Which is the best definition of mathematical proof? a/A sequence of statements that demonstrates the truth of an assertion. b/Statements that show the assertion is false using a counterexample. c/A paragraph that always has three parts and shows that an assertion is true. d/Statements in any order that show that an assertion is true.

Answer:

The correct option is b.

Step-by-step explanation:

The complete question is:

Suppose you want to study the length of time devoted to commercial breaks for two different types of television programs. Identify the types of programs you want to study (e.g., sitcoms, sports events, movies, news, children's programs, etc.). How large should the sample be for a specified margin of error.

(a) It depends only on the specified margin of error.

(b) It depends on not only the specified margin of error, but also on the confidence level.

(c) It depends only on the confidence level.

Solution:

The (1 - α) % confidence interval for population mean is:

[tex]CI=\bar x\pm z_{\alpha/2}\times \frac{\sigma}{\sqrt{n}}[/tex]  

The margin of error for this interval is:

 [tex]MOE=z_{\alpha/2}\times \frac{\sigma}{\sqrt{n}}[/tex]

Then the sample size formula is:

[tex]n=[\frac{z_{\alpha/2}\times \sigma}{MOE}]^{2}[/tex]

The sample size is dependent upon the confidence level (1 - α) %, the standard deviation and the desired margin of error.

Thus, the correct option is b.

The size of the sample 'n' depends on not only the specified margin of error, but also on the confidence level.

Given :

Suppose you want to study the length of time devoted to commercial breaks for two different types of television programs.

The following steps can be used in order to determine the size of the sample be for a specified margin of error:

Step 1 - The formula of the confidence interval is given below:

[tex]\rm CI =\bar{x}+z_{\alpha /2}\times \dfrac{\sigma }{\sqrt{n} }[/tex]

Step 2 - Now, for this interval, the formula of margin of error is given below:

[tex]\rm MOE = z_{\alpha /2}\times \dfrac{\sigma}{\sqrt{n} }[/tex]

Step 3 - Solve the above expression for sample size 'n'.

[tex]\rm n = \left(\dfrac{z_{\alpha /2}\times \sigma}{MOE}\right)^2[/tex]

From the above steps, it can be concluded that the correct option is B) It depends on not only the specified margin of error, but also on the confidence level.

For more information, refer to the link given below:

https://brainly.com/question/13990500

Answer:

true

Step-by-step explanation:

Answer:

A

Step-by-step explanation:

The height is always perpinducular to the base. The height here is perpendicular to line segment A.

Answer:

A

Step-by-step explanation:

Answer:

Step-by-step explanation:

Hello, please consider the following.

a)

[tex](2x-2)^2 - (x+4)^2 \\\\=(2x-2-(x+4))(2x-2+x+4)\\\\=(2x-2-x-4)(3x+2)\\\\=\boxed{(x-6)(3x+2)}[/tex]

b)

[tex](3x+4) (3x-4)\\\\=(3x)^2-4^2\\\\=\boxed{9x^2-16}[/tex]

Thank you.

Answer:

Kindly check explanation

Step-by-step explanation:

Taking a careful look at the graph above, the graph depicts that there is sizeable growth or increase in the rate of interest between 2008 to 2012. However, the actual increase in the rate of interest between 2008 - 2012 is (3.152% - 3.141%) = 0.011%. This change is very small compared to what is portrayed by the pictorial representation of the bar graph. This could be due to the scaling of the vertical axis which didn't start from 0, thereby exaggerating the increase in the actual rate of interest. It will thus mislead observers into thinking the increase is huge.

Answer:

1. The bottom of the ladder is 4 meters away from the wall

2. I'm not sure about this one, someone else answer please :D

Step-by-step explanation:

We can use the Pythagorean Theorem to find how far away the bottom of the ladder is.

The ladder is creating a triangle, with 5 as it's hypotenuse and 3 as one of the left.

