millionths has \bf 6 decimal places}\\\large\text{Ten millionths has \bf 7 decimal places}\)

\(\large\text{Hundred millionths has \bf 8 decimal places}\\\large\text{Billionths has \bf 9 decimal places}\)

\(\large\textsf{Tenths}\rightarrow\large\textsf{ 10.80 (also known as \bf 10.8)}\\\\\large\textsf{Hundredths}\rightarrow\large\textsf{10.75}\\\\\large\textsf{Thousandths}\rightarrow\large\textsf{10.750}\\\\\large\textsf{Ten thousandths}\rightarrow\large\textsf{10.7500}\\\\\large\textsf{Hundred thousandths}\rightarrow\large\textsf{10.7500}\\\\\large\textsf{Millionths}\rightarrow\large\textsf{10.750000}\\\\\large\textsf{Ten millionths}\rightarrow\large\textsf{10.7500000}\)

\(\large\textsf{Hundred millionths}\rightarrow\large\textsf{10.75000000}\\\\\large\textsf{Billionths}\rightarrow\large\textsf{10.750000000}\)

\(\large\textsf{YOUR NUMBERS BROKEN DOWN TO SIMPLIER TERMS}\\\large\text{1 is in the \bf TENS place}\\\large\text{0 is in the \bf ONES place}\\\large\text{.}\\\large\text{7 is in the \bf TENTHS place}\\\large\text{5 is in the \bf HUNDREDTHS place}\)

The next term in the sequence is 32. The formula for the nth term of this sequence is \(a_n = 8 + (n - 1) * 6\). The sum of the first 100 terms of the sequence is 30100.

To find the next term in the sequence, we need to determine the pattern of the sequence. By observing the given sequence 8, 14, 20, 26, we can see that each term is obtained by adding 6 to the previous term. Therefore, the common difference in this arithmetic sequence is 6.

So, the next term in the sequence is 32.

To find the formula for the nth term of an arithmetic sequence, we can use the formula:

\(a_n = a_1 + (n - 1) * d\),

where \(a_n\) represents the nth term, \(a_1\) is the first term, n is the position of the term, and d is the common difference.

In this case, the first term (\(a_1\)) is 8, and the common difference (d) is 6. Plugging these values into the formula, we can determine the nth term:

\(a_n = 8 + (n - 1) * 6\).

To find the sum of the first 100 terms of the sequence, we can use the formula for the sum of an arithmetic series:

\(S_n = (n/2) * (a_1 + a_n)\),

where \(S_n\) represents the sum of the first n terms.

In this case, we want to find the sum of the first 100 terms, so n = 100. Plugging in the values of n, \(a_1\), and \(a_n\) into the formula, we can calculate the sum:

\(S_{100} = (100/2) * (8 + a_{100})\).

Since we already have the formula for the nth term (\(a_n\)), we can substitute that into the formula for the sum:

\(S_{100} = (100/2) * (8 + (100 - 1) * 6)\).

Now we can simplify this expression to find the sum of the first 100 terms.

\(S_{100} =30100\).

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Answer:

x²-10x+34=0. here u go I'm 12

Each box will contain 1 chocolate donut, 1 strawberry donut, and 1 caramel donut, there will be a Total of 4 boxes, and each box will contain 1 chocolate donut, 1 strawberry donut, and 1 caramel donut.

The number of boxes and the distribution of donuts in each box, we need to find the greatest common divisor (GCD) of the total number of chocolate, strawberry, and caramel donuts available. The GCD will represent the maximum number of donuts that can be included in each box.

First, let's find the GCD of 36, 27, and 18. By calculating the GCD, we can determine the maximum number of donuts that can be included in each box.

GCD(36, 27, 18) = 9

Therefore, the maximum number of donuts that can be included in each box is 9.

Next, we need to determine the number of boxes. To do this, we divide the total number of each donut type by the maximum number of donuts per box.

Number of boxes for chocolate donuts = 36 / 9 = 4 boxes

Number of boxes for strawberry donuts = 27 / 9 = 3 boxes

Number of boxes for caramel donuts = 18 / 9 = 2 boxes

Since each box contains the same total number of donuts, we can conclude that there will be 4 boxes with chocolate donuts, 3 boxes with strawberry donuts, and 2 boxes with caramel donuts.

