HELPP I AM STUCK IN THIS PROBLEM AND NEED IT BY TMR!!

Answer:

Step-by-step explanation:

Round these as you see fit (these are exact values, if you're supposed to use a normal table leave a comment)

1.) 0.418045324790463

2.)0.0717364877620478

3.) 0.0192746613260287

Answer:

trump2020 I love him so much

Answer:

Sqrt 100^5

100,000

Step-by-step explanation:

Sqrt (100)^5

= sqrt 10000000

= 100000

keeping in mind that perpendicular lines have negative reciprocal slopes, let's check for the slope of the equation above

\(2y=6-2x\implies 2y=-2x+6\implies y=\cfrac{-2x+6}{2} \\\\\\ y=\stackrel{\stackrel{m}{\downarrow }}{-1}x+3\qquad \impliedby \begin{array}{|c|ll} \cline{1-1} slope-intercept~form\\ \cline{1-1} \\ y=\underset{y-intercept}{\stackrel{slope\qquad }{\stackrel{\downarrow }{m}x+\underset{\uparrow }{b}}} \\\\ \cline{1-1} \end{array} \\\\[-0.35em] ~\dotfill\)

\(\stackrel{~\hspace{5em}\textit{perpendicular lines have \underline{negative reciprocal} slopes}~\hspace{5em}} {\stackrel{slope}{ -1 \implies \cfrac{-1}{1}} ~\hfill \stackrel{reciprocal}{\cfrac{1}{-1}} ~\hfill \stackrel{negative~reciprocal}{-\cfrac{1}{-1} \implies 1}}\)

so we're really looking for the equation of a line whose slope is 1 and it passes through (6 , -7)

\((\stackrel{x_1}{6}~,~\stackrel{y_1}{-7})\hspace{10em} \stackrel{slope}{m} ~=~ 1 \\\\\\ \begin{array}{|c|ll} \cline{1-1} \textit{point-slope form}\\ \cline{1-1} \\ y-y_1=m(x-x_1) \\\\ \cline{1-1} \end{array}\implies y-\stackrel{y_1}{(-7)}=\stackrel{m}{ 1}(x-\stackrel{x_1}{6}) \implies y +7= 1 (x -6) \\\\\\ y+7=x-6\implies {\Large \begin{array}{llll} y=x-13 \end{array}}\)

The probability that a certain component lasts more than 350 hours in operation, based on the random sample of 37 components tested, is approximately 0.649.

To calculate the probability, we divide the number of components that lasted longer than 350 hours (24) by the total number of components tested (37).

Probability = Number of favorable outcomes / Total number of possible outcomes

Probability = 24 / 37 ≈ 0.649

Therefore, the probability that a certain component lasts more than 350 hours in operation, based on the given sample, is approximately 0.649.

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Step-by-step explanation:

Given: r=9cm

Volume of sphere=4/3πr^3

=4/3*3.14*9^3

=3052.08 cm^3

15 liquids that can be found in a kitchen include:

15 solids in a kitchen are:

When it comes to solids in the kitchen, we can mostly find either food ingredients, such as salt, flour, and spices, or actual food that can be prepared such as meat, vegetables, and bread.

As for liquids, these can be anything from drinks such as juice, water and wine, to liquids used for cooking such as oil, soy sauce and vinegar.

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The volume of the triangular prism is 198 cubic units.

To find the volume of a triangular prism, we need to multiply the area of the base triangle by the height of the prism.

In this case, the base of the triangular prism is a triangle with a base of 11 units and a height of 6 units, so the area of the base is:

Area of base = (1/2) × base × height

Area of base = (1/2) × 11 units × 6 units

Area of base = 33 square units

The height of the prism is given as 6 units.

Therefore, the volume of the triangular prism is:

Volume of prism = Area of base × Height

Volume of prism = 33 square units × 6 units

Volume of prism = 198 cubic units

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The result of the integral is: ∫arctan(5/x) dx = (5/x) * arctan(5/x) - [(5/x) + (5/x)^3/3] + C.

To evaluate the integral ∫arctan(5/x) dx, we can use the technique of substitution. Let's perform the substitution:

Let u = 5/x

Then, du = -5/x^2 dx

Rearranging the equation, we have dx = -x^2/(5u^2) du

Substituting these values, the integral becomes:

∫arctan(5/x) dx = ∫arctan(u) (-x^2/(5u^2)) du

Now, we can simplify the integral using this substitution. We'll also need to change the limits of integration accordingly. However, we need to be careful with the limits because the substitution involves the variable x, not u.

When x = 0, u = 5/0 (which is undefined). Thus, the lower limit of integration is not valid. We will need to determine the limit of the integral as x approaches 0 separately.