[tex]a^2 + 3^2 = 5^2\\a^2 + 9 = 25\\a^2 = 25-9\\a^2 = 16\\a = 4[/tex]

I'm sorry I couldn't answer the second one, but I hope this helped!

Answer:

a.  4m

b. 6m/s

Step-by-step explanation:

wall height = y = 3m

ladder length = L = 5m

distance from bottom of ladder to the wall = x

a.   y² + x² = L²   -----------eq.(1)

     3³ + x² = 5²        

    x = 4 m

b. How fast is the bottom of the ladder sliding away from the wall? = dx/dt

using eq.1  ---- y² + x² = L²

2y (dy/dt) + 2x (dx/dt) = 0

y (dy/dt) + 2 (dx/dt) = 0

we know that (dy/dt) = -8 m/s

3 (-8) + 4 (dx/dt) = 0

dx/dt =  -24 / -4

dx/dt = 6 m/s

Answer:

170

Step-by-step explanation:

3(4)³ - 2(4)² + 10

192 - 32 + 10 = 170

Step-by-step explanation:

( secA + 1)( sec A - 1)

Using the expansion

( a + b)( a - b) = a² - b²

Expand the expression

We have

sec²A + secA - secA - 1

That's

sec² A - 1

From trigonometric identities

So we have the final answer as

As proven

Hope this helps you

Step-by-step explanation:

Here,

LHS

= (SecA+1)(secA -1)

[tex] = {sec}^{2} A - 1[/tex]

[tex]{as{a}^{2} - {b}^{2} =(a + b)(a - b)[/tex]

Now, we have formula that:

[tex] {sec}^{2} \alpha - {tan \alpha }^{2} = 1[/tex]

[tex] {tan}^{2} \alpha = {sec }^{2} \alpha - 1[/tex]

as we got ,

[tex] = {sec}^{2} A- 1[/tex]

This is equal to:

[tex] = {tan}^{2} A[/tex]

= RHS proved.

Hope it helps....

Answer:

12 units

Step-by-step explanation:

Since all of the sides of a rhombus are congruent, JK = JM which means:

2x + 4 = 3x

-x = -4

x = 4 so 3x = 3 * 4 = 12

Answer:

.33

Step-by-step explanation:

Hey there!

1 / 3

= .333333333

.33 rounded to the nearest hundredth

Hope this helps :)

Answer:

Try to find the unit rate for bulk items that you have for these and then compare all of the prices together.

Answer:

It is, indeed, a reduction. But, the scale factor of the dilation should be a fraction instead of a whole number, since the shape has shrunk.

Instead of [tex]\frac{KN}{K'N'}[/tex], it should be [tex]\frac{K'N'}{KN}[/tex]. That would make it (4 - 2) / (8 - 4) = 2 / 4 = 1/2.

Hope this helps!

Answer:422977164210 or it could be [tex]4.2297716421(10) ^{11}[/tex]

Step-by-step explanation:

Answer:

First Attachment : Option A,

Second Attachment : Option C

Step-by-step explanation:

We are given that,

z₁ = [tex]3(\cos ((\pi )/(6))+i\sin ((\pi )/(6)))[/tex] and z₂ = [tex]4(\cos ((\pi )/(3))+i\sin ((\pi )/(3)))[/tex]

Therefore if we want to determine z₁( z₂ ), we would have to find the trigonometric form of the following expression,

[tex]3(\cos ((\pi )/(6))+i\sin ((\pi )/(6)))*4(\cos ((\pi )/(3))+i\sin ((\pi )/(3)))[/tex]

( Combine expressions )

= [tex]12(\cos ( \pi /6+\pi / 3 ) + i\sin (\pi /6 +\pi / 3 )[/tex]

( Let's now add [tex]\pi / 6 + \pi / 3[/tex], further simplifying this expression )

[tex]\frac{\pi }{6}+\frac{\pi }{3} = \frac{\pi }{6}+\frac{\pi 2}{6} = \frac{\pi +\pi 2}{6} = \frac{3\pi }{6} = \pi / 2[/tex]

( Substitute )

[tex]12(\cos ( \pi /2 ) + i\sin ( \pi /2 ) )[/tex]

And therefore the correct solution would be option a, for the first attachment.

