Mathematics High School

## Answers

**Answer 1**

There are 210 ways to **assign** grades to a class of seven students if nobody receives an R and exactly two students receive a C.

To solve this **problem**, we need to **determine** how many ways we can select 2 students out of 7 to receive a C, and then how many ways we can assign grades to the remaining 5 students.

Firstly, we need to choose 2 students out of 7 to receive a C. This can be done in 7 choose 2 ways, which is equal to (7!)/(2!*(7-2)!) = 21 ways.

Once we have **selected** the two students to receive a C, we need to assign grades to the remaining 5 students. Since nobody can receive an R, each of the remaining students can receive one of 4 grades: A, B, C (other than the two who have already received a C), or D. Therefore, there are 4^5 = 1024 ways to assign grades to the remaining 5 students.

Finally, we need to multiply the number of ways to select 2 students to receive a C by the number of ways to assign grades to the remaining 5 students to get the total number of ways to assign grades to the entire class. Therefore, the answer is 21 * 1024 = 21,504.

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## Related Questions

Assume calendar calendar = new gregoriancalendar(). ________ returns the number of days in a month

### Answers

This will return the number of days in the month represented by the **calendar object**.

In **Java**, the getActualMaximum() method can be used to return the number of days in a month for a given Calendar object. The** statement **would be:

int daysInMonth = calendar.getActualMaximum(Calendar.DAY_OF_MONTH);

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In the equation f(x) = 4000 (1.02), my a value is

and my b value is?

### Answers

In the **equation** f(x) = 4000 (1.02)ˣ, the a value is 4,000 and the b value is 2% or 0.02.

What is the equation?

The above equation depicts an **exponential growth function**.

**Exponential growth functions** are represented as f(x) = a(1 + r)ˣ, where a is the initial value, r is the growth rate, and x is the period of growth.

In the above **equation**, b represents the growth rate or pattern, usually depicted in percentages, while a is the initial or beginning value of the data.

Thus, we have represented the variables a and b in the above **exponential growth equation**.

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Complete Question:

In the equation f(x) = 4000 (1.02)ˣ my a value is and my b value is?

eddie clauer sells a wide variety of outdoor equipment and clothing. the company sells both through mail order and via the internet. random samples of sales receipts were studied for mail-order sales and internet sales, with the total purchase being recorded for each sale. a random sample of 15 sales receipts for mail-order sales results in a mean sale amount of $67.10 with a standard deviation of $28.75 . a random sample of 9 sales receipts for internet sales results in a mean sale amount of $77.80 with a standard deviation of $15.25 . using this data, find the 98% confidence interval for the true mean difference between the mean amount of mail-order purchases and the mean amount of internet purchases. assume that the population variances are not equal and that the two populations are normally distributed. step 1 of 3 : find the critical value that should be used in constructing the confidence interval. round your answer to three decimal places.

### Answers

Analysing the **data**, we can find the critical value t* for a two-tailed test with **alpha** = 0.01 and df = 17.03, which is approximately 2.898.

Since we are **constructing** a 98% confidence interval, we need to find the critical value that **corresponds** to a **two-tailed test** with alpha = 0.01. We can use a t-distribution since the sample sizes are small and the population **variances** are not assumed to be equal.

To find the critical value, we need to determine the degrees of freedom (df) using the formula df = (s1^2/n1 + s2^2/n2)^2 / [(s1^2/n1)^2 / (n1 - 1) + (s2^2/n2)^2 / (n2 - 1)], where s1 and s2 are the sample standard deviations, n1 and n2 are the sample sizes.

Plugging in the values, we get df = 17.03. Using a t-table or calculator, we can find the critical value t* for a two-tailed test with alpha = 0.01 and df = 17.03, which is approximately 2.898.

This means that if we were to repeat this study many times, we would expect the true mean difference between mail-order and internet sales to fall within 2.898 standard errors of the sample mean difference 98% of the time.

In other words, we can be 98% confident that the true mean difference between mail-order and internet sales falls within the range of (x1 - x2) ± t* × SE, where x1 and x2 are the sample means, SE is the standard error of the difference, and t* is the critical value we just found.

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4. This water bottle has a base with area B square inches and a height of h inches. Tyler

thinks the volume of the water bottle is Bh. Elena thinks the volume is less than Bh.

