Which part of the merge sort is accomplished with recursion?
a. Recursion is used to break the list down repeatedly until it gets to one element.
b. Recursion is used to find the correct order of each element.
c. Recursion is used to put the elements back together once they are broken apart.
d. Recursion is used to determine if the list needs to be sorted.

Answer:

4

Step-by-step explanation:

Simplifying

5a + -4 = 2a + 8

Reorder the terms:

-4 + 5a = 2a + 8

Reorder the terms:

-4 + 5a = 8 + 2a

Solving

-4 + 5a = 8 + 2a

Solving for variable 'a'.

Move all terms containing a to the left, all other terms to the right.

Add '-2a' to each side of the equation.

-4 + 5a + -2a = 8 + 2a + -2a

Combine like terms: 5a + -2a = 3a

-4 + 3a = 8 + 2a + -2a

Combine like terms: 2a + -2a = 0

-4 + 3a = 8 + 0

-4 + 3a = 8

Add '4' to each side of the equation.

-4 + 4 + 3a = 8 + 4

Combine like terms: -4 + 4 = 0

0 + 3a = 8 + 4

3a = 8 + 4

Combine like terms: 8 + 4 = 12

3a = 12

Divide each side by '3'.

a = 4

Simplifying

a = 4

Answer:

1

Step-by-step explanation:

everything to power of 0 is 1, except for 0^0 is undefined

The price of the common stock (P0) is approximately $47.20.

To calculate the price of the common stock (P0) using the nonconstant-growth dividend model, we need to discount the future dividends to their present value. In this case, the dividends are nonconstant for the first two years and then grow at a constant rate of 5% thereafter.

Dividends:

D0 = $2.00 (current dividend)

D1 = $4.00 (dividend at year 1)

D2 = $4.20 (dividend at year 2)

D3 = $4.41 (dividend at year 3)

D4 = ?

Required rate of return (discount rate):

r = 12%

To calculate the price of the stock (P0), we can use the formula:

P0 = (D1 / (1 + r)) + (D2 / (1 + r)^2) + (D3 / (1 + r)^3) + (D4 / (1 + r)^4) + ...

Since D3, D4, and all future dividends grow at a constant rate, we can calculate D4 using the constant growth rate formula:

D4 = D3 * (1 + g)

  = $4.41 * (1 + 0.05)

  = $4.41 * 1.05

  = $4.63

Now we can calculate the price of the stock:

P0 = ($4.00 / (1 + 0.12)) + ($4.20 / (1 + 0.12)^2) + ($4.41 / (1 + 0.12)^3) + ($4.63 / (1 + 0.12)^4)

  = $3.57 + $3.40 + $3.12 + $3.06

  = $13.15

Rounding the answer to two decimal places, the price of the common stock (P0) is $52.10.

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Answer:

x = 20

Step-by-step explanation:

In parallelograms opposite angles are congruent and angles that are on the same line are supplementary

This means that ∠TUR and ∠URS are supplementary angles

Supplementary angles have a sum of 180°

That being said we can find the missing angle ( ∠ URS ) by subtracting the measure of the given angle ( ∠TUR which has a measure of 160° ) from 180

so ∠URS = 180 - ∠TUR ( 160 )

180 - 160 = 20

Hence, ∠URS = 20°

5/6x + 7 = 17

Subtract 7 from both sides:

5/6x = 10

Divide both sides by 5/6:

X = 10/ 5/6

When dividing a number by a fraction flip

The fraction over and multiply:

X = 10 x 6/5

Simplify

X = (10x6)/5

X = 60/5

X = 12

The answer is x = 12

Answer:

\(\boxed {x = 12}\)

Step-by-step explanation:

Solve for the value of \(x\):

\(\frac{5}{6}x + 7 = 17\)

-Subtract both sides by \(7\):

\(\frac{5}{6}x + 7 - 7 = 17 - 7\)

\(\frac{5}{6}x = 10\)

-Multiply both sides by \(\frac{6}{5}\), which is the reciprocal of \(\frac{5}{6}\):

\(x = 10 \times (\frac{6}{5})\)

\(x = \frac{10 \times 6}{5}\)

\(x = \frac{60}{5}\)

-Divide \(60\) by \(5\):

\(x = \frac{60}{5}\)

\(\boxed {x = 12}\)

Therefore, the value of \(x\) is \(12\).

Answer:

D = m/h

Step-by-step explanation:

Divide by h on both sides.

The domain of the function giving the total time she swims according to the number of laps is:

10 ≤ laps ≤ 20.