To find the distribution of donuts in each box, we divide the maximum number of donuts per box by the GCD:

Distribution in each box: 9 = 1 × 9

Therefore, each box will contain 1 chocolate donut, 1 strawberry donut, and 1 caramel donut, there will be a total of 4 boxes, and each box will contain 1 chocolate donut, 1 strawberry donut, and 1 caramel donut.

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Answer:

6/5

Step-by-step explanation:

rise/run

I hope this is right! :D

Answer:

The slope would be 6/5

Answer:

$94.00

Step-by-step explanation:

First, do \(79.9*\frac{100}{85}\)

Then multiply to get your answer of 94.

Answer:

Step-by-step explanation:

im asuming you want to find x.

25 + 16 = 41.

the length of the rectangle is sqrt41

64-41 = 23

the width of the rectangle is sqrt23

21.The probability that a class of 10 students will have more than 4 males is 0.3222

22.The probability that a number drawn from this distribution will have a value between -1 and 2 is 0.8185.

The probability that a class of 10 students will have more than 4 males is given by:

P(X > 4) = 1 - P(X ≤ 4)

The probability that a student is male is

P(male) = 1 - P(female)

            = 1 - 0.80

            = 0.20

The number of male students in a class of 10 students follows a binomial distribution with n = 10 and p = 0.20.

The probability that a class of 10 students will have exactly 4 males is:

P(X = 4) = nCₓpˣ(1 - p)ⁿ⁻ˣ

where n = 10,

p = 0.20, and

x = 4

nCₓ = n! / x!(n - x)!

      = 10! / 4!(10 - 4)!

      = 210P(X = 4)

      = 210 × (0.20)⁴ × (0.80)⁶

      = 0.0880

The probability that a class of 10 students will have 4 or fewer males is:

P(X ≤ 4) = Σ P(X = x) where

x = 0, 1, 2, 3, 4

P(X ≤ 4) = P(X: 0) + P(X: 1) + P(X :2) + P(X: 3) + P(X:4)

P(X ≤ 4) = (0.80)¹⁰ + 10 × (0.20)¹ × (0.80)⁹ + 45 × (0.20)² × (0.80)⁸ + 120 × (0.20)³ × (0.80)⁷ + 210 × (0.20)⁴ × (0.80)⁶

             = 0.6778P(X > 4)

             = 1 - P(X ≤ 4)

             = 1 - 0.6778

             = 0.3222

Therefore, the probability that a class of 10 students will have more than 4 males is 0.3222

22.Let X be the random variable of the normal probability distribution having a mean of 0 and a standard deviation of 1.

The probability that a number drawn from this distribution will have a value between -1 and 2 is given by:

P(-1 ≤ X ≤ 2) = Φ(2) - Φ(-1) where

Φ(z) is the standard normal cumulative distribution function, which gives the probability that a number drawn from a standard normal distribution is less than or equal to z.

Using a standard normal distribution table or calculator, we can find that:

Φ(2) ≈ 0.9772Φ(-1)

       ≈ 0.1587

Therefore,P(-1 ≤ X ≤ 2) = Φ(2) - Φ(-1)

                                     = 0.9772 - 0.1587

                                     = 0.8185

The probability that a number drawn from this distribution will have a value between -1 and 2 is 0.8185.

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The formula for calculating the required sample size to estimate the population mean with a 90% confidence level is given by:

n = ((z_(α/2)×σ) / E)²Here, z_(α/2) is the z-value for the given level of confidence (90% in this case), σ is the population standard deviation (6.8 in this case), and E is the maximum error we can tolerate (0.02 in this case).

Substituting the given values in the formula, we get:

n = ((z_(α/2)×σ) / E)²n = ((1.645×6.8) / 0.02)²n = 1910.96

Rounding up to the nearest whole number, we get the required sample size to be 1911.

Therefore, a sample size of 1911 is required to estimate the population mean with a 90% confidence level and an error of less than 0.02.

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