Do you agree with either of them? Explain

your reasoning.

### Answers

**Tyler's** statement is **correct** for the theoretical volume of the water bottle, but **Elena** is also **correct** in that the actual volume will be slightly less than Bh due to the curved shape of the bottle.

We have,

The volume of a **cylinder** (such as a water bottle) is given by the formula

V = Bh

Where B is the area of the base and h is the height.

So,

**Tyler's** statement is correct.

However, it's important to note that a water bottle has a curved shape, which means that its actual volume is slightly less than Bh.

This is because the curved sides of the bottle take up some space that is not included in the calculation of Bh.

So,

**Elena** is also correct in that the actual volume is less than Bh.

Thus,

**Tyler's** statement is **correct** for the theoretical volume of the water bottle, but **Elena** is also **correct** in that the actual volume will be slightly less than Bh due to the curved shape of the bottle.

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a square has an area of 49 square inches. if the same amount is added to the length and removed from the width, the resulting rectangle has an area of 45 square inches. find the dimensions of the rectangle.

### Answers

**Answer:**

length = 9 in

width = 5 in

**Step-by-step explanation:**

Algebraic equations and solving:First **find** the **side of square** from the given area.Frame the **expression** for **length** and **width** with the help of the side of square.Frame the equation for finding the area of rectangle.Now, by **solving** the equation, we can find the **dimensions** of the rectangle.

Area of square =side * side

Side² = 49

Side = √49

= 7 inches

Let the measurement added to the length and removed from the width be 'x'.

length = (7 + x) in

width = (7 -x )in

**Area of rectangle = length *width**

length * width = 45 square inches

(7 + x)(7-x) = 45

{**Identity: (a + b)(a -b) = a² - b² where a = 7 & b = x**}

7² - x² = 45

49 - x² = 45

-x² = 45 - 49

-x² = - 4

x² = -4 ÷ (-1)

x² = 4

x = √4

x = 2

**Dimension of the rectangle:**

** **length =7 + 2 = 9 in

width = 7 - 2 = 5 in

a town recently dismissed 6 employees in order to meet their new budget reductions. the town had 8 employees over 50 years of age and 17 under 50 . if the dismissed employees were selected at random, what is the probability that less than 5 employees were over 50 ?

### Answers

To find the probability that less than 5 of the 6 dismissed employees were over 50 years of age, we can use the **binomial distribution** with n=6 (the number of trials) and p=8/25 (the probability of success, i.e. being over 50). We need to calculate the **probability** of getting 0, 1, 2, 3, or 4 successes in 6 trials, and then sum these probabilities to get the total probability of less than 5 successes.

Let X be the number of dismissed employees who are over 50. Then X follows a **binomial distribution** with n=6 and p=8/25, since there are 8 over-50 employees out of a total of 25 employees. We want to find P(X < 5), which is the probability of getting less than 5 successes in 6 **trials**.

To calculate this probability, we can use the **cumulative distribution function** (CDF) of the binomial distribution:

P(X < 5) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4)

Using the binomial probability formula, we can calculate each of these probabilities:

P(X = k) = (6 choose k) * (8/25)^k * (17/25)^(6-k)

where (6 choose k) is the number of ways to choose k successes out of 6 trials. Plugging in k=0, 1, 2, 3, and 4, we get:

P(X = 0) = (6 choose 0) * (8/25)^0 * (17/25)^6 = 0.124

P(X = 1) = (6 choose 1) * (8/25)^1 * (17/25)^5 = 0.329

P(X = 2) = (6 choose 2) * (8/25)^2 * (17/25)^4 = 0.329

P(X = 3) = (6 choose 3) * (8/25)^3 * (17/25)^3 = 0.170

P(X = 4) = (6 choose 4) * (8/25)^4 * (17/25)^2 = 0.042

Summing these probabilities, we get:

P(X < 5) = 0.124 + 0.329 + 0.329 + 0.170 + 0.042 = 0.994

Therefore, the** probability** that less than 5 of the 6 dismissed employees were over 50 is approximately 0.994, or 99.4%.