Let's proceed with the upper limit:

When x approaches infinity, u approaches 0. Thus, the upper limit of integration is u = 0.

Now, we can rewrite the integral and proceed with the evaluation:

∫arctan(5/x) dx = ∫arctan(u) (-x^2/(5u^2)) du

The indefinite integral of arctan(u) can be evaluated as follows:

∫arctan(u) du = u * arctan(u) - ∫(1 + u^2) du

= u * arctan(u) - (u + u^3/3) + C

Substituting back the value of u:

= (5/x) * arctan(5/x) - [(5/x) + (5/x)^3/3] + C

So, the result of the integral is:

∫arctan(5/x) dx = (5/x) * arctan(5/x) - [(5/x) + (5/x)^3/3] + C

Note that we had to handle the limit as x approaches 0 separately and that the result includes the constant of integration, C.

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Answer:

y = 2x + 5

Step-by-step explanation:

the equation of a line in slope- intercept form is

y = mx + c ( m is the slope and c the y- intercept )

calculate m using the slope formula

m = \(\frac{y_{2}-y_{1} }{x_{2}-x_{1} }\)

with (x₁, y₁ ) = A (1, 7 ) and (x₂, y₂ ) = B (- 3, - 1 )

m = \(\frac{-1-7}{-3-1}\) = \(\frac{-8}{-4}\) = 2 , then

y = 2x + c ← is the partial equation

to find c substitute either of the 2 points into the partial equation

using (1, 7 )

7 = 2(1) + c = 2 + c ( subtract 2 from both sides )

5 = c

y = 2x + 5 ← equation of line

The density of air in the inflated basketball is approximately 1.17 kg/m³, rounded to two decimal places.

Density of air in the inflated basketball, we can follow these steps:
Calculate the mass of air added to the basketball by finding the difference in weight between the inflated and deflated basketballs.
Inflated weight: 0.618 kg
Deflated weight: 0.615 kg
Mass of air added = Inflated weight - Deflated weight = 0.618 kg - 0.615 kg = 0.003 kg
Calculate the volume of the basketball, assuming it is perfectly spherical.

We can use the formula for the volume of a sphere:
Volume = (4/3) × π × r³, where r is the radius of the sphere.
The basketball has a diameter of 0.17 meters, so its radius is half of that:
Radius (r) = 0.17 m / 2 = 0.085 m
Now, we can calculate the volume:
Volume = (4/3) × π × (0.085 m)³ ≈ 0.00257 m³
Finally, calculate the density of the air in the ball using the formula:
Density = Mass / Volume
Density = 0.003 kg / 0.00257 m³ ≈ 1.17 kg/m³

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Answer: To find the discount, simply multiply the original selling price by the %discount:

ie:  39 x 33/100= $12.87

So, the discount is $12.87.

Step-by-step explanation: To find the sale price, simply minus the discount from the original selling price:

ie: 39- 12. 87=  26.13

So, the sale price is $26.13

Answer:

Step-by-step explanation:

b

The population proportion who participated in voting (63%) is higher than the sample proportion from this survey (58.91%).

To determine whether the population proportion who participated in voting or the sample proportion from the survey is higher, we need to compare the percentages.

The population proportion who participated in voting is given as 63% of all registered voters.

This means that out of every 100 registered voters, 63 participated in voting.

In the survey of retired voters, 162 out of 275 participants voted. To calculate the sample proportion, we divide the number of retired voters who participated (162) by the total number of retired voters in the sample (275) and multiply by 100 to get a percentage.

Sample proportion = (162 / 275) \(\times\) 100 ≈ 58.91%, .

Comparing the population proportion (63%) with the sample proportion (58.91%), we can see that the population proportion who participated in voting (63%) is higher than the sample proportion from this survey (58.91%).

Therefore, based on the given data, the population proportion who participated in voting is higher than the sample proportion from this survey.

It's important to note that the sample proportion is an estimate based on the surveyed retired voters and may not perfectly represent the entire population of registered voters.

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Answer:

(4,-1)

x=4

y=-1

Step-by-step explanation:

Answer:

x = 4 and y = -1

Step-by-step explanation:

hope this helps!

In mathematics, the point below which a specified percentage of the observations fall is commonly referred to as the percentile.

A percentile is a measure that indicates the relative position of a particular value within a dataset.

For example, if a value is at the 80th percentile, it means that 80% of the observations in the dataset are below that value, and only 20% of the observations are above it.

Percentiles are often used to analyze and understand the distribution of data, especially in fields such as statistics, probability, and data analysis. They provide insights into how individual data points compare to the overall dataset and help identify outliers or extreme values.

A percentile is a statistical measure that indicates the relative standing of a particular value within a dataset. It represents the point below which a specified percentage of the observations or data points fall. Percentiles are primarily used to understand the distribution of data and analyze how individual values compare to the overall dataset.