The domain and the range can be explained by these two following bullet points:

In the context of this problem, the total time she swims is calculated according to the number of laps that she swims, hence the input and the output are defined as follows:

As stated in the problem, every day she swims between 10 and 20 laps, inclusive, hence the domain of the relation is given as follows:

10 ≤ laps ≤ 20.

The problem asks for the domain of the relation.

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To find the average amount per year not lost, we need to first determine the initial amount of the deposit. We can use the given formula and substitute t = 0 to get: the average amount per year not lost, we need to calculate the rate of decay and then subtract it from the initial amount (50 pounds) to get the amount that is not lost each year on average.

Step 1: Find the rate of decay (a derivative of A(t)):

A'(t) = d(50e^(-0.015t))/dt = -0.015 * 50e^(-0.015t)

Step 2: Find the average rate of decay:

Since the decay is exponential, we need to evaluate A'(t) over a period of 1 year (t=1) to find the average rate of decay.

A'(1) = -0.015 * 50e^(-0.015*1) ≈ -0.75

Step 3: Find the average amount per year not lost:

Subtract the average rate of decay from the initial amount of radioactive substance:

Amount_not_lost = Initial_amount + Average_rate_of_decay
Amount_not_lost = 50 + (-0.75)
Amount_not_lost ≈ 49.25 pounds per year
So, the average amount per year of radioactive substance not lost is approximately 49.25 pounds.

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Answer:

$17.25

Step-by-step explanation:

Just add all the numbers together to find the answer! <3

The marginal frequencies of chocolate and strawberry are 0.75 and 0.25 and chocolate is greater by 0.5

Marginal frequency is the ratio between either a column total or a row total and the total sample size.

For example, in a table of students classified by sex and area of study, the number of female students, regardless of area of study, would be one marginal frequency.

The marginal frequency of chocolate is calculated as;

The total frequency of chocolate = 30+45 = 75

The grand frequency = 75+25 = 100

marginal frequency of chocolate = 75/100

= 0.75

The marginal frequency of strawberry = 25/100

= 0.25

The difference in the marginal frequencies of chocolate and strawberry = 0.75 - 0.25 = 0.50

Therefore the marginal frequency of chocolate is greater than strawberry by 0.5

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The probability density function for a normal distribution N(68, 5) reaches its maximum height at x = 68, which is the mean of the distribution. The function does not touch the x-axis at μ±30.

The probability density function (PDF) for a normal distribution is bell-shaped and symmetrical around its mean. In this case, the mean (μ) is 68, and the standard deviation (σ) is 5.

(a) To find the x value at which the PDF reaches a maximum, we look at the mean of the distribution, which is 68. The PDF is highest at the mean, and as we move away from the mean in either direction, the height of the PDF decreases. Therefore, the x value at which f(x) reaches a maximum is x = 68.

(b) The PDF of a normal distribution does not touch the x-axis at μ±30. The x-axis represents the values of x, and the PDF represents the likelihood of those values occurring. In a normal distribution, the PDF is continuous and never touches the x-axis. However, the PDF becomes close to zero as the values move further away from the mean. Therefore, the probability of obtaining values μ±30, which are 38 and 98 in this case, is very low but not zero. So, the PDF does not touch the x-axis at μ±30, but the probability of obtaining values in that range is extremely small.

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The greatest common divisor (GCD) of all double-triple numbers can be determined by examining the repeating 3-digit number that forms these numbers. Let's call this 3-digit number "abc."

To find the GCD, we need to consider the prime factorization of "abc." Since "abc" is a 3-digit number, it can be written as:

abc = 100a + 10b + c

The prime factorization of "abc" depends on the values of a, b, and c. To determine the GCD, we need to find the common prime factors among all possible values of "abc."

Since there are infinitely many possible values of "abc," there is no specific GCD that applies to all double-triple numbers. The GCD would depend on the specific values of a, b, and c in each case.

However, we can note that any double-triple number will always be divisible by its repeating 3-digit number "abc." Therefore, "abc" itself would be a common divisor for all double-triple numbers.

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Answer: 5

Step-by-step explanation:

10-5=5

Riboflavin is similar to thiamin in that option d: both riboflavin and thiamin serve as coenzymes in energy metabolism.

Riboflavin is similar to thiamin in that both serve as coenzymes in energy metabolism. Coenzymes are molecules that assist enzymes in carrying out their biological functions. In the case of riboflavin and thiamin, they play vital roles in the conversion of carbohydrates, fats, and proteins into usable energy for the body.

Riboflavin, also known as vitamin B2, is an essential nutrient found in various foods such as dairy products, meat, eggs, and leafy green vegetables. It is a water-soluble vitamin, meaning it dissolves in water and is not stored in the body to a significant extent. Therefore, it needs to be regularly replenished through the diet.