______________________________________________

For this second attachment, we would have to solve for the following expression,

[tex]\frac{3\left(\cos \left(\frac{\pi \:}{6}\right)+i\sin \left(\frac{\pi \:}{6}\right)\right)}{4\left(\cos \left(\frac{\pi \:}{3}\right)+i\sin \left(\frac{\pi \:}{3}\right)\right)}[/tex]

From which we know that cos(π/6) = √3 / 2, sin(π/6) = 1 / 2, cos(π/3) = 1 / 2, and sin(π/3) = √3 / 2. Therefore,

[tex]\:\frac{3\left(\cos \left(\frac{\pi }{6}\right)+i\sin \left(\frac{\pi }{6}\right)\right)}{4\left(\cos \left(\frac{\pi }{3}\right)+i\sin \left(\frac{\pi }{3}\right)\right)}:\quad \frac{3\sqrt{3}}{8}-i\frac{3}{8}[/tex]

[tex]\frac{3\sqrt{3}}{8}-i\frac{3}{8} = \frac{3\sqrt{3}}{8}-\frac{3}{8}i[/tex]

Our solution for the second attachment will be option c.

Answer:

[tex]\boxed{Probability=\frac{1}{2} }[/tex]

Step-by-step explanation:

The probability of flipping at least 1 head from flipping 2 coins is:

=> Total sides of the coins = 4

=> Sides which are head = 2

=> Probability = 2/4 = 1/2

Answer:

False

Step-by-step explanation:

We also need to show that it is true for n=1

and for n=k+1

Answer: 7 mi

Step-by-step explanation: since the distance from bluepoint to Manchester is 12 9/10 mi and you know that bluepoint to Silverstone is 5 9/10 subtract that and you get 7 mi as your answer

Answer:

7 miles

Step-by-step explanation:

Hey there!

Well given BM and BS, we need to subtract them.

12 9 /10 - 5 9/10

9/10 - 9/10 = 0

12 - 5 = 7

Silvergrove to Manchester is 7 miles.

Hope this helps :)

Answer:16

Step-by-step explanation:

Answer: {5 ± 2√10, 5 - 2√10}

Step-by-step explanation: First isolate the binomial squared by adding 40 to both sides to get (x - 5)² = 40.

Next, square root both sides to get x - 5 = ± √40.

Notice that root of 40 can be broken down to 2√10.

So we have x - 5 = ± 2√10.

To get x by itself, add 5 to both sides to get x = 5 ± 2√10.

So our answer is just {5 ± 2√10, 5 - 2√10}.

As a matter of form, the number will always come before the

radical term in your answer to these types of problems.

In other words, we use 5 ± 2√10 instead of ± 2√10 + 5.

Answer:

Hey there!

We can simplify this by combining like terms.

-5g+10+7g-3

-5g+7g+10-3

2g+7

Let me know if this helps :)

Answer: [tex]2g+7[/tex]

Combine Like Terms

[tex]-5g+10+7g+-3\\(-5g+7g)+(10+-3)\\2g+7[/tex]

Assuming you meant to write [tex]A = \pi r^2[/tex], then the answer is C) circle

On your keyboard, you can say A = pi*r^2 to mean the same thing as above.

Answer:

The point that lies on the line parallel to line KL would be ( 8, - 10 )

Step-by-step explanation:

Line KL passes through the points ( - 6, 8 ) and ( 6, 0 ) while it's respective parallel line passes through point M, ( - 4, - 2 ).

Our approach here is to first determine the slope of KL such that the slope of it's parallel line will be the same, and hence we can determine a second point on this line.