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Of the following choices of 8, which is the largest that could be used successfully with an arbitrary e in an epsilon-delta proof of lim (9x +36) = 9? -3 A. 8 = B. 8 = C. 8 = 2€ D. 8 = 3€ DO E. 8 = 9€

### Answers

**Answer: **Among the given choices, the largest one that could be used successfully in an epsilon-delta proof of lim (9x + 36) = 9 is:

C. 8 = 2ε

**Step-by-step explanation:**

To determine the largest choice of 8 that could be used successfully in an **epsilon-delta** proof, we need to find the maximum value of epsilon (ε) that satisfies the given limit statement.

In an **epsilon-delta** proof, for a given limit statement lim (9x + 36) = 9, we want to show that for any epsilon (ε) greater than 0, there exists a corresponding delta (δ) such that whenever 0 < |x - a| < δ (where 'a' is the value of x approaching), it implies |(9x + 36) - 9| < ε.

Let's consider each choice of 8:

A. 8 = B: This choice doesn't provide any information about epsilon (ε) and is unrelated to the **limit** statement.

B. 8 = C: Similarly, this choice doesn't provide any relevant information.

C. 8 = 2ε: Here, we have epsilon (ε) defined as 2ε. In this case, we can choose delta (δ) such that whenever 0 < |x - a| < δ, it implies |(9x + 36) - 9| < 2ε. Therefore, this choice of 8 could be used successfully in the epsilon-delta proof.

D. 8 = 3ε: With this choice, we would need to find a delta (δ) such that |(9x + 36) - 9| < 3ε. However, since the limit statement only requires |(9x + 36) - 9| < ε, this choice is unnecessarily **restrictive**. Therefore, it is not the largest choice that could be used successfully.

E. 8 = 9ε: Similar to choice D, this choice is overly restrictive and not necessary for the **limit** statement.

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Melissa has 16 stamps 3/4 of the stamps are from the United States. How many are from the United States and how many are not. Answer

### Answers

**Answer:**

**12** from the U.S, and **4** are not from the U.S

**Step-by-step explanation:**

**16 divided by 4 is 4**

**4 times 3 is 12**

**12 minus 16 is 4**

Use the information below to write the equation of the circle.

Center: (-15,1)

Radius: 1

### Answers

The equation of the **circle **is [tex](x + 15)^2 + (y - 1)^2 = 1.[/tex]

The equation of a circle with **center **(a, b) and radius r is:

[tex](x - a)^2 + (y - b)^2 = r^2[/tex]

In this case, the **center **is (-15, 1) and the radius is 1. Therefore, substituting these values into the equation, we get:

[tex](x - (-15))^2 + (y - 1)^2 = 1^2[/tex]

Simplifying and **expanding **the equation, we get:

[tex](x + 15)^2 + (y - 1)^2 = 1[/tex]

Therefore, the equation of the circle is [tex](x + 15)^2 + (y - 1)^2 = 1.[/tex]

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perform the gram-schmidt process on the following sequence of vectors. -6 -3 -6 , 3 3 0 , -7 -2 -1 . , , .

### Answers

The Gram-Schmidt process was applied to the vectors [-6, -3, -6], [3, 3, 0], and [-7, -2, -1], resulting in the **orthogonal vectors** [-0.577, -0.289, -0.577], [0.526, 0.526, -0.667], and [-0.648, 0.729, -0.220].

To perform the **Gram-Schmidt process** on the given sequence of vectors, let's denote the vectors as v1, v2, and v3

v1 = [-6, -3, -6]

v2 = [3, 3, 0]

v3 = [-7, -2, -1]

Set the first orthogonal vector, u1, equal to the normalized v1:

u1 = v1 / ||v1||

Find the **projection **of v2 onto u1:

proj_v2_u1 = (v2 ⋅ u1) * u1

where ⋅ represents the dot product.

Find the orthogonal vector v2_orthogonal by subtracting the projection from v2:

v2_orthogonal = v2 - proj_v2_u1

Set the second orthogonal vector, u2, equal to the normalized v2_orthogonal:

u2 = v2_orthogonal / ||v2_orthogonal||

Find the projection of v3 onto u1 and u2:

proj_v3_u1 = (v3 ⋅ u1) * u1

proj_v3_u2 = (v3 ⋅ u2) * u2

Find the** orthogonal vector** v3_orthogonal by subtracting the projections from v3:

v3_orthogonal = v3 - proj_v3_u1 - proj_v3_u2

Set the third orthogonal vector, u3, equal to the normalized v3_orthogonal:

u3 = v3_orthogonal / ||v3_orthogonal||

The resulting orthogonal vectors u1, u2, and u3 will form an orthonormal basis. Remember to normalize the vectors by dividing by their respective norms (||v||) at each step.

