A cylindrical tank is one-fifth full of oil. The cylinder has a base radius of 80 cm. The height of the cylinder is 200 cm. 1 litre 1000 cm3 How many litres of oil are in the tank? Round your answer to the nearest litre

Step-by-step explanation:

The new square has a scale factor of 50%, which means that its dimensions will be 50% of the original dimensions.

The original dimensions were 12 inches long and 12 inches wide, so the new square would be 6 inches long and 6 inches wide.

Answer:

Where is your question?

Answer:

B : 279.46 ( 24871.94 ÷89)

= 279.46 good for the hot tub

(B) 279.46

Answ

Step-by-step explanation:

its d i got wrong

The volume of the cuboid to be approximately 1.8cm³.

In this case, we are given that the volume of the small cube is 1cm³. Therefore, using the formula, we can solve for the length of its side as follows:

1cm³ = s³

Taking the cube root of both sides: ∛(1cm³) = ∛(s³) 1cm = s

Hence, the length of the side of the small cube is 1cm.

Then, we can estimate the other two dimensions of the cuboid as follows:

   • The width (w) could be a little more than the length of the side of the small cube, say 1.2a.

   • The height (h) could also be a little more than the length of the side of the small cube, say 1.5a.

Therefore, using the formula for the volume of a cuboid, we can estimate the volume of the cuboid as follows:

V = lwh V = a × 1.2a × 1.5a V = 1.8a³

Substituting the value of 'a' as 1cm, we get:

V ≈ 1.8cm³

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The rectangle is a closed figure having 4 sides, 4 vertices, and 4 angles.

The opposite sides of the rectangle are parallel and equal to each other. The measure of the angles of the rectangle is 90° i.e. they are right angles.

A rectangle is a 2-dimensional figure.  

All rectangles can be squares but all squares cannot be rectangles. A square is a special type of rectangle. This is possible when all the sides of the rectangle are equal.

The various objects in our day-to-day life which has rectangular shape are -

box, mobile, book, blackboard or whiteboard, table, etc.

The visual representation of the rectangle is as follows-

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Answer:

Step-by-step explanation:

Side "h" ( wall height ) is 35

The height of the wall "h".

\({\boxed{\mathcal{\red{The\:height\:of\:the\:wall\:"h"\:is\:35.71\:ft.}}}}\)

\(\sf\purple{Using\:Pythagoras \:theorem, \:we\:have}\)

\(( {Perpendicular})^{2} + ( {Base})^{2} = ( {Hypotenuse})^{2} \\ ➡ \: {h}^{2} + ({35 \: ft})^{2} = ({50 \: ft})^{2} \\ ➡ \: {h}^{2} + 1225 \: {ft}^{2} = 2500 \: {ft}^{2} \\ ➡ \: {h}^{2} = 2500 {ft}^{2} - 1225 \: {ft}^{2} \\ ➡ \: {h}^{2} = 1275 \: {ft}^{2} \\ ➡ \: h \: = \sqrt{1275 \: {ft}^{2} } \\ ➡ \: h = 35.707\: ft \: \\ ➡ \: h = 35.71\: ft\)

\(\sf\red{Therefore\:the\:height\:of\:the\:wall\:"h"\:is\:35.71\:ft.}\)

\(( {35.71 \: ft})^{2} + ( {35 \: ft})^{2} = ({50 \: ft})^{2} \\ ✒ \: 1275 \: {ft}^{2} + 1225 \: {ft}^{2} \: = 2500 \: {ft}^{2} \\ ✒ \: 2500 \: {ft}^{2} = 2500 \: {ft}^{2} \\ ✒ \: L.H.S.=R. H. S\)

Hence verified.

\(\circ \: \: { \underline{ \boxed{ \sf{ \color{green}{Happy\:learning.}}}}}∘\)

Answer:

∠1 is 33°

∠2 is 57°

∠3 is 57°

∠4 is 33°

Step-by-step explanation:

First off, we already know that ∠2 is 57° because of alternate interior angles.