Slope of KL : ( y₂ - y₁ ) / ( x₂ - x₁ ),

( 0 - 8 ) / ( 6 - ( - 6 ) ) = - 8 / 6 + 6 = - 8 / 12 = - 2 / 3

Slope of respective Parallel line : - 2 / 3,

Another point on Parallel line : ( 8, - 10 )

How can we check if this point really belongs to the parallel line? Let's take the slope given the points ( - 4, - 2 ) and ( 8, - 10 ), and check if it is - 2 / 3.

( y₂ - y₁ ) / ( x₂ - x₁ ),

( - 10 - ( - 2 ) ) / ( 8 - ( - 4 ) ) = ( - 10 + 2 ) / ( 8 + 4 ) = - 8 / 12 = - 2 / 3

And therefore we can confirm that this point belongs to line KL's parallel line, that passes through point M.

Answer:

D

Step-by-step explanation:

Answer:

1260

Step-by-step explanation:

(7!)/ (2!times 2!)

7 factorial divided by 2factorial times 2 facotiral

The number of distinguishable ways the letters of the following word can be arranged 'MILLION' is 1260.

The different arrangements which can be made out of a given number of objects by taking out some or all at a time are called permutations.

The number of different permutations of n objects with m₁ repeated items, m₂ repeated items,...,mₙ repeated items can be calculated as;

m!/(m₁!)(m₂!)...(mₙ!)

Here, the letter of the word 'MILLION' is a total of 7 letters.

So, the number of possible arrangements will be

(7!)/ (2!times 2!)

= 1260

Therefore the number of distinguishable ways the letters of the following word can be arranged 'MILLION' is 1260.

Learn more about permutations here -

brainly.com/question/4301655

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Answer:

[tex]|R_n (x)| \leq \dfrac{|x - \dfrac{\pi}{2}|^{n+1}}{(n+1)!}[/tex]

Step-by-step explanation:

From the given question; the objective is to show that :

[tex]\lim_{n \to \infty} R_n (x) = 0[/tex] for all x in the interval of convergence f(x)=cos x, a= π/2

Assuming for the convergence f the taylor's series , f happens to be the derivative on an open interval I with a . Then the Taylor series for the convergence of f , for all x in I , if and only if  [tex]\lim_{n \to \infty} R_n (x) = 0[/tex]  

where;

[tex]\mathtt{R_n (x) = \dfrac{f^{(n+1)} (c)}{n+1!}(x-a)^{n+1}}[/tex]

is a remainder at x and c happens to be between x and a.

Given that:

a= π/2

Then; the above equation can be written as:

[tex]\mathtt{R_n (x) = \dfrac{f^{(n+1)} (c)}{n+1!}(x-\dfrac{\pi}{2})^{n+1}}[/tex]

so c now happens to be the points between π/2 and x

If we recall; we know that:

[tex]f^{(n+1)}(c) = \pm \ sin \ c \ or \ cos \ c[/tex] (as a result of the value of n)

However, it is true that for all cases that  [tex]|f ^{(n+1)} \ (c) | \leq 1[/tex]

Hence, the remainder terms is :

[tex]|R_n (x)| = | \dfrac{f^{(n+1)}(c)}{(n+1!)}(x-\dfrac{\pi}{2})^{n+1}| \leq \dfrac{|x - \dfrac{\pi}{2}|^{n+1}}{(n+1)!}[/tex]

If [tex]\lim_{n \to \infty} R_n (x) = 0[/tex]   for all x and x is fixed, Then

[tex]|R_n (x)| \leq \dfrac{|x - \dfrac{\pi}{2}|^{n+1}}{(n+1)!}[/tex]

Answer:

[tex]\pi[/tex] radian

Step-by-step explanation:

We know that angle for a full circle is 2[tex]\pi[/tex]

In the given figure shape is semicircle

hence,

angle for semicircle will be half of angle of full circle

thus, angle for given figure = half of  angle for a full circle = 1/2 * 2[tex]\pi[/tex] = [tex]\pi[/tex]

Thus, answer is [tex]\pi[/tex] radian

alternatively, we also know that angle for a straight line is 180 degrees

and 180 degrees is same as [tex]\pi[/tex] radian.