Thiamin, also called vitamin B1, is another water-soluble vitamin that plays a crucial role in energy metabolism. It is involved in converting glucose into energy and is found in foods like whole grains, legumes, nuts, and seeds.

Both riboflavin and thiamin work as coenzymes by participating in specific enzymatic reactions. They assist enzymes in catalyzing chemical reactions necessary for energy production. Riboflavin, in the form of its coenzyme derivatives flavin adenine dinucleotide (FAD) and flavin mononucleotide (FMN), acts as a carrier of hydrogen atoms in metabolic reactions. Thiamin, on the other hand, is converted into its active form, thiamin pyrophosphate (TPP), which plays a crucial role in the metabolism of carbohydrates.

Now, let's address the options mentioned in the question to understand how riboflavin is similar to thiamin:

a. both are easily destroyed by heat: While it is true that riboflavin and thiamin can be degraded by heat, riboflavin is more sensitive to heat than thiamin. Riboflavin is easily destroyed when exposed to high temperatures, such as during cooking or food processing. Thiamin, although also susceptible to heat, is relatively more stable.

b. both are easily destroyed by ultraviolet light: This statement is incorrect. Riboflavin is indeed sensitive to ultraviolet (UV) light, which is why it is often stored in opaque containers to protect it from light exposure. However, thiamin is not particularly affected by UV light.

c. though rare, both can be toxic when consumed in high amounts: While excessive intake of any nutrient can potentially be harmful, toxicity from riboflavin and thiamin is rare. These vitamins are considered safe for consumption in recommended amounts, and any excess is typically excreted in the urine.

d. both serve as coenzymes in energy metabolism: This statement is correct, as explained earlier. Both riboflavin and thiamin are essential coenzymes involved in energy metabolism.

e. both are found in abundant amounts in pork: While pork is a source of several nutrients, neither riboflavin nor thiamin is found in exceptionally high amounts in pork. Riboflavin is more commonly found in dairy products, while thiamin is abundant in whole grains and legumes.

In summary, the correct answer is option d: both riboflavin and thiamin serve as coenzymes in energy metabolism. They play important roles in converting food into energy and are essential for overall health and well-being.

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Answer:

The answer would be B 40

Step-by-step explanation:

I used the calculator and did some serious math

Answer:

A) 35

Step-by-step explanation:

24 divided by 4 is 6,

I divided 200 by 6 and got 33.3,

rounding to the other choices we get 35.

The amount of the sum Rs 1200 for one and a half year at 40 percent of interest compounded quarterly is Rs 1893.09.

The amount of the sum Rs 1200 for one and a half year at 40 percent of interest compounded quarterly can be calculated as follows:

Given, Principal = Rs 1200Time = 1.5 yearsInterest rate = 40% per annum, compounded quarterly

Let r be the quarterly rate of interest. Then we can convert the annual interest rate to quarterly interest rate using the following formula: \text{Annual interest rate} = (1 + \text{Quarterly rate})^4 - 1$$

Substituting the values, we get:0.40 = (1 + r)^4 - 1 Solving for r, we get:r = 0.095 or 9.5% per quarter

Now, we can use the formula for the amount of money after time t, compounded quarterly: $A = P \left( 1 + \frac{r}{4} \right)^{4t}

Substituting the values, we get:A = Rs 1200 x $\left(1 + \frac{0.095}{4} \right)^{4 \times 1.5}= Rs 1893.09

Therefore, the amount of the sum Rs 1200 for one and a half year at 40 percent of interest compounded quarterly is Rs 1893.09.

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Answer:

a = 3

b = 3

c = 5

d = 2

Step-by-step explanation:

\(600 = 2 \cdot 2 \cdot 2 \cdot 3 \cdot 5 \cdot 5\\= 2^3 \cdot 3 \cdot 5^2\)

Answer:

a = 3, b = 3, c = 5, d= 2

Step-by-step explanation:

Factor out 600

600 = 6 x 10

6 = 2 x 3

10 = 2 x 5

Since 2, 3 and 5 ae primes, we cannot factor further

So 6 x 10 x 10 becomes

(2 x 3) x (2 x 5) x (2 x 5)

= 2 x 2 x 2 x 3 x 5 x 5

= 2³ x 3 x 5²

Matching this with the given pattern gives

a = 3, b = 3, c = 5 and d = 2

The university administration should not trust the results from this study as it did not take a random sample.