Second, it's important to know that rhombus' diagonals bisect each other; meaning they form 90° angles in the intersection. Another cool thing is that the diagonals bisect the existing angles in the rhombus. Therefore, 57° is just half of something.

Then, you basically just do some other pain-in-the-butt things after.

Since that ∠2 is just the bisected half from one existing angle, that means that ∠3 is just the other half; meaning that ∠3 is 57°, as well.

Next is to just find the missing angle ∠1. Since we already know ∠3 is 57°, we can just add that to the 90° that the diagonals formed at the intersection.

57° + 90° = 147°

180° - 147° = 33°

∠1 is 33°

Finally, since that ∠4 is just an alternate interior angle of ∠1, ∠4 is 33°, too.

In a two-way factorial design analysis, there are two main effects tested.

A two-way factorial design involves the simultaneous manipulation of two independent variables, each with multiple levels, to study their individual and combined effects on a dependent variable. The main effects in such a design represent the effects of each independent variable independently, ignoring the influence of the other variable.

When conducting a two-way factorial design analysis, there are two main effects tested, corresponding to each independent variable. The main effect of one variable is the difference in the means across its levels, averaged over all levels of the other variable. Similarly, the main effect of the other variable is the difference in the means across its levels, averaged over all levels of the first variable.

Testing the main effects allows researchers to determine the individual impact of each independent variable on the dependent variable, providing insights into their overall influence. By analyzing the main effects, researchers can assess the significance and directionality of the effects, aiding in the interpretation of the experimental results and understanding the relationship between the independent and dependent variables in the factorial design.

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Answer:

4 days

Step-by-step explanation:

because Mariana drank 2 pints a day 8 pints are in a gallon so 8 ÷ 2 = 4

Answer:

4/13

Step-by-step explanation:

Thus, the solutions of the equation sin 2x cos x = 0 are given by x = nπ/2 and x = π/2 + nπ, where n is any integer.

To solve the equation sin 2x cos x = 0 for x, we need to use some trigonometric identities. Let's begin by factoring out the common term cos x from the left side of the equation. Hence, sin 2x cos x = 0 cos x (sin 2x) = 0Now, we can set each factor to zero since the product of two factors is zero if and only if at least one of them is zero. Thus, either cos x = 0 or sin 2x = 0. Solving for cos x = 0, we get x = π/2 + nπ for any integer n. Solving for sin 2x = 0, we get 2x = nπ for any integer n. Hence, x = nπ/2 for any integer n. Thus, the solutions of the equation sin 2x cos x = 0 are given by x = nπ/2 and x = π/2 + nπ, where n is any integer.

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The first option :

Multiply the first equation by 8 and the second equation by 2

would not result in the elimination of the variables in this system

There were 1.4 pounds of honey in each jar.

Since unitary method is a method for solving a problem by the first value of a single unit and then finding the value by multiplying the single value.

Given that Jasmine extracted 5.6 pounds of honey from a beehive. She divided the honey evenly into 4 jars.

Thus, divided 5.6/4

In each jar there would be; 1.4 pounds.

Now multiplying 1.4 by 4

1.4/4 =  5.6.

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The proper way to report the groups' mean and standard deviation of independent sample t test is  

Group 1 (M = 11.6, SD = 3.1) Group 2 (M = 13.24, SD = 2.9).

The APA style guide provides detailed instructions for referencing statistical test results, so in addition to following the basic format correctly, you must pay attention to details like punctuation, bracket placement, italicization, and other formatting elements.

Given, that the independent samples t test is conducted, where we have two groups.

Group 1- mean= 11.6 and SD=3.1

Group 2- mean=19 and SD=2.9

We have to write the values in the APA format for a report.

The APA formatting to represent the mean and SD of two groups is as follows, Mean and standard deviation are denoted by the letters M and SD, respectively.

Group 1 (M = 11.6, SD = 3.1) Group 2 (M = 13.24, SD = 2.9).

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Answer:

33.3%

Step-by-step explanation:

10/ 30= 0.3(3 would be a reoccurring decimal)

0.3*100=33.3

10% of students take the bus

Answer:

nope

Step-by-step explanation:

The given ratios  5:3 and 6:4 are not equivalent.