Answer:

a number, r, added to 6

Step-by-step explanation:

a number, r, added to 6

Answer:

a number, r, added to 6

Step-by-step explanation:

a number, r, added to 6

Answer:

√32

Step-by-step explanation:

Answer:

$16.43.

Step-by-step explanation:

At the grocery store, you spent $485.72. With 2% cashback, you would get 485.72 * 0.02 = 9.7144 dollars worth of cashback.

At other places, you spend $671.28. With 1% cashback, you would get 671.28 * 0.01 = 6.7128 dollars worth of cashback.

9.7144 + 6.7128 = 16.4272, which is about $16.43 of cashback.

Hope this helps!

The amount of cashback that you earned will be $16.42.

The amount of any product is given as though it was a proportion of a hundred. The ratio can be expressed as a quarter of 100. The phrase % translates to one hundred percent. It is symbolized by the character '%'.

Suppose you earn 2% cash back at grocery stores and 1% on all other purchases. If you spent $485.72 at the grocery store and $671.28 on all other purchases.

The total cashback is calculated as,

⇒ 0.02 x $485.72 + 0.01 x $671.28

⇒ $9.71 + $6.71

⇒ $16.42

More about the percentage link is given below.

https://brainly.com/question/8011401

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Answer:

B. z Subscript alpha divided by 2 zα/2 = 1.96.

Step-by-step explanation:

We are given that we want to construct a confidence interval. For this, the summary statistics for randomly selected weights of newborn​ girls:

n = 236​, [tex]\bar x[/tex] = 30.3 ​hg, s = 7.2 hg. The confidence level is 95​%.

As we can clearly see here that the population standard deviation is unknown and the sample size is also very large.

It has been stated that when the population standard deviation is unknown, we should use t-distribution but since the sample size is very large so we can use z distribution also as it is stated that at very large samples; the t-distribution corresponds to the z-distribution.

Here, [tex]\alpha[/tex] = level of significance = 1 - 0.95 = 0.05 or 5%

         [tex]\frac{\alpha}{2}=\frac{0.05}{2}[/tex] = 0.025 or 2.5%

So, the value of [tex]Z_(_\frac{\alpha}{2} _)[/tex] in the z table is given as 1.96 with a 2.5% level of significance.

Answer:

[tex]\displaystyle \large \boxed{ \lim_{x \rightarrow +\infty} {x\left(\sqrt{x^2-1}-x\right)}=-\dfrac{1}{2}}[/tex]

Step-by-step explanation:

Hello, please consider the following.

[tex]\sqrt{(x^2-1)}-x\\\\=\sqrt{x^2(1-\dfrac{1}{x^2})}-x\\\\=x\left( \sqrt{1-\frac{1}{x^2}}-1\right)[/tex]

For x close to 0, we can write

[tex]\sqrt{1+x}=1+\dfrac{1}{2}x-\dfrac{1}{8}x^2+o(x^2)\\\\\ \text{x tends to } +\infty \text{ means }\dfrac{1}{x} \text{ tends to 0}\\\\\text{So, when }\dfrac{1}{x}\text{ is close to 0, we can write.}\\\\\sqrt{1-\dfrac{1}{x^2}}=1-\dfrac{1}{2}\dfrac{1}{x^2}-\dfrac{1}{8}\dfrac{1}{x^4}+o(\dfrac{1}{x^4})[/tex]

So,

[tex]x\left( \sqrt{1-\frac{1}{x^2}}-1\right)\\\\=x(1-\dfrac{1}{2}\dfrac{1}{x^2}+o(\dfrac{1}{x^2})-1)\\\\=-\dfrac{1}{2x}+o(\dfrac{1}{x})[/tex]

It means that

[tex]\displaystyle \lim_{x \rightarrow +\infty} {x\left(\sqrt{x^2-1}-x\right)}\\\\=\lim_{x \rightarrow +\infty} {-\dfrac{x}{2x}}=-\dfrac{1}{2}[/tex]

Thank you