g 6. what can you conclude based on the results of your test (hint: what does the p-value tell you?)? do you reject the null hypothesis, or fail to reject the null hypothesis? does the chis sample population have a statistically significant different average height than 68 inches tall?

Answer with Step-by-step explanation:

Regular price=$39.79

Rounded price=$40

Discount percent=25%

Estimated Discount=25% of $40

$10

Estimated sale price= $40-$10

$30

Answer:

u fking  b

ur bad

Answer:

I think it is 0 solutions

Step-by-step explanation:

it should be 0 because no matter what number you substitie x for, the right side will always be 9 higher than it, so x never equals x

May I have brainliest please? :)

Answer:

If he earns 5$ an hour then just multiply how long he works over time by 5 and you get you answer C:     for ex: if he worked 24h he would get 120$ because 24 x 5=120

Step-by-step explanation:

The depth of the water increasing when the water is 13 feet deep is approximately 1.68 feet per minutes.

We know that the conical water tank has a radius of 12 feet and is 26 feet high.

We also know that water is flowing into the tank at a rate of 30ft³/min. In other words, our derivative of the volume with respect to time t is:

                                   \(\frac{dV}{dt} = \frac{10 ft^3}{min}\)

We want to find how fast the depth of the water is increasing when the water is 13 feet deep. So, we want to find dh/ dt.

First, remember that the volume for a cone is given by the formula:

                V = 1/3 π r² h    

We want to find dh/dt. So, let's take the derivative of both sides with respect to the time t. However, first, let's put the equation in terms of h.

We can see that we have two similar triangles. So, we can write the following proportion:

         \(\frac{r}{h} = \frac{12}{36}\)

Multiply both sides by h:

    ⇒ \(r = \frac{12}{36} h\)

So, let's substitute this in r:

\(V = \frac{1}{3} \pi \frac{12}{36} h^{2} h\)

Square:

\(V = \frac{1}{3} \pi \frac{144}{3888} (h^{2} ) h\)

Simplify:

\(V = \frac{144}{3888} \pi h^{2}\)

Now, let's take the derivative of both sides with respect to t:

\(\frac{d}{dt} (V) = \frac{d}{dt} [\frac{144}{3888} ] \pi h^{3}\)

Simplify:

\(\frac{dV}{dt} =\frac{144}{3888} \pi (3h^{2} ) \frac{dh}{dt}\)

We want to find dh/dt when the water is 13 feet deep. So, let's substitute 13 for h. Also, let's substitute 10 for dV/dt. This yields:

\(10 = \frac{144}{3888} \pi (3(13^{2} ) \frac{dh}{dt}\)

\(10 = \frac{144}{3888} \pi (507) \frac{dh}{dt}\)

\(10 = \frac{73008}{3888} \pi \frac{dh}{dt}\)

\(38880 = 73008 \pi \frac{dh}{dt}\)

\(\frac{dh}{dt} = \frac{38880}{73008} \pi\)

\(\frac{dh}{dt} = \frac{38880}{73008} X\frac{22}{7}\)

≈ 1.6737109 feet / min

The water is rising at a rate of approximately 1.68 feet per minute.

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Answer: A, Purpose

Step-by-step explanation: Nothing else makes sence In the sentence, and  certain purpose makes you have a goal, which matches up with a speech being effective.

Answer:  The correct answer is a. purpose

Step-by-step explanation:

the solution to the equation is x=12

The distributive property is defined as

Let us now solve the equation.

5(x - 6) = 2(x + 3)

Using the distributive property, simplify the parentheses on both sides.

5(x - 6) = 2(x + 3)

5x - 30 = 2x + 6

The parenthesis was simplified. However, we have one x term on one side and another x term on the other.

5x - 30 - 2x = 2x + 6 - 2x

3x - 30 = 6

We only have an x term on one side of the equation now.

To begin, reverse the subtraction by adding 30 to either side of the equation.

To keep an equation balanced or true, we do the identical operation on both sides.

3x - 30 + 30 = 6 + 30

3x = 36

We're getting close to an answer. Only one more step.

The x term is multiplied by 3.

We can undo the multiplication operation by dividing both sides by 3.

3x / 3 = 36 / 3

x = 12

So, x is equal to 12.

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Alex can find the product by shading 5 sections in each row. In this way, each shaded row represents 5/6.

Now, to find the product, Alex has to add the shaded values, that is,

Answer: False

Step-by-step explanation: It literally says false.

The statement "The population will be normally distributed if the sample size is 30 or more" is false.

A normal distribution is a probability distribution that is bell-shaped and symmetrical around the mean. When we measure a characteristic of a large population, such as the height of adult men in the United States, the distribution of those measurements follows a normal distribution. The normal distribution is used to model a wide range of phenomena in fields like statistics, finance, and physics.

Sample size is the number of observations in a sample. The larger the sample size, the more reliable the results, which is why researchers typically aim for large sample sizes.

Therefore, it is false to say that if the sample size is 30 or more, the population will be normally distributed.

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The odds of randomly generating an even number with the first two odd numbers and unique numbers =  \(\frac{10}{16}= \frac{5}{8}\).

Sum will be even if either both numbers are odd or both numbers are even .

So, Total no. of way to select numbers for making sum even

= 5C2 ​+ 4C2 ​= 10 + 6 = 16

And, Total numbers of ways to select two odd numbers such that sum is even = 5C2 ​= 10

Probability that both numbers are odd = \(\frac{10}{16}= \frac{5}{8}\)

Hence the answer is the odds of randomly generating an even number with the first two odd numbers and unique numbers =  \(\frac{10}{16}= \frac{5}{8}\).

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Answer:

This is because, a relation could be any set of ordered pairs where as a function is a set of ordered pairs where there is only ONE value of y for every value of x.

Another way of explaining it is:

In fact, every function is a relation. However, not every relation is a function. In a function, there cannot be two lists that disagree on only the last element. This would be tantamount to the function having two values for one combination of arguments. By contrast, in a relation, there can be any number of lists that agree on all but the last element.

The statement "All functions are relations but some relations are not functions" highlights the relationship between functions and relations and the distinction between the two concepts.

In mathematics, a relation is a set of ordered pairs that establishes a connection between elements of two sets. It can be thought of as a collection of inputs and outputs.

On the other hand, a function is a specific type of relation in which each input value (x-coordinate) is associated with exactly one output value (y-coordinate).

So, when we say "all functions are relations," we are acknowledging that functions are a subset of relations. This is because functions satisfy the property of assigning a unique output for each input, making them a special kind of relation.

However, the statement also recognizes that there are relations that do not meet the criteria of a function. This occurs when an input value is associated with multiple output values or when an input value has no corresponding output value.

In other words, some relations may have more than one output value for a given input, or they may lack a well-defined output for certain inputs. Such relations are not considered functions.

Therefore, while all functions can be classified as relations, not all relations can be classified as functions due to the specific requirement of having a unique output for each input.

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Translate this to english

Por favor ayúdame, no puedo tener constante y> -2x + 1

y≤ x + 3 caca es 6 pies arriba en la jaula de mi perro

Answer: 0.7

Step-by-step explanation: hope it helps

Answer:

0.7

Step-by-step explanation:

0.7

The number of boys in the class can be found by finding the solution of equation as 6.

A linear equation for the given case can be written by assuming any variable as the unknown quantity. Then, as per the given data the required operations are done and it is equated to some value.

The ratio of boys and girls are given as 3 : 1.

As per the question the number of girls are 18.

Suppose, the number of boys be x.

Then, the following equation can be written as,

3/1 = 18/x

⇒ x = 18/3

⇒ x = 6

Hence, the number of boys in the class is given as 6.

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Answer:

34 boxes

Step-by-step explanation:

Please see attached picture for full solution.