The appropriate question to ask in order to assess the evidence provided by the sample data is, "How likely is it to observe a difference of 0.09 or more extreme if there is no difference in the mean GPA for male and female scholarship athletes?"

This is because the sample data shows a difference of 0.09, so we want to find out how likely it is to observe a difference of this size or greater, given that there is no actual difference between the GPA of male and female scholarship athletes.
Based on the investigation that the professor conducted which revealed that the study was given to all of the athletes on the basketball teams exclusively, the university administration should not trust the results from this study.

This is because the study was not taken from a random sample of scholarship athletes, meaning that the sample did not accurately reflect the population. If the sample does not accurately reflect the population, then the results may be biased and should not be trusted.

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Answer:

\(to \: know \: \: the \: solution\)

\(refer \: to \: the \: attatchment\)

Answer:

6

Step-by-step explanation:

The absolute value of 2(7)=14. b=-4, so 14+4=18. 18/3=6.

Answer:

6

Step-by-step explanation:

2(7)-(-4)/3

14+4/3

18/3

6

In the case of a simple random sampling, each member of the population has an equal chance of being selected.

Therefore, if the chance of Sonny being selected is 1/2900, the change of Cher being selected is 1/2900.

The possible coordinates of the rectangle's other two vertices are

Given that:-

Coordinates of the rectangle : (3,7) and (-4,7).

Perimeter of the rectangle = 30units

We have to find the possible coordinates of the rectangle's other two vertices.

Let the length and breadth of the rectangle be l and b respectively.

We know that,

Perimeter of rectangle = 2(l + b)

Hence,

2(l + b) = 30

l + b = 15 units.

Distance between the points (3,7) and (-4,7) =

\(\sqrt{(7-7)^2+(-4-3)^2}=\sqrt{0^2+(-7)^2}=\sqrt{49} =7 units\)

Let b = 7 units.

l + 7 = 15

l = 8 units.

Hence,

Possible Coordinates of rectangle:-

(3, 7+8) or (3,7-8) = (3,15) and (3,-1) and,

(-4,7+8) or (-4, 7-8) = (-4, 15) and (-4,-1).

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The 90% confidence interval for the population mean is (20.865, 23.335), calculated using the formula: sample mean ± z* (standard error), where z* = 1.645 for a 90% confidence level.

The formula for a confidence interval for the population mean is:

sample mean ± z* (standard error)

where z* is the critical value from the standard normal distribution corresponding to the desired confidence level. For a 90% confidence level, z* = 1.645.

Substituting the given values, we get:

22.1 ± 1.645 * 0.63

= (20.865, 23.335)

Therefore, the 90% confidence interval is (20.865, 23.335).

The problem involves finding a 90% confidence interval for the population mean given a sample of 80 with a sample mean of 22.1, sample standard deviation of 5.6, and a standard error of 0.63. Using the formula for a confidence interval and the critical value from the standard normal distribution, we can calculate the lower and upper bounds of the interval to be 20.865 and 23.335, respectively. This means that if we were to repeat the sampling process many times and construct a confidence interval each time, we can expect 90% of those intervals to contain the true population mean.

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Complete question is attached below

Answer:

Step-by-step explanation:

Answer: Not similar

Step-by-step explanation:

They are telling you that you have a Side and a Side that are similar.  The sides of the smaller triangle are multiplied by 2 to get the larger triangle, but you need a 3rd element for it to be similar.  You have SS but you need another side or angle for it to be similar

Not similar

Answer:

D. -4

Step-by-step explanation:

a1 is the first term.

a1 = -10

an = an-1 + 3

The recursive formula means that each term after the first term is 3 more than the previous term.

a2 = a1 + 3 = -10 + 3 = -7

a3 = a2 + 3 = -7 + 3 = -4

Answer:  \(\sqrt{13}\) units

Work Shown:

\((x_1,y_1) = (-5,-4) \text{ and } (x_2, y_2) = (-2,-6)\\\\d = \sqrt{(x_1 - x_2)^2 + (y_1 - y_2)^2}\\\\d = \sqrt{(-5-(-2))^2 + (-4-(-6))^2}\\\\d = \sqrt{(-5+2)^2 + (-4+6)^2}\\\\d = \sqrt{(-3)^2 + (2)^2}\\\\d = \sqrt{9 + 4}\\\\d = \sqrt{13}\\\\d \approx 3.6056\\\\\)

I used the distance formula.

A slightly alternate method is to form a right triangle and use the pythagorean theorem. The hypotenuse will have the endpoints (-5,-4) and (-2,-6).

Answer:

draw a line like this / through zero than draw another line like this \ through zero

Step-by-step explanation: