A bike rental company rents beach bikes. They charge an initial fee plus $1 for each hour the bike is rented. Chad rented a bike for six hours and was charged $10. How much is the initial fee?-I need help with the first step and I NEED HELP ASP!!

The value of the given expression [\(87^3+ 13^3/87^2 -87 * 13 + 13^2\)] by simplifying the numerator and denominator using suitable identities is 100.

We will first calculate the numerator:

As  (\(a^3\) + \(b^3\)) = (a + b)(\(a^2\) - ab + \(b^2\)) :

\(87^3\) + \(13^3\) = (87 + 13)(\(87^2\) - \(87 * 13\) + \(13^2\))

= 100(\(87^2\) - 87 * 13 + \(13^2\))

Now, calculate the denominator:

\(87^2 - 87 * 13 + 13^2\)

As,(\(a^2 -2ab +b^2\)) =\((a - b)^2\):

\(87^2 - 87 * 13 + 13^2 = (87 - 13)^2\)

\(= 74^2\)

So by solving the equation further:

\((87^3+13^3) / (87^2- 87 * 13+13^2) = 100*(87^2- 87 *13 + 13^2)/(87^2 - 87 * 13 + 13^2)\)

As we can see the numerator and denominator are the same expressions (\(87^2 - 87 * 13 + 13^2\)). so, they cancel each other:

\((87^3 + 13^3) / (87^2 - 87 * 13 + 13^2) = 100\)

So, the value of the given expression is 100.

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(a) high in fat, (b) high in protein, and (c) unspecified composition. The data collected reveals variations in body weight, with a mean of 194 and a standard deviation that differs among the diets.

The study aimed to investigate the effects of different macronutrient compositions on body weight. All diets had the same calorie content, ensuring that any observed differences were not due to variations in total energy intake. The summary statistics indicate that the mean body weight across the three diets was 194. However, it is important to note that the standard deviation varied among the diets. This suggests that the different macronutrient compositions influenced the variability in body weight outcomes. The second paragraph of the answer would provide a more detailed explanation of the potential reasons behind the observed variations and their implications.

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Answer:

15

Step-by-step explanation:

Answer:

Quotient = 3x-5

Remainder = 3

Step-by-step explanation:

For any long division, I use the DMS principle, divide, multiply, subtract.

(x+6) | 3x² + 13 x - 27

First of all, divide the first part of the equation by x. (which is 3x²)

You get 3x.

Now take your 3x and multiply it by (x+6)

You get 3x²+18x. You need to subtract this from your equation.

(3x² + 13 x - 27) - (3x² + 18x) = -5x - 27

One round of DMS complete

Now,

(x+6) | - 5x - 27

We again divide the first part only by x again. You get -5

We multiply -5 by (x+6)

-5x-30.

Let's subtract it from the equation.

(-5x-27) - (-5x-30) = - 5x - 27 + 5x + 30 = 3

Second round of DMS complete.

We cannot divide 3 further by (x+6) so this is it.

The remainder is 3.

Now we take our answers, which is 3x and - 5 and add them to get 3x-5. This is our quotient.

Answer:

Use M A T H W AY

trust me its goated

The surface density of the lights on the Rockefeller Center Christmas Tree is 9.38 lights per square foot.

The surface density of the lights on the Rockefeller Center Christmas Tree can be found by dividing the total number of lights by the surface area of the tree.

To find the surface area of the tree, we need to first find the slant height of the cone. We can use the Pythagorean theorem to find the slant height, which is the square root of the sum of the height squared and the radius squared:

The slant height:

= sqrt(72^2 + (45/2)^2)

= 75.43 feet

The lateral area of a right cone can be found using the formula: πrs

r = 45/2 = 22.5 feet

Substituting the given values into the formula for the lateral area, we get:

Lateral area = π(22.5)(75.43)

Lateral area = 5329.1295 square feet

To find the surface density of the lights, we can divide the total number of lights by the lateral area:

Surface density = 50,000 / 5329.1295

Surface density = 9.38239537996

Surface density= 9.38 lights/square foot.

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The given function is f(x) = x² + 28.We need to find f''(x).Formula to find second derivative of a function: If y = f(x), theny'' = d²y/dx²To find the second derivative of f(x),

we first find the first derivative of the function and then the second derivative of the function.The first derivative of the function is:f'(x) = d/dx [x² + 28]f'(x) = d/dx (x²) + d/dx (28)f'(x) = 2x + 0f'(x) = 2xThe second derivative of the function is:f''(x) = d/dx [2x]f''(x) = 2d/dx(x)f''(x) = 2 * 1f''(x) = 2Thus, f''(x) = 2.

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Answer:

12

Step-by-step explanation:

20-(5+(9-6)) = 20-(5+(3)) = 20-(8) = 12.

Alternatively, rewrite the question without parenthesis.

20-(5+(9-6)) = 20-(5+9-6) = 20-5-9+6 = 12.

Answer:

|t - 350| ≤5

Step-by-step explanation:

t be the actual temperature

We want to be within 5 degrees of  350  ( that includes 5 degrees)

|t - 350| ≤5

Answer:

14 : -17 : 2

Step-by-step explanation:

8 + (9 · 2 – 3) : 25 – 4 2 : 2

= 14 : -17 : 2

Answer:

- 12.4

Step-by-step explanation:

\(8 + \frac{9 * 2 - 3}{25} - \frac{42}{2}\)

\(8 + \frac{18 - 3}{25} - 21\)

\(8 + \frac{15}{25} - 21\)

\(8 + \frac{3}{5} - 21\)

\(8 \frac{3}{5} - 21\)

\(-12 \frac{2}{5}\)

The exponential distribution may be used to predict the failure rate of certain items over time. An LED light bulb has an expected lifetime of 25000 hours. Assuming that the lifetime of the LED light bulb follows the exponential distribution, the probability that it will last more than 3 years is closest to (C) 53%. Correct answer is option C

This is because the lifetime of an LED light bulb can be estimated using the following equation : P(x > 3 years) = 1 - P(x ≤ 3 years)where x is the lifetime of the LED light bulb.If we convert 3 years to hours, we get 3 * 365 * 24 = 26280 hours. As a result, P(x ≤ 3 years) = P(x ≤ 26280 hours)

Using the formula for exponential distribution, the probability of the LED light bulb failing after 26280 hours is : Probability = 1 - e^{-λx} Where λ is the failure rate per hour and x is the length of time in hours.We can now calculate the value of λ by dividing the expected lifetime of the bulb by the total number of hours.

 λ = 1/25000 hours  This implies that the probability of the LED light bulb failing after 26280 hours is : P (x ≤ 26280 hours)

Therefore, the probability that the LED light bulb will last more than 3 years is approximately 53 percent. The Correct answer is option C

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Answer:

\(n^{2}\) × \(p^{2}\) × \(r^{3}\)

Step-by-step explanation:

\(n^{2}\) × \(p^{2}\) × \(r^{3}\)

Pls consider marking my answer as Brainliest! It would mean a lot!

The final expression in terms of exponent are

\(n^{2} .p^{2} .r^{3}\).

It is required to write  the given expression using exponents.

A quantity representing the power to which a given number or expression is to be raised, usually expressed as a raised symbol beside the number or expression. An exponent refers to the number of times a number is multiplied by itself.

Given that:

The expression are

n•n•p•p•r•r•r

there are 2 n , 2 p and 3 r is given with a multiplication sign

So, we need to multiply with the same character, we get

\(n.n=n^{2}\)

\(p.p=p^{2} \\\\\\r.r.r= r^{3}\)

So, n has the power of 2 ,p has the power of 2 and r has the power of 3.

∴The final expression in terms of exponent are

\(n^{2} .p^{2} .r^{3}\).

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Answer:

Line A

Step-by-step explanation:

Any line facing towards the left side of the coordinate plane will have a negative slope.

Line C is facing towards the right

Line D and B doesn't have a slope

A is the only right answer

The given statement "calculated fields must always contain at least one constant" is false because calculated fields do not necessarily have to contain a constant value.

They can involve variables, formulas, functions, or expressions that perform calculations based on the given data. The purpose of calculated fields is to derive new values or perform computations based on existing data within a system or application.

While constants can be used in calculated fields, they are not a requirement. The specific requirements and structure of calculated fields may vary depending on the context and the software or system being used.