Just after the funnel is clogged height of the liquid remains constant at 25 cm

The given problem can be solved by differentiating and integrating the cone formulas and then applying the concept of related rates.

We will also need to use the concept of similar triangles.Let's solve the given problem -1. At 15 cm deep, the radius and height of the cone can be found as follows

We have a cone with radius r and height h.

The radius of the cone is equal to its height.

So,r = h

Volume of the cone,

V = (1/3)πr²h

⇒ V = (1/3)πh²h

⇒ V = (1/3)πh³

Since the liquid is being poured into the cone at a steady rate of 200 cm³/s, the volume of liquid in the cone is given by

V = 200t cm³ where t is the time in seconds.

Volume of the liquid at height 15 cm
Using similar triangles, we can write,

(h - 15)/h = r/R

⇒ r = (h/2)

At height h = 15 cm, radius r = h/2 = 7.5 cm

Volume of the liquid at height 15 cm = (1/3)π(7.5)²(15 - 0) cm³

                                                             = 1767.85 cm³

The volume of the liquid that has flowed out through the hole in the bottom of the cone is given by

V = 20t cm³

Equating the two volumes, we get

200t = 1767.85 + 20t

⇒ t = 8.83925 s

Differentiating the equation V = (1/3)πh³ with respect to time t, we get

dV/dt = πh² dh/dt

At the instant when the height of the liquid is 15 cm, the height of the cone is also 15 cm.

Therefore, at this instant, the rate of change of volume of the liquid with respect to time is

dV/dt = 200 cm³/s

We need to find the rate at which the height of the liquid is changing, i.e., dh/dt when h = 15 cm.

Using the relation,dV/dt = πh² dh/dt

we get, dh/dt = (dV/dt) / (πh²)

⇒ dh/dt = (200) / (π(15)²) cm/s

⇒ dh/dt = 0.02856 cm/s

So, the height of the liquid is increasing at a rate of 0.02856 cm/s when the liquid in the funnel is 15 cm deep.

2. At 25 cm deep, the radius and height of the cone can be found as follows

Using similar triangles, we can write,

(h - 25)/h = r/R

⇒ r = (h/3)

At height h = 25 cm, radius r = h/3 = 8.333 cm

Volume of the liquid at height 25 cm = (1/3)π(8.333)²(25 - 0) cm³

                                                              = 1458.33 cm³

Since the funnel becomes clogged at the bottom when the height of the liquid is 25 cm, the volume of liquid in the cone remains constant at 1458.33 cm³.

Differentiating the equation V = (1/3)πh³ with respect to time t, we get

dV/dt = πh² dh/dt

At the instant when the height of the liquid is 25 cm, the rate of change of volume of the liquid with respect to time is

dV/dt = 0

We need to find the rate at which the height of the liquid is changing just after the funnel gets clogged.i.e., dh/dt when h = 25 cm and dV/dt = 0.

From the equation,dV/dt = πh² dh/dt

we get,

dh/dt = (dV/dt) / (πh²)

⇒ dh/dt = 0 cm/s

Therefore, just after the funnel gets clogged, the height of the liquid remains constant at 25 cm.

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If the actual length of the living room is 8 meters, according to the scale drawing living room in the sketching should be 2 millimeters

Scale drawings are images that depict objects at a scale different from their actual size. Depending on the size of the object being represented and how the sketching will be used, they can depict objects at either a bigger or smaller scale than full size. The scale gives the proportion of a distance at full size to the distance that, at the scale in use, would have the same length.

Scale drawings are used to illustrate objects when sketching them at their true scale would not be practical or beneficial. This could be because depicting the object as its whole would be difficult or difficult to fit on a single sheet of paper.

since 4meter = 1mm

then 8meter = 2mm

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The location of the test score associated with the third quartile is the value that corresponds to the 123rd rank in a class of 163 students.

