how many ways are there to choose a president, vice president, and treasurer of a 7- member club, if no person can hold more than one oce?

Answer:

4 centimetres per hour.

Step-by-step explanation:

Answer:

7.1437 is greater

Step-by-step explanation:

Answer:

7.133 < 7.1437

Step-by-step explanation:

Line up the numbers till you find one that is greater

Answer:

12 I believe

Step-by-step explanation:

[tex]12\sqrt{2}[/tex] x sin(45) = 12

I am pretty positive I am correct, I am sorry if I am off a little

Answer:

4. 50,000 square miles

5. The office would round to $800,000 instead of $700,000 because they would lose that needed $35,495. If they rounded down, their budget would be cut short, and they won’t have enough money for the office.

6. Five numbers that could also be the population: 577,777; 575,555; 578,999: 578,787; 581,545.

7. 600,000

8.584,000

9. 260,000

Step-by-step explanation:

4. In this question, we’re rounding the number, “54,555” go the nearest ten thousand. When rounding numbers you look at the number next to the digit your rounding. In this case, since the second number is not 5 or higher, we round it to 50,000 as that’s closer than 60,000.

5. Like I stated above: if they round to the lowest amount, they won’t have enough money to but what they need. Therefore, even if according to rounding it needs to be lower, it will not be beneficial to the company.

6. This are all possible numbers cause they can all by round to 580,000 AND 600,000

7. Since the number next to six is not five or higher, we round it to the closest number which is 600,000.

8. Since the number next to 3 is 5, we round it to the higher number. So it’s 584,000

9. Since the number next to the underlined nunber is 6, we round it to the closest number which is 260,000

If you have any more questions, you can ask me :)

The variable "pounds of popcorn" served at a movie theater is a continuous variable that can take any value within a certain range. It is not restricted to specific discrete values or categories.

A continuous variable is one that can take any value within a certain range. In the case of "pounds of popcorn" served at a movie theater, it can have fractional values and can vary continuously from very small amounts to large amounts.

It is not restricted to specific discrete values or categories. In contrast, discrete variables can only take on specific values, while ratio variables have a meaningful zero point and allow for comparisons of ratios. Since the amount of popcorn served can be measured with any level of precision, it falls under the category of a continuous variable.

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[tex]\(\sqrt{k+1}\)[/tex] is irrational.

By the principle of strong induction, we can conclude that the square root of 18 is irrational.

What is irrationality?

In mathematics, irrationality refers to a property of certain numbers that cannot be expressed as a fraction of two integers or as a terminating or repeating decimal.

To prove that the square root of 18 is irrational using strong induction, we need to show that for every positive integer [tex]\(n\), if \(n > 1\) and \(\sqrt{n}\) is irrational, then \(\sqrt{n+1}\)[/tex] is also irrational.

[tex]\textbf{Base Case:}[/tex]

For [tex]\(n = 2\)[/tex], we have [tex]\(\sqrt{2}\).[/tex] It is a known fact that [tex]\(\sqrt{2}\)[/tex] is irrational. Thus, the base case holds true.

[tex]\textbf{Inductive Step:}[/tex]

Assume that for some positive integer k, if[tex]\(2 \leq k\) and \(\sqrt{k}\)[/tex] is irrational, then [tex]\(\sqrt{k+1}\)[/tex] is also irrational.

Now, consider the case for [tex]\(n = k+1\)[/tex]). We want to prove that [tex]\(\sqrt{k+1}\)[/tex] is irrational.

Since [tex]\(k \geq 2\)[/tex], we have [tex]\(k+1 > 2\)[/tex]. Therefore,[tex]\(\sqrt{k+1}\) is greater than \(\sqrt{2}\).[/tex]

Assume, for contradiction, that [tex]\(\sqrt{k+1}\)[/tex] is rational. Then, we can write \[tex](\sqrt{k+1}\)[/tex] as a fraction [tex]\(\frac{p}{q}\),[/tex] where p and q are positive integers with no common factors other than 1.

By squaring both sides, we get [tex]\(k+1 = \left(\frac{p}{q}\right)^2 = \frac{p^2}{q^2}\).[/tex]

Rearranging the equation, we have[tex]\(p^2 = (k+1)q^2\).[/tex]

Since [tex]\(k+1\)[/tex] is a positive integer and [tex]\(q^2\)[/tex] is also a positive integer, [tex]\(p^2\)[/tex]must be a multiple of [tex]\(k+1\)[/tex].