Performing the calculations, the orthogonal vectors are:

u1 = [-0.577, -0.289, -0.577]

u2 = [0.526, 0.526, -0.667]

u3 = [-0.648, 0.729, -0.220]

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an investigator was interested in the relationship between color preference and number of siblings. a test of independence produced a c2 that allowed the null hypothesis to be rejected. the proper conclusion is

### Answers

The proper conclusion when a test of independence produces a chi-square statistic ([tex]\chi^2[/tex]) that allows the **null hypothesis** to be rejected is that there is evidence of a relationship between the variables being studied.

In statistical hypothesis testing, the **chi-square test** of independence is used to determine whether there is a significant association between two categorical **variables**.

The null hypothesis assumes that there is no relationship between the variables, while the alternative hypothesis suggests that there is a relationship.

When the test of independence yields a chi-square statistic that is large enough to reject the null hypothesis, it means that the **observed data** provides evidence against the assumption of independence.

In other words, the results suggest that there is a relationship between the variables being studied.

The rejection of the null hypothesis does not provide information about the nature or **strength** of the relationship.

It simply indicates that the observed data is unlikely to occur if the variables were truly independent.

Therefore, the proper conclusion in this case is that there is evidence of a relationship between color preference and the number of siblings based on the results of the test of independence.

Further analysis may be needed to explore the nature and significance of this relationship.

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Which of the following adjustments to the game would make the game fair? Select ALL that apply.

a) Remove the payout for rolling a 5. b)Increase the cost to play by $1. c) Remove the payout for rolling a 9. d)Include a payout of $6 for rolling a 7. e)Decrease the cost to play by $2.

Dice roll sumNumber of waysProbabilityPayouts

2$72

3$36

4$18

5$9

60

70

80

9$9

10$18

11$36

12$72

### Answers

To make the game fair, we would need to adjust the payouts so that the expected value of playing is zero. This means that the sum of the products of each possible **outcome** and its probability should equal the cost to play.

Removing the payouts for **rolling** a 5 and 9 would both make the game fair, as these outcomes have non-zero payouts but their probabilities are relatively low. Increasing the cost to play by $1 would also make the game fair, as it would decrease the expected value of playing.

However, **decreasing** the cost to play by $2 would make the game less fair, as it would increase the expected value of playing. Including a payout of $6 for rolling a 7 would also make the game less fair, as it would increase the expected value of playing.

Therefore, the **adjustments** that would make the game fair are: removing the payouts for rolling a 5 and 9, and possibly increasing the cost to play by $1.

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What is the probability that you choose a prime number or a double digit in the month of august?

### Answers

Answer:

30/31 or approximately 0.9677 (rounded to four decimal places).

Step-by-step explanation:

There are 31 days in the month of August, and we need to determine the probability of selecting a prime number or a double-digit number.

Prime numbers less than 31 are: 2, 3, 5, 7, 11, 13, 17, 19, 23, 29. Therefore, there are 10 prime numbers less than 31.

Double-digit numbers less than 31 are: 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23, 24, 25, 26, 27, 28, 29, 30.

Therefore, there are 21 double-digit numbers less than 31.

However, we need to make sure that we don't count the number 1 as a prime number.

Therefore, the total number of favorable outcomes (prime numbers or double-digit numbers) is 10 + 21 - 1 = 30.The total number of possible outcomes is 31 (since there are 31 days in August).

Thus, the probability of selecting a prime number or a double-digit number in August is 30/31 or approximately 0.9677 (rounded to four decimal places).

Determine the measures of the arcs and angles:

HE

HD

DG

EG

DEG

HGE

NEED ASAP!!!!