The third quartile, also known as the upper quartile, is a statistical measure that divides a set of data into quarters. It is the point at which 75% of the data lies below it and 25% of the data lies above it. To find the location of the test score associated with the third quartile, we need to arrange the test scores in ascending order and then find the value that corresponds to the 75th percentile.

Given that there are 163 students in the class, we can determine the rank of the score associated with the third quartile as follows:

Rank of the third quartile = 0.75 x 163 = 122.25

Since we cannot have a fractional rank, we need to round up to the next whole number. Therefore, the rank of the score associated with the third quartile is 123.

Next, we need to find the value of the test score that corresponds to the 123rd rank. We can do this by using a percentile rank calculator or by manually counting the test scores from the lowest to the highest until we reach the 123rd score.

In summary, the location of the test score associated with the third quartile in a class of 163 students is the value that corresponds to the 123rd rank.

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The iterated integral to find the area is:\[A = \int_{0}^{3} \int_{x^2 - 4x}^{10x - x^2} dy \, dx\]

To find the area of the region bounded by the graphs of the equations \(y = 10x - x^2\) and \(y = x^2 - 4x\), we can set up an iterated integral.

First, let's find the x-values at which the two curves intersect. Setting the equations equal to each other, we have:

\[10x - x^2 = x^2 - 4x\]

Simplifying:

\[2x^2 - 6x = 0\]

Factoring out an x:

\[x(2x - 6) = 0\]

So we have two solutions: x = 0 and x = 3. These are the x-values where the curves intersect.

To set up the iterated integral, we need to determine the limits of integration for both x and y.

For x, the limits of integration will be from x = 0 to x = 3, as these are the x-values where the curves intersect.

For y, the limits will be determined by the equations of the curves. The lower curve is \(y = x^2 - 4x\) and the upper curve is \(y = 10x - x^2\). Therefore, the limits for y will be from y = \(x^2 - 4x\) to y = \(10x - x^2\).

The iterated integral to find the area is:

\[A = \int_{0}^{3} \int_{x^2 - 4x}^{10x - x^2} dy \, dx\]

Evaluating this iterated integral will give us the area of the region bounded by the two curves.

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Answer:

15x - 6

Step-by-step explanation:

(12x -2) -1 (3x+4)

(12x - 2 + 3x-4

12x-2+3x -4

12x -6 + 3x

15x - 6

We will try to leave the letter \(r\) alone by modifying the given equation.

This equation is defined with the following expressions.

Ans: \(r=\frac{3t}{t-1}\)

The most appropriate choice for percentage and fraction will be given by-

The process of conversion of fraction to percentage has been explained.

What is fraction and percentage?

Suppose there is a whole collection of objects and some parts of the objects are taken from the whole collection. Fraction represents those parts which are taken. In other words, part of a whole is called fraction. The upper part of the fraction is the numerator and the lower part of the fraction is the denominator.

If a number is taken as fraction of 100 then the fraction is known as percentage.

To convert a fraction into percentage, we have to simply multiply the fraction with 100.

For example.

Suppose we have a fraction \(\frac{1}{4}\), we need to convert  \(\frac{1}{4}\) to percentage.

So on multiplying \(\frac{1}{4}\) by 100, we obtain = \(\frac{1}{4} \times 100\) = 25%

So, \(\frac{1}{4}\) is equivalent to 25%

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The angles of the triangle is :

A = 0.506 , B = 3.692 and C  = 1.850

We have the following information is:

A triangle is defined by the three points A=(3,10), B=(6,9), and C=(5,2).

We have to find the:

Determine all angles theta, theta, and thetaθA, θB, and θC in the triangle.

Now, According to the question:

The first thing we need to do, is find the length of the sides a , b and c. We can do this by using the Distance Formula.

The Distance Formula states, where d is the distance, that:

\(d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}\)

So,

\(a=\sqrt{(6-5)^2+(9-2)^2}\)\(=\sqrt{50}\)

\(b=\sqrt{(3-5)^2+(10-2)^2} =\sqrt{66}\)

\(c=\sqrt{(6-3)^2+(9-10)^2}=\sqrt{10}\)

We now know all 3 sides, but since we don't know any angles, we will have to use the Cosine Rule.