This implies that p must also be a multiple of [tex]\(k+1\).[/tex] Let \(p = m(k+1)\), where m is a positive integer.

Substituting this into the equation, we have [tex]\((m(k+1))^2 = (k+1)q^2\)[/tex].

Simplifying, we get [tex]\(m^2(k+1) = q^2\).[/tex]

This implies that [tex]\(q^2\)[/tex] is a multiple of [tex]\(k+1\)[/tex], which means [tex]\(q\)[/tex] is also a multiple of [tex]\(k+1\).[/tex]

However, this contradicts our assumption that p and q have no common factors other than 1.

Hence, our assumption that  [tex]\(\sqrt{k+1}\)[/tex] is rational must be false.

Therefore, [tex]\(\sqrt{k+1}\)[/tex] is irrational.

By the principle of strong induction, we can conclude that the square root of 18 is irrational.

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To evaluate dy/dt given the equation y³ = 2x² + 14, along with dx/dt = 3, x = 5, and y = 4, we can use implicit differentiation and substitute the given values.

dy/dt is approximately equal to 0.51 (rounded to two decimal places).

To explain the solution, let's start by differentiating both sides of the equation implicitly with respect to t:

3y²(dy/dt) = 4x(dx/dt)

Now, we substitute the given values x = 5, y = 4, and dx/dt = 3 into the equation:

3(4)²(dy/dt) = 4(5)(3)

Simplifying this expression, we have:

48(dy/dt) = 60

Dividing both sides by 48, we find:

dy/dt = 60/48 = 1.25

Therefore, dy/dt is approximately equal to 1.25. Rounded to two decimal places, the value is 0.51.

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Answer:

Step-by-step explanation:

Given Information :

Diameter : 10.2 units

Radius (r) = ?

[tex]r = \frac{d}{2} [/tex]

Input the values

[tex]r = \frac{10.2}{2} \\ \\ r = 5.1[/tex]

Answer: -7

Because if y is -7 then the equation would be 4 - 7 = -3 which is true :)

Answer:

Don't click the link its a virus

The reasonable estimate for limStartFraction 2 x squared minus x + 15 Over x cubed minus 5 x minus 12 EndFraction as x approaches 3 is [tex]\dfrac{1}{2}[/tex]

Given the limit of a function expressed as:

[tex]\lim_{x \to 3}\frac{2x^2-x+15}{x^3-5x-12}[/tex]

First, we need to substitute x = 3 into the function to have:

[tex]=\frac{2(3)^2-3+15}{3^3-5(3)-12}\\=\frac{18-3+15}{27-15-12}\\=\frac{0}{0} (indeterminate)[/tex]

Apply l'hospital rule on the function:

[tex]=\lim_{x \to 3}\frac{\frac{d}{dx} (2x^2-x+15)}{\frac{d}{dx} (x^3-5x-12)}\\=\lim_{x \to 3}\frac{4x-1}{3x^2-5}\\[/tex]

Subtitute x = 3 into the result

[tex]=\frac{4(3)-1}{3(3)^2-5}\\=\frac{12-1}{27-5}\\=\frac{11}{22}\\=\frac{1}{2}[/tex]

Hence the reasonable estimate for limStartFraction 2 x squared minus x + 15 Over x cubed minus 5 x minus 12 EndFraction as x approaches 3 is [tex]\dfrac{1}{2}[/tex]

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Answer:

1/2

Step-by-step explanation:

i am in your walls

154 yd ²

the formula is (b • h) / 2 = A

so plug it in (14 • 22) / 2 = 154

Answer:

32.5 units²

Step-by-step explanation:

We can find the area by dividing the figure into two shapes. If we solve for the area of a square and a triangle, we can add them together.

First, we can solve for the area of the 5x5 square. To find the area, we can multiply the two dimensions.

5 · 5 = 25 units²

Next, we can find the area of the triangle. It has an equal height to the square, so the height of the triangle is 5 units. The width isn't specified. Instead, we are shown that the width of both the square and the triangle equals 8 units. If we subtract the width of the square from the total, we can find the width of the triangle, too.

8 - 5 = 3

So, the height and the width of the triangle are 5x3. To find the area we can multiply these together, and then divide the product by two.

5 · 3 = 15

[tex]\frac{15}{2}[/tex] = 7.5

The area of the triangle is 7.5 units².

Finally, we can add the area of the triangle and the square together.

25 + 7.5 = 32.5

The area of the figure, then, is 32.5 units².

The answer is the last option.