To calculate a percentile, the dataset is arranged in ascending order, from the smallest value to the largest value. The position of a specific percentile is then determined based on the percentage of data points below it. For example, the 75th percentile represents the value below which 75% of the data points fall.

Percentiles are commonly used in various fields, including statistics, probability, and data analysis. They provide valuable insights into the spread, variability, and distribution of data. Here are a few key points to consider:

1. Median: The median is the 50th percentile, representing the value that divides the dataset into two equal halves. It is a measure of central tendency and provides information about the middle point of the distribution.

2. Quartiles: Quartiles divide the dataset into four equal parts. The first quartile, or the 25th percentile (Q1), represents the value below which 25% of the data points fall. The third quartile, or the 75th percentile (Q3), represents the value below which 75% of the data points fall. The difference between Q3 and Q1 is known as the interquartile range (IQR) and provides insights into the spread of the middle 50% of the data.

3. Percentile Ranks: Percentile ranks indicate the percentage of data points that are below a specific value. For example, if a student scores in the 80th percentile on a standardized test, it means they performed better than 80% of the test-takers.

4. Outliers: Percentiles can be useful in identifying outliers, which are data points that significantly deviate from the rest of the dataset. Extremely high or low percentiles may indicate unusual or extreme values that warrant further investigation.

5. Normal Distribution: In a normal distribution, the 50th percentile (median) coincides with the mean and mode, and specific percentiles have known standard deviations from the mean (e.g., the 68-95-99.7 rule).

Percentiles are versatile tools for summarizing and analyzing data, providing valuable insights into the distribution and relative positions of individual values. They enable comparisons and help make informed decisions based on the characteristics of a dataset.

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Answer:

Step-by-step explanation:

3^2*5^2= 225

9*25=225

Three to the 2nd power times 5 to the 2nd power.

West Game Rental is cheaper for renting more games monthly; South Games is cheaper for renting fewer games monthly.

Part A: The statement is correct when the total number of games rented per month is less than or equal to 14.

Part B: The statement is incorrect when the total number of games rented per month is greater than 14.

Part C: To find the number of games rented per month in which the cost of renting games is the same for both companies, we can set the total cost for each company equal to each other:

14 + 1g = 3g

Simplifying this equation, we get:

14 = 2g

Thus, the number of games rented per month in which the cost of renting games is the same for both companies is 7.

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Answer:

3,247,315

thats three million, two hundred forty-seven thousand, three hundred fifteen in number form

The given statement "For any sampled population, the population of all sample means is approximately normally distributed" is true. This is a result of the Central Limit Theorem, which states that when sample sizes are large enough, the means of all samples will be approximately normally distributed.

A sample is a smaller portion of the population that has been taken to draw conclusions about the characteristics of the population as a whole. The sample needs to be representative of the population for the data analysis to be valid.

A sample mean is a statistical analysis in which the sum of all data points in a sample is divided by the number of observations. It gives an idea of the average value of the data points in the sample population.

For any sampled population, the population of all sample means is approximately normally distributed. The central limit theorem states that when the sample size is sufficiently large, the distribution of the sample means will be approximately normal, regardless of the underlying distribution of the population.

The central limit theorem is an important concept in inferential statistics because it provides a theoretical foundation for the use of statistical inference methods to draw conclusions about population parameters based on sample statistics.

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Answer: line ZX is congruent to line VX (option 4)

Step-by-step explanation:

We already know <X is congruent to <X, we also know that line YX is congruent to line XW. Now all we need is one more line adjacent to X which is going to be ZX ad VX

Answer:

hhjbvft

Step-by-step explanation:

jojo giving kbjvuvuvvuvugubibbibiogg

To find the union and intersection of the intervals I₁ = (-2, 2] and I₂ = [1, 5), let’s consider the overlapping values and the combined range.

The union of two intervals includes all the values that belong to either interval. Taking the union of I₁ and I₂, we have:

I₁ U I₂ = (-2, 2] U [1, 5)

To find the union, we combine the intervals while considering their overlapping points:

I₁ U I₂ = (-2, 2] U [1, 5)
= (-2, 2] U [1, 5)

So the union of the intervals I₁ and I₂ is (-2, 2] U [1, 5).

Now let’s find the intersection of the intervals I₁ and I₂, which includes the values that are common to both intervals:

I₁ ∩ I₂ = (-2, 2] ∩ [1, 5)

To find the intersection, we consider the overlapping range between the two intervals:

I₁ ∩ I₂ = [1, 2]

Therefore, the intersection of the intervals I₁ and I₂ is [1, 2].