Equivalent ratios are the ratios that are the same when we compare them. Example 1/2 and 2/4 are equivalent ratios.

Two or more ratios can be compared with each other to check whether they are equivalent or not.

The given two ratios are 5:3 and 6:4

We can write them as 5/3 and 6/4

Let us apply subtraction, if we get 0 then these are equivalent.

6/4-5/3

The LCM of 4 and 3 is 12

=(18-20)/12

The value of 18-20 is -2

=-2/12

We have to divide 2 by 12 we get -1/6.

=-1/6 not equal to 0

The ratios 5:3 and 6:4 are not equivalent.

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Answer:

1/216

Step-by-step explanation:

1/6 times 1/6 times 1/6

or 6^-3

Answer:

Step-by-step explanation:

There is only 1 way you can throw 18 with 3 dice. There are 6 * 6 * 6 = 216 ways of throwing 3 dice. Only 1 will give you  18. So the answer is 1/216

Answer:

C

Step-by-step explanation:

we want to "isolate" h (height). that means "h" needs to be alone on one side of the equation, and all other factors need to be brought to the other side.

so, yes, we need to divide both sides by all factors of h : pi and r². only then will we have the isolated h on one side.

therefore, dividing both sides by pi is a good first step.

The answer is (A) \(\angle A \cong \angle A'\)

Both P₄(x) and P₅(x) are represented by the same expression.

\($P_5(x) = P_4(x) = \frac{2}{5} - \frac{2}{5}(x-6\pi) - \frac{4}{5}\left(\frac{1}{3}\right)(x-6\pi)^2 + \frac{8}{5}(x-6\pi)^3 + \frac{48}{5}\left(\frac{1}{3}\right)(x-6\pi)^4$\)

What is the polynomial equation?

A polynomial equation is an equation in which the variable is raised to a power, and the coefficients are constants. A polynomial equation can have one or more terms, and the degree of the polynomial is determined by the highest power of the variable in the equation.

To find the Taylor polynomials P₄ and P₅ centered at a = π/6  for f(x)= 5cos(x), we'll need to calculate the derivatives of f(x) and evaluate them at x= π/6.

We can then use these derivatives to construct the Taylor polynomials.

Let's start by finding the derivatives of f(x):

​

\(f(x) = 5\cos(x)$\\\\f'(x) = -5\sin(x)$\\$\\f''(x) = -5\cos(x)$\\\\$f'''(x) = 5\sin(x)$\\$\\f''''(x) = 5\cos(x)$\)

Now, let's evaluate these derivatives at x= π/6:

\(f\left(\pi/6\right) = 5\cos\left(\pi/6\right) = \frac{2}{5}$\\\\$f'\left(\pi/6\right) = -5\sin\left(\pi/6\right) = -\frac{2}{5}$\\\\$f''\left(\pi/6\right) = -5\cos\left(\pi/6\right) = -\frac{5}{3}$\\\\$f'''\left(\pi/6\right) = 5\sin\left(\pi/6\right) = \frac{2}{5}\\$\\$f''''\left(\pi/6\right) = 5\cos\left(\pi/6\right) = \frac{5}{3}$\)

Now, we can construct the Taylor polynomials:

The fourth degree Taylor polynomial P₄(x) centered at a = π/6 is given by:

\($P_4(x) = f\left(6\pi\right) + f'\left(6\pi\right)(x-6\pi) + \frac{2!}{2}f''\left(6\pi\right)(x-6\pi)^2 + \frac{3!}{3}f'''\left(6\pi\right)(x-6\pi)^3 + \frac{4!}{4}f''''\left(6\pi\right)(x-6\pi)^4$\)

Substituting the evaluated derivatives, we have:

Simplifying this expression will give you the specific form of P₄(x).