I hope this helps ^^

Answer:

1h=20m

60/4

0,25min+1h

1+0,25

=1,2

He will complete the race in 1.2 hours time.

Hence, option B is correct.

A measurement unit is a standard quality used to express a physical quantity. Also it refers to the comparison between the unknown quantity with the known quantity.

Given that,

The distance completing in 1 hour = 20 miles

So 1 miles of distance is covered in = 1/20 hours

It is also given that,

The distance of rally = 24 miles.

Therefore,

The time taken to cover 24 miles = 24/20

                                                        = 1.2

Thus,

The required time = 1.2 hours

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Answer:

250

Step-by-step explanation:

i did cross multiplication 40/100 = 100/x

100*100 and divide by 40

The correct answer is c) an outlier and a leverage point.A data point far from the mean of both the x's and y's is both an outlier and a leverage point.

A data point that is far from the mean of both the x-values and y-values can be considered an outlier and a leverage point. An outlier is a data point that significantly deviates from the overall pattern of the data. It lies far away from the majority of the data points and can have a significant impact on statistical analysis.

On the other hand, a leverage point is a data point that has an extreme value in terms of its x-value. It has the potential to influence the regression line and can greatly affect the regression model's fit. Therefore, a data point far from the mean of both x's and y's can be considered both an outlier and a leverage point.

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Answer:

(3x-2)(2x-3)

Step-by-step explanation:

Good luck!

Answer:

Step-by-step explanation:

Answer:

4225

Step-by-step explanation:

65x65=4225

P(X < 250) = P(X ≤ 249) = 0 (approximately) Hence, P(X ˃ 310) = 0 and P(X ˂ 250) = 0.

Given a random sample size of n = 900 from a binomial probability distribution with P = 0.30. The probability that the number of successes is greater than 310 and the probability that the number of successes is fewer than 250 are to be found.

Solution: a)We know that P(X > 310) can be found using normal approximation.

We have to check whether np and nq are greater than or equal to 10 or not, where p=0.30, q=0.70 and n=900.

Here, np = 900*0.30 = 270 and nq = 900*0.70 = 630. Since np and nq are greater than or equal to 10, we can use normal approximation for this binomial distribution.

Using the normal approximation formula, z = (X - μ) / σwhere X = 310, μ = np and σ = √(npq), we getz = (310 - 270) / √(900*0.30*0.70)z = 4.25

Using the z-table, the probability of z being greater than 4.25 is almost zero.

Therefore, P(X > 310) = P(X ≥ 311) = 0 (approximately)

b)We know that P(X < 250) can be found using normal approximation. We have to check whether np and nq are greater than or equal to 10 or not, where p=0.30, q=0.70 and n=900.  

Here, np = 900*0.30 = 270 and nq = 900*0.70 = 630.

Since np and nq are greater than or equal to 10, we can use normal approximation for this binomial distribution.

Using the normal approximation formula,z = (X - μ) / σwhere X = 250, μ = np and σ = √(npq), we getz = (250 - 270) / √(900*0.30*0.70)z = -4.25Using the z-table, the probability of z being less than -4.25 is almost zero.

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Given data: n = 900, P = 0.30.

a. The probability that the number of successes is greater than 310 is 0.0000.

b. The probability that the number of successes is fewer than 250 is 0.0174.

a. The formula for finding probability of binomial distribution is:

P(X > x) = 1 - P(X ≤ x)

P(X > 310) = 1 - P(X ≤ 310)

Mean μ = np

= 900 × 0.30

= 270

Variance σ² = npq

= 900 × 0.30 × 0.70

= 189

Standard deviation

σ = √σ²

= √189

z = (x - μ) / σ

z = (310 - 270) / √189

z = 4.32

Using normal approximation,

P(X > 310) = P(Z > 4.32)

= 0.00001673

Using calculator, P(X > 310) = 0.0000(rounded to four decimal places)

b. P(X < 250)

Mean μ = np

= 900 × 0.30

= 270

Variance σ² = npq

= 900 × 0.30 × 0.70

= 189

Standard deviation

σ = √σ²

= √189

z = (x - μ) / σ

z = (250 - 270) / √189

z = -2.12

Using normal approximation, P(X < 250) = P(Z < -2.12) = 0.0174.

Using calculator, P(X < 250) = 0.0174(rounded to four decimal places).

Therefore, the probability that the number of successes is greater than 310 is 0.0000 and the probability that the number of successes is fewer than 250 is 0.0174.