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Answer:

C. about $15.00

Step-by-step explanation:

To solve, you take the initial amount, which is $100, and multiply it by the percentage in decimal form: 15% ---> 0.15

100 x .15 = 15

Answer:

109

Step-by-step explanation:

Three consecutive integers:

Sum of them:

The smallest integer out of three is 109

━━━━━━━━━━━━━━━ ♡ ━━━━━━━━━━━━━━━  

3 consecutive integers = x+(x+1)+(x+2) or 3x+3 (x is the smallest number)

(Consecutive means one after another)

So 3x+3 = 330.

Now subtract 3 from both sides.

(3x-3)=3x

(330-3)=327

The left side is 3x and the right side is 327.

The equation now is 3x=327.

Now divide each side by 3 (so x is isolated)

3x÷3=x

327÷3=109

The left side is x and the right side is 109.

The final equation is x=109.

Since x is the smallest of these integers we don't have to do anything else.

Okay, so let's check the answer now. 109+110+111=330.

So the smallest of the 3 consecutive numbers is 109.

━━━━━━━━━━━━━━━ ♡ ━━━━━━━━━━━━━━━  

Answer:

As x approaches 6:

\( \frac{6 - x}{ {x}^{2} - 36 } = - \frac{x - 6}{(x - 6)(x + 6)} = - \frac{1}{x + 6} = - \frac{1}{6 + 6} = - \frac{1}{12} \)

The limit of the given function as x approaches 6 is -1/12. This is achieved by factoring and revising the original function, and then substituting into the revised function.

The student is asking for the limit of the function (6-x) / (x²-36) as x approaches 6. In mathematics, this is a problem of calculus and specifically involves limits. Let's solve this by first factoring the denominator to get (6-x) / ((x-6)(x+6))

By realizing we can revise the numerator as -(x-6), we make it obvious that the limit can be directly computed by substituting x=6 after canceling out the (x-6) terms. The result is -1/12, therefore the limit of the function as x approaches 6 is -1/12.

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The general solution to the given differential equation is:

r = ±Ce^θ

To find the general solution of the given differential equation:

dr/dθ - rsec(θ) = cos(θ)

We can solve this differential equation by separating the variables and integrating:

1/(rsec(θ)) dr = cos(θ) dθ

Multiplying both sides by sec(θ) gives:

1/r dr = cos(θ)sec(θ) dθ

Integrating both sides:

∫ (1/r) dr = ∫ (cos(θ)sec(θ)) dθ

ln|r| = ∫ (cos(θ)/cos(θ)) dθ

ln|r| = ∫ dθ

ln|r| = θ + C

where C is the constant of integration.

Exponentiating both sides:

|r| = e^(θ + C)

|r| = e^θ * e^C

|r| = Ce^θ

where C = ±e^C (a constant of integration).

Therefore, the general solution to the given differential equation is:

r = ±Ce^θ

where C is an arbitrary constant.

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Using these eigenvalues and eigenvectors, we can write the solution to the system of differential equations in the form: [z; y] = A₁ e^(Y₁t) [v₁₁; v₁₂] + A₂ e^(Y₂t) [v₂₁; v₂₂], where A₁ and A₂ are constants. the values of A1, A2, Y1 and Y2 are:

A1= -(1/2), A2= -1, Y1=1, Y2=6. The solution is obtained in the form of a linear combination involving constants A₁ and A₂. The specific values of A₁, A₂, Y₁, and Y₂ are determined to complete the solution.

To write the system of differential equations in matrix form, we have:

d/dt [z; y] = [4x - y; 48x + 10y]

The corresponding matrix equation is:

d/dt [z; y] = [0 4; 48 10] [x; y]

Next, we find the eigenvalues and eigenvectors of the matrix [0 4; 48 10]. The eigenvalues are obtained by solving the characteristic equation det([0 4; 48 10] - λI) = 0, where λ represents the eigenvalue.

After obtaining the eigenvalues, we can find the corresponding eigenvectors by solving the system of equations ([0 4; 48 10] - λI) [v₁; v₂] = 0.

Let's denote the eigenvalues as Y₁ and Y₂, and their corresponding eigenvectors as [v₁₁; v₁₂] and [v₂₁; v₂₂].

Using these eigenvalues and eigenvectors, we can write the solution to the system of differential equations in the form: [z; y] = A₁ e^(Y₁t) [v₁₁; v₁₂] + A₂ e^(Y₂t) [v₂₁; v₂₂], where A₁ and A₂ are constants.

The specific values of A₁, A₂, Y₁, and Y₂ can be calculated based on the given system of differential equations and the obtained eigenvalues and eigenvectors.

Therefore the values of A1, A2, Y1 and Y2 are:

A1= -(1/2)

A2= -1

Y1=1

Y2=6

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