### Answers

The **measures **of the **arcs **and **angles **are

HE = 62 degreesHD = 118 degreesEG = 118 degreesDG = 62 degreesDEG = 31 degreesHGE = 31 degrees

Determining the measures of the arcs and angles:

From the question, we have the following parameters that can be used in our computation:

The circle

The **arc **HE **subtends **the angle HAE with a measure of 62 degrees at the **center **of the **circle**

This means that

HE = 62 degrees

By corresponding **angle theorem**, we have

DG = 62 degrees

Next, we have

HD = EG = 180 - 62

HD = EG = 118 degrees

The **angle **DEG **subtends **the arc DG with a measure of 62 degrees at the **circumference **of the **circle**

This means that

HGE = DEG = 1/2 * 62 degrees

HGE = DEG = 31 degrees

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Latonya needs to purchase 5 notebooks at $1.99 each and 3 pens at $0.89 each. About how much will she spend?

### Answers

The total amount that Latonya will spend is **$12.62**

The cost of **each** notebook is **$1.99**

Latonya needs to purchase **5** notebooks

∴ The cost of 5 notebooks = **Cost of each notebook × 5**

= $1.99 × 5

∴ The **total **cost** **of 5 notebooks = **$9.95**

The cost of each pen is **$0.89**

Latonya needs to purchase **3** pens

∴ The cost of 3 pens = **Cost of each pen × 3**

= $0.89 × 3

∴ The **total** cost of 3 pens =** $2.67**

Total amount that Latonya will spend on notebooks and pens = **Total cost of 5 notebooks + Total cost of 3 pens**

Total amount = $9.95 + $2.67

∴ Total amount = $12.62

Therefore, the total amount that Latonya will spend is $12.62.

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Stellar spherical gas clouds (20 points) The Emden Equation is xy00 + 2y 0 + xy = 0 where x is proportional to the distance from the stellar gas cloud center and y is the density at that distance, normalized such that the density at the center is one. Substitute the series into the Emden Equation and determine a0, a1, a2, a3, a4 and a5. Write down the series you have constructed. Do you recognize it?

### Answers

The** power series** constructed is y(x) = a₀ + (-a₁/3)x + (-a₀/2)x² + (-a₁/3)x³ + (a₀/8)x⁴ + (a₁/15)x⁵ + ...

The **Emden Equation** and determine the coefficients a₀, a₁, a₂, a₃, a₄, and a₅, substitute a power series into the equation and match the coefficients term by term.

Let's assume the power series solution for y(x) is:

y(x) = a₀ + a₁x + a₂x² + a₃x³ + a₄x⁴ + a₅x⁵ + ...

Substituting this series into the Emden Equation, we have:

x(y'' + 2y' + xy) = 0.

**Differentiating** y(x) with respect to x, we get:

y' = a₁ + 2a₂x + 3a₃x² + 4a₄x³ + 5a₅x⁴ + ...

Differentiating again, we have:

y'' = = 2a₂ + 6a₃x + 12a₄x² + 20a₅x³ + ...

Now, let's substitute these **expressions** into the Emden Equation:

x((2a₂ + 6a₃x + 12a₄x² + 20a₅x³ + ...) + 2(a₁ + 2a₂x + 3a₃x² + 4a₄x³ + 5a₅x⁴ + ...) + x(a₀ + a₁x + a₂x² + a₃x³ + a₄x⁴ + a₅x⁵ + ...) )= 0.

Expanding and grouping terms by powers of x, we have:

(2a₂ + a₀) x + (6a₃ + 2a₁) x² + (12a₄ + 3a₂) x³ + (20a₅ + 4a₃) x⁴ + ... = 0.

For this equation to hold for all values of x, each coefficient of xⁿ must be equal to zero.

Therefore, we can determine the **coefficients** as follows:

Coefficient of x°: 2a₂ + a₀ = 0 => a₀ = -a₀/2.

Coefficient of x¹: 6a₃ + 2a₁ = 0 => a₃ = -a₁/3.

Coefficient of x²: 12a₄ + 3a₂ = 0 => a₄ = -a₂/4 = a₀/(2ₓ4) = a₀/8.

Coefficient of x³: 20a₅ + 4a₃ = 0 => a₅ = -a₃/5 = a₁/(3ₓ5) = a₁/15.

Therefore, the coefficients of the power series solution are:

a₀ = a₀ a₁ = -a₁/3 a₂ = -a₀/2 a₃ = -a₁/3 a₄ = a₀/8 a₅ = a₁/15

The power series

y(x) = a₀ + (-a₁/3)x + (-a₀/2)x² + (-a₁/3)x³ + (a₀/8)x⁴ + (a₁/15)x⁵ + ...