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Answer:

That's pretty much correct. A continuous function is one that has no breaks or holes. For example, 1/x+5 is not continuous (or discontinuous) because the domain does not include -5. However, I don't think it needs to necessarily have to have curves because all linear functions are continuous. Hope this helps! (I learned about this a few weeks ago in my pre-calc class lol)

Step-by-step explanation:

Answer:

He deposited 16 $5 bills.

Step-by-step explanation:

State your variables

let x be the number of $1 bills

let y be the number of $5 bills

Create a system of equations

x + 5y = 200       (eq'n 1 -- for amount of money)

x + y = 136           (eq'n 2 -- for number of bills)

Solve the system for y

I will solve using substitution.  Rearrange eq'n 2 to isolate variable x.

x + y = 136

x = 136 - y           (eq'n 3)

Substitute eq'n 3 into eq'n 1.

x + 5y = 200

136 - y + 5y = 200

136 + 4y = 200

4y = 64

y = 16

Solve for x to check answer

Substitute y = 16 into eq'n 2.

x + y = 136

x + 16 = 136

x = 120

Substitute x = 120 into eq'n 1.

x + 5y = 200

120 + 5(16) = 200

120 + 80 = 200

200 = 200

LS = RS          Both sides are equal, so the solution is correct.

Therefore, Jack deposited 16 five dollar bills.

Answer:

no solution

Step-by-step explanation:

h - 4h + 10 = - 3h , that is

- 3h + 10 = - 3h ( subtract 10 from both sides )

- 3h = - 3h - 10 ( add 3h to both sides )

0 = - 10 ← false statement

this false statement indicates the equation has no solution

The value of the lettered sides of the triangle are n=4, m=3

In geometry, two figures or objects are congruent if they have the same shape and size, or if one has the same shape and size as the mirror image of the other. More formally, two sets of points are called congruent if, and only if, one can be transformed into the other by an isometry

From the Triangles, <BAC = <DBC                         given

⇒ΔBAD ≅ ΔDBC       (SAS)

/BA/ =/BD/ = 3 units

Also, /BC/ = /AD/ = 4 units

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Answer:

look i think is -5.005 I don't really know but I hope you get it right. Have a good day

we have

using a graphing tool

see the attached figure

This is a quadratic equation (vertical parabola) open up

The vertex is a minimum---------> Vertex

The domain of the function is all real numbers-------> interval (-∞,∞)

The range of the function is the interval----------> [-6.125,∞)

The function has two x-intercepts----->  and

The function is increasing over the interval-----> (0.25,∞)

The function is decreasing over the interval-----> (-∞,0.25)  

Answer:

the range of function is all real numbers,

and the function has 2 x intercepts

Step-by-step explanation:

super easy way to do it, look for an i in your equation, if there is none, and you dont take the square root of a negative number, all answers are real. then you can see that you function crosses the x-axis twice, meaning you have two x-intercepts

Answer:

900 in a minute

Step-by-step explanation:

2700/3

=900

The probability histogram to the right represents the number of live births by a mother 51 to 54 years old who had a live birth in 2012 to have had 1.75 live births (rounded to one decimal place). This means that on average, we would expect these mothers to have had between one and two live births.

(a) To find the probability that a randomly selected 51- to 54-year-old mother who had a live birth in 2012 has had her fourth live birth, we need to look at the histogram and locate the bar for the fourth live birth. We can see that the height of this bar is 0.1. Therefore, the probability is 0.1 or 10% (as a decimal).

(b) To find the probability that a randomly selected 51- to 54-year-old mother who had a live birth in 2012 has had her fourth or fifth live birth, we need to add the heights of the bars for the fourth and fifth live births. The height of the fourth live birth bar is 0.1 and the height of the fifth live birth bar is 0.05. Therefore, the probability is 0.1 + 0.05 = 0.15 or 15% (as a decimal).

(c) To find the probability that a randomly selected 51- to 54-year-old mother who had a live birth in 2012 has had her sixth or more live birth, we need to add the heights of the bars for the sixth or more live births. The height of the bar for six or more live births is 0.05. Therefore, the probability is 0.05 or 5% (as a decimal).