The fifth-degree Taylor polynomial P₅(x) centered at a = π/6 ​can be obtained by including one more term with the fifth derivative:

\($P_4(x) = \frac{2}{5} - \frac{2}{5}(x-6\pi) - \frac{4}{5}\left(\frac{1}{3}\right)(x-6\pi)^2 + \frac{8}{5}(x-6\pi)^3 + \frac{48}{5}\left(\frac{1}{3}\right)(x-6\pi)^4$\)

\($P_5(x) = P_4(x) = \frac{2}{5} - \frac{2}{5}(x-6\pi) - \frac{4}{5}\left(\frac{1}{3}\right)(x-6\pi)^2 + \frac{8}{5}(x-6\pi)^3 + \frac{48}{5}\left(\frac{1}{3}\right)(x-6\pi)^4$\)

Let's calculate the fifth derivative and complete the expression for  P₅(x):

(since the derivative of sin(x) evaluated at π/6 is 0)

Hence, both P₄(x) and P₅(x) are represented by the same expression.

\($P_5(x) = P_4(x) = \frac{2}{5} - \frac{2}{5}(x-6\pi) - \frac{4}{5}\left(\frac{1}{3}\right)(x-6\pi)^2 + \frac{8}{5}(x-6\pi)^3 + \frac{48}{5}\left(\frac{1}{3}\right)(x-6\pi)^4$\)

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It appears that the answer is my fat coeck

The radii and the centers are explained below.

Given that are equations of circles we need to find the center and the radius of each circle.

1) x² + 6x + y² - 4y = -9

x² + 2·3·x + y² - 2·2·y = -9

Add 13 to both side,

x² + 2·3·x + y² - 2·2·y + 13 = -9 + 13

x² + 2·3·x + y² - 2·2·y + 9 + 4 = 4

(x+3)² + (y-2)² = 4

The center = (-3, 2) and the radius = 2

2) x² + 10x + y² - 8y = -31

x² + 2·5·x + y² - 2·4·y = -31

Add 41 to both sides,

x² + 2·5·x + y² - 2·4·y + 41 = -31 + 41

x² + 2·5·x + y² - 2·4·y + 25 + 16 = 10

(x+5)² + (y-4)² = 10

The center = (-5, 4) and the radius = √10

3) x² - 2x + y² + 4y -11 = 0

x² - 2x + y² + 4y = 11

x² - 2·1·x + y² - 2·2·y = 11

Add 5 to both sides,

x² - 2·1·x + y² - 2·2·y + 5 = 11 + 5

x² - 2·1·x + y² - 2·2·y + 4 + 1 = 16

(x-1)² + (y-2)² = 4²

The center = (1, 2) and the radius = 4

4) x² + 9x + y² = 0

\(\left(x-\left(-\frac{9}{2}\right)\right)^2+\left(y-0\right)^2=\left(\frac{9}{2}\right)^2\)

The center = (-9/2, 0) and the radius = 9/2

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There are 92 bit strings of length 8 that start with a 11 or end with 000.

We can solve this problem using the principle of inclusion-exclusion. Let A be the set of bit strings of length 8 that start with 11, and let B be the set of bit strings of length 8 that end with 000. We want to find the size of the union of A and B.

The number of bit strings of length 8 that start with 11 is 2^6, since there are 6 remaining bits that can be either 0 or 1. The number of bit strings of length 8 that end with 000 is also 2^5, since there are 5 remaining bits that can be either 0 or 1.

However, we have counted the bit strings that both start with 11 and end with 000 twice. To correct for this, we need to subtract the number of bit strings of length 8 that start with 11000, which is 2^2.

Therefore, the number of bit strings of length 8 that start with a 11 or end with 000 is:

|A ∪ B| = |A| + |B| - |A ∩ B|

= 2^6 + 2^5 - 2^2

= 64 + 32 - 4

= 92

So there are 92 bit strings of length 8 that start with a 11 or end with 000.

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There are 88 bit strings of length 8 that start with "11" or end with "000."

To determine the number of bit strings of length 8 that start with "11" or end with "000," we can use the principle of inclusion-exclusion.

Let's consider the two conditions separately:

Bit strings that start with "11":

In this case, the first two bits are fixed as "11." The remaining 6 bits can be either 0 or 1, giving us 2^6 = 64 possibilities.