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The equations of the images are after the transformations are

a. y = -3x

b. y = x

c. x = 0

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

a. y = 3x

b. y = -x

c. x = 0.

The rule of the lines when reflected in the x-axis is

(x, y) = (x, -y)

This means that the functions are negated

So, we have the images to be

a. y = -3x

b. y = x

c. x = 0

Hence, the equations of the images are

a. y = -3x

b. y = x

c. x = 0

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Complete question :

The latter is leaning against the building so that the distance from the ground to the top of the ladder is 9 feet less than the length of the ladder finally put the ladder in the distance from the bottom of the ladder to the building 15 feet. Find length of the ladder.

Answer:

17

Step-by-step explanation:

From Pythagoras :

Hypotenus = sqrt(opposite² + adjacent²)

From the diagram

l = sqrt((l-9)² + 15²)

l² = (l² - 18x + 81 + 225)

Square both sides

l² = l² - 18x + 306

l² - l² = 306 - 18x

18l = 306

l = 306/18

l = 17

Answer:

7/12

Step-by-step explanation:

Basically to find the boys fraction I subtracted 40 from 96 and got 56. Your answer for the number of boys would be 56/96. I divided 4 from each and got 14/24. lastly I divide by 2 and got 7/12.

Answer: x= 12/103

alternate form x =0.116505

Step-by-step explanation:

Try Photo math! It explains step by step!

Hope this helps!!!!

Answer:

4c + 5b + 11

Step-by-step explanation:

4c + 3 + 5b + 8

4c + 5b + 11

Answer:

A rhombus is a special type of parallelogram

Step-by-step explanation:

Similarities in a rhombus and parallelogram:

A rhombus is a special type of parallelogram in that

____________________________

So your answer would be C, [tex]\bold{a=5}[/tex].

____________________________

The estimator V₁ = 1/2(X₁ + X₂) is preferred over V₂ = 1/2(X₁ + 2X₂) because it is unbiased and has a smaller variance.

To determine which estimator to use, consider the properties of unbiased estimators and compare the variances of V₁ and V₂.

Unbiasedness:

For V₁:

E(V₁) = E(1/2(X₁ + X₂)) = 1/2(E(X₁) + E(X₂)) = 1/2(µ + µ) = µ

For V₂:

E(V₂) = E(1/2(X₁ + 2X₂)) = 1/2(E(X₁) + 2E(X₂)) = 1/2(µ + 2µ) = 3/2µ

From the calculations, we can see that V₁ is an unbiased estimator of the population mean µ, while V₂ is biased since E(V₂) ≠ µ

Therefore, V₁ is preferred in terms of unbiasedness.

Variance:

Var(V₁) = Var(1/2(X₁ + X₂)) = 1/4(Var(X₁) + Var(X₂) + 2Cov(X₁, X₂))

Var(V₁) = 1/4(δ² + δ² + 2*0) = 1/2δ²

Var(V₂) = Var(1/2(X₁ + 2X₂)) = 1/4(Var(X₁) + 4Var(X₂) + 2Cov(X₁, 2X₂))

Var(V₂) = 1/4(δ² + 4δ² + 2*0) = 5/4δ²

Comparing the variances, we can see that Var(V₁) = 1/2δ², while Var(V₂) = 5/4δ².

Since Var(V₁) is smaller than Var(V₂), V₁ is preferred in terms of variance as well.

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The solution to the system is {eq}(x, y) = (2, 1){/eq}. For the given system, we have the solutions (-3, 0) and (0, 6).

To solve a system of equations, we must find the value of each variable that satisfies both equations. One of the most common methods for solving systems of equations is called substitution.

In substitution, we solve for one variable in one equation and then plug that expression into the other equation.

For example, if we have the system {eq}2x + y = 5,

\quad 4x - y = 7 {/eq},

we can solve for y in the first equation: {eq}y = 5 - 2x {/eq} Then we substitute this expression for y into the second equation:

{eq}4x - (5 - 2x) = 7 {/eq}

Simplifying gives {eq}6x = 12 {/eq}, or {eq}x = 2 {/eq} Once we have a value for one variable, we can substitute it into either equation to find the value of the other variable.

Using {eq}y = 5 - 2x {/eq}, we have {eq}y = 5 - 2(2) = 1 {/eq}These solutions represent the values of x and y that satisfy both equations in the system. To check,

we can substitute each solution into both equations to ensure they are valid. If both equations are satisfied, then we have found the correct solutions.

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