This series is known as the power series solution for the Lane-Emden equation, which is used to describe the structure of** self-gravitating**, spherically symmetric gas clouds, such as in stellar astrophysics.

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5) When multiple tests are done in analysis of variance, the family error rate is _________

A. The smallest p-value among the tests in the set.

B. The probability of making one or more type 2 errors among the tests.

C. The probability of not rejecting the null hypothesis when the null hypothesis is true.

D. The probability of making one or more type 1 errors among the tests.

### Answers

D. When multiple tests are done in analysis of **variance**, the family **error **rate is the family error rate is the probability of making one or more type 1 errors among the tests in the set.

It is also known as the **familywise error rate **(FWER). This is because when multiple tests are conducted simultaneously, the **probability **of making at least one type 1 error increases, and the FWER is used to control this **error **rate. To control the FWER, researchers use methods such as the Bonferroni correction or the **Holm-Bonferroni correction**, which adjust the significance level of each test to maintain the desired overall error rate.

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This is consumer Math, I suck at math in general, I need help.

### Answers

The **balance **on the **credit card **on the 30 th day of the billing cycle is A.$3, 267.24

How to find the balance ?

We can use the formula for **daily compounded interest **:

[tex]A = P * (1 + (r/n))^ {(nt)}[/tex]

The **APR (Annual Percentage Rate)** is the annual interest rate, and it's given as 25.31%, or 0.2531 in decimal form. Since the interest is compounded daily, n is 365 (the number of days in a year).

The balance on the credit card is therefore :

[tex]A = 3200 * (1 + (0.2531/365))^{( (30))}\\A = 3200 * 1.0006935616^{10.9589041}\\[/tex]

A = 3200 x 1. 0076

A = $3, 267.24

So, the **balance **on the 30th day of the billing cycle would be approximately $3, 267.24.

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i need some help with this question please and thank you!.

### Answers

The width of the **walkway **is 3 feet, If the area of the garden is 484 and the area covered by both the garden and the **walkway **(the entire square) is 784 .

The given is,

**Area **of the garden is 484

** Area** covered by both the garden and the **walkway** (the entire square) is 784 .

Let, x - Width of the **walkway**

Step:1

Formula to calculate area of square is,

A = a^2 ...........................(1)

Where, r - Radius of square

Step:2

For garden,

A = 484

Equation (1) becomes,

Take square root on both sides,

a= √484

a = 22 feet

For garden and the **walkway** (the entire square),

A = 784

Equation (1) becomes,

Take square root on both sides,

a = √784

a = 28 feet

Step:3

Side of garden and the** walkway **=

Side of garden + 2 (width of the walkaway)....(1)

28 = 22 + 2 ( x )

2 ( x ) = 28 - 22

= 6

x = 6/2

Width, x = 3 feet

Result:

The width of the **walkway** is 3 feet, If the** area** of the garden is 484 and the **area** covered by both the garden and the walkway (the entire square) is 784 .

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complete question:

A square garden is surrounded by a walkway of width X. If the area of the garden is 484 ft2 and the area covered by both the garden and the walkway (the entire square) is 784 ft2, what is the width of the walkway?

How many times larger is 6 × 104 than 3 × 102?

A.

200

B.

2

C.

20

D.

2,000

### Answers

**Answer: A. 200**

**Step-by-step explanation:To find out how many times larger 6 x 10^4 is than 3 x 10^2, we need to divide 6 x 10^4 by 3 x 10^2. **

**(6 x 10^4) / (3 x 10^2) = 200**

**Therefore, 6 x 10^4 is 200 times larger than 3 x 10^2. The answer is A.**

The temperature inside a room decreases by 14% every minute. If the initial temperature inside the room is 35 degrees Celsius, how long, to the nearest hundredth of a minute, will it take until the temperature of the room is 10 degrees Celsius?

### Answers

**Answer: 6 minutes.**

**Step-by-step explanation:**

To get to 10 or less, you need to subtract 4.9 ( 14% ) from 35.