Bit strings that end with "000":

Similarly, the last three bits are fixed as "000." The first 5 bits can be either 0 or 1, resulting in 2^5 = 32 possibilities.

However, we have counted some bit strings twice because they satisfy both conditions (start with "11" and end with "000"). These bit strings have a length of at least 5 (3 bits in the middle), and there are 2^3 = 8 possibilities for these middle bits.

By using the principle of inclusion-exclusion, we can compute the total number of bit strings satisfying either condition as follows:

Total = Bit strings starting with "11" + Bit strings ending with "000" - Bit strings satisfying both conditions

= 64 + 32 - 8

= 88

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Answer:

y = -3/4x + 5/4

Step-by-step explanation:

Points: (7, -4) and ( -1, 2)

Slope:

m=(y2-y1)/(x2-x1)

m=(2+4)/(-1-7)

m=6/(-8)

m= -3/4

Slope-intercept:

y - y1 = m(x - x1)

y + 4 = -3/4(x - 7)

y + 4 = -3/4x + 21/4

y = -3/4x + 5/4

Answer:

y = -3/4x + 5/4

Step-by-step explanation:

Hope this helps

= 3 × 4

= 3 × 2²

x = 1

= 3 × 15 = 5x

= 45 /5 = x

= 9 = x

Answer:

Step-by-step explanation:

1. 4

2. 9

3. 7  ,14 , 21 , 28 ,35 , 42 , 49

4. 7 , 3

5. 527 , 752 , 572 , 725

6. 6 = d ( a / 2 )

   7 = e ( 2a )

   8 = a ( a * a )

   9 = b  ( 3a )

  10 = c ( 0 )

Answer:

See below for answers and explanations

Step-by-step explanation:

Problem 4:

The line of regression is \(\hat y = a+bx\) where:

\(a=\overline y-b \overline x\)

\(b=\frac{r*s_y}{s_x}\)

We are given that \(\overline y=75\), \(s_y=8\), \(\overline x=280\), \(s_x=30\), and \(r=0.60\), therefore our slope, \(b\), is:

\(b=\frac{r*s_y}{s_x}\)

\(b=\frac{(0.60)(8)}{30}\)

\(b=0.16\)

Therefore, the slope of the regression line is 0.16, which can be used along with the values of \(\overline y\) and \(\overline x\) to find the constant \(a\):

\(a=\overline y-b \overline x\)

\(a=75-(0.16)(280)\)

\(a=30.2\)

This means our final regression line is \(\hat y = 30.2 + 0.16x\)

Problem 5:

The slope, \(b=0.16\), means that for every 1 point earned for a student's pre-exam total, their final exam score will increase by 0.16 points for each point they earned on the pre-exam.

Problem 6:

The y-intercept (or constant), \(a=30.2\), means that if a student's pre-exam total were 0, then they would expect to get a 30.2 on the final exam.

Problem 7:

\(R^{2}=(0.60)^{2}=0.36=36\%\)

Problem 8:

The fraction of variability (aka. coefficient of determination), \(R^2\), means that a certain proportion (or percentage) of the variance in the response variable can be explained by the explanatory variable. In context, this means that 36% of the variance in final exam scores can be explained by the pre-exam scores.

Answer:

pic one is rock cycle, pic 2 crystaline structure, and pic three and 4 are the same pictures as question 2

Step-by-step explanation:

Answer:

I have the same homework and I could do it either

Step-by-step explanation:

The required number line is typically drawn from left to right to show all of the possible solutions that make the inequality true.

A number line is defined as the number marked on the line calibrated into an equal number of units. For example -1, 0, 1, and so on.

Here,
The number line is typically drawn from left to right to show all of the possible solutions that make the inequality true.

The inequality may involve a variable x, and the number line would represent all possible values of x that satisfy the inequality. The solutions could be, for example, all values of x greater than 3, in which case the number line would start at 3 and extend to the right to represent all values of x greater than 3. Alternatively, the solutions could be all values of x less than or equal to -2, in which case the number line would start at -2 and extend to the left to represent all values of x less than or equal to -2.

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