The Cosine Rule states that:

\(a^2=b^2+c^2-2bc.cos(A)\)

Plug all the values:

\((\sqrt{50} )^2=(\sqrt{66} )^2+(\sqrt{10} )^2-2(\sqrt{66} )(\sqrt{10} ).cosA\)

50 = 66 + 10 -2\(\sqrt{66}.\sqrt{10} cosA\)

cos (A) = 50-66-10/ -2\(\sqrt{66}.\sqrt{10}\)

cos (A) = 13/25.69

A = \(cos ^ -^1 \, (cos(A))=cos^-^1\)(13/25.69) = 0.506

We rearrange the formula for angle B.

\(b^2=a^2+c^2-2bc.cos(A)\)

Angle B:

\((\sqrt{66} )^2=(\sqrt{50} )^2+(\sqrt{10} )^2-2(\sqrt{66} )(\sqrt{10} ).cosA\)

66 = 50 + 10 -2\(\sqrt{66}.\sqrt{10} cosA\)

cos (A) = 66 -50 -10/ -2\(\sqrt{66}.\sqrt{10}\)

cos(A) = 6/ -2\(\sqrt{66}.\sqrt{10}\)

cos(A) = 3.692

A = \(cos ^ -^1 \, (cos(A))=cos^-^1\)3.692

Angle C:

\(\pi -(\frac{\pi }{4} +0.506)\) = 1.850

The angles of the triangle is :

A = 0.506 , B = 3.692 and C  = 1.850

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Answer:

My life

Step-by-step explanation:

\( \sin(45) = \frac{opposite}{hypotenuse \\ } \\ \\ \sin(45) = \frac{8}{x} \\ \frac{ \sqrt{2} }{2} = \frac{8}{x} \\ x \sqrt{2} = 16 \\ \frac{x \sqrt{2} }{ \sqrt{2} } = \frac{16}{ \sqrt{2} } \\ x = \frac{16 \times \sqrt{2} }{ \sqrt{2 \times \sqrt{2} } } \\ x = \frac{16 \sqrt{2} }{2} \\ 8 \sqrt{2} \)

D is the correct answer

PLEASE GIVE BRAINLIEST

Step-by-step explanation:

if the goal is to have both numbers with denominator 6, then the first one is totally simple :

3 1/6 = 3 1/6

what else could it be ... ?

and the second, well, how many thirds are in a whole ?

right, 3, that's why we call them "thirds".

how many 6ths are in a whole ?

right, 6.

do you notice something ? it is easy to get from 3 to 6 by multiplying 3 by 2.

in order to keep the overall value of a fraction unchanged, when we multiply one part of the fraction by something we need to multiply the other part by the same number too.

so,

5 2/3 = 5 2/2×2/3 = 5 4/6

in short : yes, every third contains 2 6ths.

therefore, 2/3 = 4/6

Answer:

Linear

Step-by-step explanation:

Performing a Chi-Square test is a statistical tool used to determine if there is a significant difference between observed and expected data. The test helps to analyze categorical data by comparing observed frequencies to the expected frequencies. The p-value in a Chi-Square test refers to the probability of obtaining the observed results by chance alone.

If a p-value lower than 0.01 is obtained in a Chi-Square test, it means that the results are statistically significant. In other words, there is strong evidence to reject the null hypothesis, which states that there is no significant difference between the observed and expected data. This means that the observed data is not due to chance alone, but rather to some other factor or factors.

The mean, or average, is not directly related to the Chi-Square test or the p-value. The Chi-Square test is specifically used to determine the significance of the observed data. However, the mean can be used as a measure of central tendency for continuous data, but it is not applicable to categorical data.

In conclusion, obtaining a p-value lower than 0.01 in a Chi-Square test means that there is strong evidence to reject the null hypothesis, and that the observed data is statistically significant.

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