(d) To find how many live births we would expect a randomly selected 51- to 54-year-old mother who had a live birth in 2012 to have had, we need to find the mean of the distribution. We can do this by multiplying each number of live births by its corresponding probability and adding up the products.

Expected value = (0 x 0.2) + (1 x 0.15) + (2 x 0.15) + (3 x 0.2) + (4 x 0.1) + (5 x 0.05) + (6 x 0.05)
Expected value = 1.75

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Answer:

Inequality:

83 ≤ (82 + 86 + 84 + x)/4 ≤ 87

Step-by-step explanation:

You must collect no less than 80 and no more than 96 pop tabs.

x=19

add 45 to both sides 3

a. Two populations.

b. Qualitative (categorical) data.

c. Proportion stats.

d. Null hypothesis: p1 = p2 (The proportion of seniors who bring a bag lunch is equal to the proportion of sophomores who bring a bag lunch). Alternative hypothesis: p1 > p2 (The proportion of seniors who bring a bag lunch is greater than the proportion of sophomores who bring a bag lunch).

e. Z-test for two proportions.

f. P-value.

g. A small P-value provides convincing evidence to support Phoebe's hunch that older students are more likely to bring a bag lunch than younger students.

h. 95% confidence interval to estimate the difference in proportions.

What is hypothesis?

A hypothesis is a proposed explanation or assumption that is tested through research and data analysis. It is a statement that suggests a relationship or difference between variables or phenomena and serves as a basis for scientific investigation.

a. There are two populations in this problem: the population of seniors and the population of sophomores.

b. This is a problem about qualitative (categorical) data since we are examining whether students bring a bag lunch or not.

c. We will use the proportion stats option in StatCrunch to complete this problem.

d. Null hypothesis (H0): The proportion of seniors who bring a bag lunch is equal to the proportion of sophomores who bring a bag lunch.

Alternative hypothesis (Ha): The proportion of seniors who bring a bag lunch is greater than the proportion of sophomores who bring a bag lunch.

e. The test statistic is the z-test for two proportions.

f. The P-value is the probability of observing a test statistic as extreme as the one calculated, assuming the null hypothesis is true.

g. The strength of the P-value determines the level of evidence against the null hypothesis. If the P-value is small (below the significance level), it provides strong evidence against the null hypothesis. In this case, a P-value less than 0.05 (assuming significance level α = 0.05) would suggest convincing evidence to support Phoebe's hunch.

h. To construct a 95% confidence interval to estimate the difference in proportions, we use the formula:

Confidence interval = (p1 - p2) ± z * sqrt((p1 * (1 - p1) / n1) + (p2 * (1 - p2) / n2))

where p1 and p2 are the sample proportions, n1 and n2 are the sample sizes, and z is the critical value corresponding to the desired confidence level (in this case, z value for a 95% confidence level).

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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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Component 3 must be started on day 197 to meet the delivery date of day 200.

To illustrate the backward schedule, we need to start from the delivery date (day 200) and work our way backward, taking into account the lead times and dependencies of each operation.

(a) Backward schedule:

Operation    |  Work Center  |  Time (hours)  |  Start Day

--------------------------------------------------------

Final Assembly |   Work Center 1  |        1       |     200

Sub-assembly 1 |   Work Center 2  |        2       |     199

Component 4     |   Work Center 3  |        3       |     197

Component 2     |   Work Center 4  |        4       |     196

Component 3     |   Work Center 5  |        3       |     ????

(b) To identify when component 3 must be started to meet the delivery date, we need to consider its dependencies and lead times.

From the backward schedule, we see that component 3 is required for sub-assembly 1, which is scheduled to start on day 199. The time required for sub-assembly 1 is 2 hours, which means it should be completed by the end of day 199.

Since component 3 is needed for sub-assembly 1, we can conclude that component 3 must be started at least 2 hours before the start of sub-assembly 1. Therefore, component 3 should be started on day 199 - 2 = 197 to ensure it is completed and ready for sub-assembly 1.

Hence, component 3 must be started on day 197 to meet the delivery date of day 200.

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Answer:

Step-by-step explanation:

Answer:

- 4

Step-by-step explanation:

x + 4 - 4 ( 4 - 4 = 0 )

= x