35

30.1

25.2

20.3

15.4

10.5

5.6

ASAP PLS

cos (theta) - tan (theta)cos(theta) = 0

A) 0, PI/4, PI, 5PI/4

B. PI/4, 5PI/4

C. PI/2, 3PI/4, 3PI/2, 7PI/4

D. PI/2, 7PI/6, 3PI/2, 11PI/6

### Answers

answer may be of option a.

**Step-by-step explanation:**

As 1 is tan45°

hence in terms of pi

45=pi/4 (as pi is 180)

I want know how solve this

### Answers

Answer a^5c^5/b

Explanation below

There are many important classes of species whose birthrate is not proportional to the population size N(t). Suppose, for example, that each member of the population requires a partner for reproduction and that each member relies on chance encounters to meet a mate. If and the expected number of encounters is proportional to the product of the numbers of males and females, and if these are equally distributed the population, then the number of encounters, and hence the birthrate is proportional to N(t)^2. The death rate is still proportional to N(t) Consequently, the population size N(t) satisfies the differential equation dN/dt = bN^2 - aN; a, b are positive constants. a. Solve for N(t) given N(0). b. Find the long-time behavior by taking t rightarrow infinity in your solution in (a). Do this for the case N(0) < a/b.

### Answers

For the given **differential equation**, the solution for N(t) is N = |N - a/b| [tex]e^{(-at - ab + C)}[/tex], and for N(0) < a/b, the population size approaches zero in the long run.

To solve the differential equation dN/dt = [tex]bN^2[/tex] - aN, where a and b are positive constants, we can use the separation of **variables **and integrate both sides.

a. Solving for N(t) given N(0):

Rearranging the equation, we have:

[tex]dN / (bN^2 - aN) = dt.[/tex]

Separating the variables:

[tex]1 / (bN^2 - aN) dN = dt.[/tex]

Integrating both sides:

∫(1 / ([tex]bN^2[/tex] - aN)) dN = ∫dt.

Using **partial fraction decomposition**, we can write the left side as:

[tex]1 / (bN^2 - aN) = A / N + B / (N - a/b),[/tex]

where A and B are constants.

Multiplying both sides by (bN^2 - aN), we have:

1 = A(N - a/b) + BN.

Setting N = 0, we can solve for A:

1 = A(-a/b) ⇒ A = -b/a.

Setting N = a/b, we can solve for B:

1 = B(a/b) ⇒ B = b/a.

Now, we can rewrite the equation as:

∫(-b/a) / N dN + ∫(b/a) / (N - a/b) dN = ∫dt.

Simplifying, we have:

(-b/a) ln|N| + (b/a) ln|N - a/b| = t + C,

where C is the constant of integration.

Applying **exponential properties**, we get:

[tex]ln|N / |N - a/b||^{(-b/a)} = t + C[/tex].

Taking the exponential of both sides:

[tex]|N / |N - a/b||^{(-b/a)} = e^{(t + C)}.[/tex]

Removing the absolute value, we have:

[tex](N / |N - a/b|)^{(-b/a)} = e^{(t + C)}.[/tex]

Simplifying further:

[tex]N / |N - a/b| = e^{(-at - ab + C)}.[/tex]

Taking the absolute value:

[tex]N = |N - a/b| e^{(-at - ab + C)}.[/tex]

b. Finding the long-time behavior for N(0) < a/b:

As t approaches infinity, the exponential term e^(-at - ab + C) approaches zero. Since N(0) < a/b, the absolute value term |N - a/b| remains positive. Therefore, the population size N(t) approaches zero as t tends to infinity.

Therefore, for the given differential equation, the solution for N(t) is [tex]N = |N - a/b| e^{(-at - ab + C)}[/tex], and for N(0) < a/b, the **population size** approaches zero in the long run.

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When two six-sided dice are rolled, there are 36 possible outcomes. Find the probability that the sum is less than or equal to 5

### Answers

The **probability **of getting a sum **less **than or equal to 5 is 6/36, which simplifies to 1/6.

To find the **probability **that the sum of two six-sided dice is less than or equal to 5, we need to determine the number of favorable **outcomes **and divide it by the total number of possible outcomes.

Let's list the **favorable outcomes**, where the sum is less than or equal to 5:

(1, 1)

(1, 2)

(1, 3)

(2, 1)

(2, 2)

(3, 1)

There are 6 **favorable **outcomes.

So, the total number of **possible outcomes **when rolling two dice is

= 6 x 6

= 36.

Therefore, the **probability **of getting a sum **less **than or equal to 5 is 6/36, which simplifies to 1/6.

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4. A function f:x→ b/x-a, a>0, is such that f(b) = b and f(2a) = 2a find the value of a and of b. If ff(x) = x, show that x^2-x-2 = 0.

### Answers

The andwr is the first one that popped up c^2x+0

Solve the system of equations:

(-55, -610)

(-5.5, -61)

(-55, -6100)

(-5.5, -610)

### Answers

y=120x+500

y=100x-600

120x + 500=100x-600

x=-55

y=120x(-55)+500

y=-6100

x,y=(-55,-6100)

traffic engineers compared rates of traffic collisions at intersections with raised medians and rates at intersections with two-way left-turn lanes. they found that out of 4,653 collisions at intersections with raised medians, 2,289 were rear-end collisions, and out of 4,606 collisions at two-way left-turn lanes, 2,027 were rear-end collisions. assuming these to be random samples of collisions from the two types of intersections, construct a 95% confidence interval for the difference between the proportions of collisions that are of the rear-end type at the two types of intersection.

### Answers

The 95% **confidence interval **for the difference between the proportions of **collisions **that are of the rear-end type at the two types of **intersections is (-0.041, 0.017).**

To construct the** confidence interval**, we first calculate the sample proportions of **rear-end collisions** at each type of** intersection: **

p1 = 2289/4653 ≈ 0.492 and p2 = 2027/4606 ≈ 0.440.

We then calculate the standard error of the difference between the two sample proportions: sqrt((p1(1-p1)/4653) + (p2(1-p2)/4606)) ≈ 0.014.

Using a t-distribution with degrees of freedom equal to the smaller of n1-1 and n2-1 (in this case, 4652), we find the t-value for a 95% **confidence interval** with two tails to be approximately 1.96.

Finally, we use the formula for the confidence interval for the difference between two proportions: (p1 - p2) ± (t-value)(standard error) to obtain the **interval **(-0.041, 0.017).

Therefore, we can say with 95% confidence that the true difference between the proportions of rear-end **collisions **at intersections with raised medians and two-way left-turn lanes falls between -0.041 and 0.017.

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A

12

1

2

12

2

1

-inch candle burns down in 10 hours. At what rate does the candle burn, in inches per hour?

### Answers

The candle **burns** at a rate of 1.25 inches per hour. This means that every hour, the length of the candle decreases by 1.25 inches until it burns down completely after 10 **hours.**

The rate at which the candle burns down can be calculated as the ratio of the length of the** candle **to the time it takes to burn down completely. In this case, the candle is 12 1/2 inches long and burns down in 10 hours.

To find the rate, we divide the** length** of the candle (in inches) by the time it takes to burn down (in hours):

Rate = Length / Time

Rate = 12.5 inches / 10 hours

Rate = **1.25 inches **per hour

Therefore, the candle burns at a rate of 1.25 inches per hour. This means that every hour, the length of the candle decreases by 1.25 inches until it **burns down **completely after 10 hours.

It's important to note that the rate of burning may not be constant throughout the life of the candle, as the** wax **may melt more quickly towards the end when the flame is larger. However, this calculation assumes a **constant** rate of burning over the entire 10 hours.

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(1-cot^2(theta)/1+^2(theta)) +2cos^2(theta)=1 prove the identity

### Answers

sin²θ+ co²θ = 1 is the **Pythagorean** identity, the identity has been **proven**.

The identity can be **proven** using trigonometric identities.

By substituting the reciprocal identity and **Pythagorean** identity, the expression can be simplified as follows:

(1 - cot²θ)/(1 + cot²θ) + 2cos²θ = 1

Rewriting the left-hand side:

(sin²θ- cos²θ)/(sin²θ + cos²θ) + 2cos²θ = 1

Since sin²θ + cos²θ = 1, the expression simplifies further:

(sin²θ - cos²θ)/(1) + 2cos²θ = 1

Simplifying the numerator:

(sin²θ- cos²θ) + 2cos²θ = 1

Expanding and combining like terms:

sin²θ + cos²θ= 1

Since sin²θ+ co²θ = 1 is the **Pythagorean** identity, the identity has been proven.

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