HELP ME ASAP!!! I WILL GIVE MANY POINTS

Answer:

0

Step-by-step explanation:

54m² - 54m² = 0

Water is 0

The answer is B.

Since ΔQRS ~ ΔXYZ, the value of tan(Q) is :

Answer:

Option D

Step-by-step explanation:

Using the surd law :

[tex]\sqrt{ab} = \sqrt{a}\sqrt{b}[/tex]

We can find the largest square number that goes into 75 :

Let's write the multiples of 75 :

1 , 75

3 , 25

5 , 15

The only square number is 25

So using the law mentioned above we split √75 into :

√25√3

The square root of 25 is 5

Now we have our final answer of 5√3

Hope this helped and have a good day

The simplified form of expression √75 is 5√3.

Option D is the correct answer.

We have,

To simplify √75, we can factor it into its prime factors and then take the square root:

√75 = √(3 * 5 * 5)

= √(3 x 5²)

Take out the perfect square factor from under the square root:

= √3 x √5²

= √3 x 5

= 5√3

Thus,

The simplified form of expression √75 is 5√3 which is option D.

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No,the population mean is not equal to 9.2 and the value in t statistic is -1.02.

Given sample size of 49,sample mean of 8.5,standard deviation of 1.5, significance level of 0.01.

We are required to find out whether the population mean is equal to 9.2 and the value of t in test statistic.

We have to first make the hypothesis for this.

[tex]H_{0}[/tex]:μ≠9.2

[tex]H_{1}[/tex]:μ=9.2

We have to use z statistic because the sample size is more than 30.

Z=(X-μ)/σ

We have been given sample mean but we require population mean in the formula so we will use sample mean.

Z=(8.5-9.2)/1.5

=-0.7/1.5

=-0.467

P value of -0.467 is 0.67975.

P value is greater than 0.01 so we will accept the hypothesis means population mean is not equal to 9.2.

t=(X-μ)/s/[tex]\sqrt{n}[/tex]

=(8.5-9.2)/1.5/[tex]\sqrt{49}[/tex]

=-0.7/0.21

=-1.02

Hence it is concluded that no,the population mean is not equal to 9.2 and the value in t statistic is -1.02.

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Answer:

see explanation

Step-by-step explanation:

total parts = 3 + 18 + 6 + 3 = 30

3 grams of silver = [tex]\frac{3}{30}[/tex] = [tex]\frac{1}{10}[/tex]

18 grams of copper = [tex]\frac{18}{30}[/tex] = [tex]\frac{3}{5}[/tex]

6 grams of aluminium = [tex]\frac{6}{30}[/tex] = [tex]\frac{1}{5}[/tex]

3 grams of zinc = [tex]\frac{3}{30}[/tex] = [tex]\frac{1}{10}[/tex]

Answer:

[tex]\displaystyle{\sin \theta = \dfrac{2ab}{a^2+b^2}}\\\\\displaystyle{\cos \theta = \dfrac{a^2-b^2}{a^2+b^2}}\\\\\displaystyle{\csc \theta = \dfrac{a^2+b^2}{2ab}}\\\\\displaystyle{\sec \theta = \dfrac{a^2+b^2}{a^2-b^2}}\\\\\displaystyle{\cot \theta = \dfrac{a^2-b^2}{2ab}}[/tex]

Step-by-step explanation:

We are given that:

[tex]\displaystyle{\tan \theta = \dfrac{2ab}{a^2-b^2}}[/tex]

To find other trigonometric ratios, first, we have to know that there are total 6 trigonometric ratios:

[tex]\displaystyle{\sin \theta = \sf \dfrac{opposite}{hypotenuse} = \dfrac{y}{r}}\\\\\displaystyle{\cos \theta = \sf \dfrac{adjacent}{hypotenuse} = \dfrac{x}{r}}\\\\\displaystyle{\tan \theta = \sf \dfrac{opposite}{adjacent} = \dfrac{y}{x}}\\\\\displaystyle{\csc \theta = \sf \dfrac{hypotenuse}{opposite} = \dfrac{r}{y}}\\\\\displaystyle{\sec \theta = \sf \dfrac{hypotenuse}{adjacent} = \dfrac{r}{x}}\\\\\displaystyle{\cot \theta = \sf \dfrac{adjacent}{opposite} = \dfrac{x}{y}}[/tex]

Since we are given tangent relation, we know that [tex]\displaystyle{y = 2ab}[/tex] and [tex]\displaystyle{x = a^2-b^2}[/tex], all we have to do is to find hypotenuse or radius (r) which you can find by applying Pythagoras Theorem.

[tex]\displaystyle{r=\sqrt{x^2+y^2}}[/tex]

Therefore:

[tex]\displaystyle{r=\sqrt{(a^2-b^2)^2+(2ab)^2}}\\\\\displaystyle{r=\sqrt{a^4-2a^2b^2+b^4+4a^2b^2}}\\\\\displaystyle{r=\sqrt{a^4+2a^2b^2+b^4}}\\\\\displaystyle{r=\sqrt{(a^2+b^2)^2}}\\\\\displaystyle{r=a^2+b^2}[/tex]

Now we can find other trigonometric ratios by simply substituting the given information below:

Hence:

[tex]\displaystyle{\sin \theta = \dfrac{y}{r} = \dfrac{2ab}{a^2+b^2}}\\\\\displaystyle{\cos \theta = \dfrac{x}{r} = \dfrac{a^2-b^2}{a^2+b^2}}\\\\\displaystyle{\csc \theta = \dfrac{r}{y} = \dfrac{a^2+b^2}{2ab}}\\\\\displaystyle{\sec \theta = \dfrac{r}{x} = \dfrac{a^2+b^2}{a^2-b^2}}\\\\\displaystyle{\cot \theta = \dfrac{x}{y} = \dfrac{a^2-b^2}{2ab}}[/tex]

will be other trigonometric ratios.

The AD's length of 3m will enable the angle at B to be divided in half.

Angle Bisector Theorem: What is it?

The angle bisector of a triangle divides the opposing side into two portions that are proportional to the other two sides, according to the angle bisector theorem, in simpler words the angle bisector theorem is concerned with the relative lengths of the two segments that a triangle's side is divided into by a line that bisects the opposite angle. It equates their relative lengths to the relative lengths of the other two sides of the triangle.

The triangle has sides of 5, 7, and (x+4) m.

Angle B's angle bisector will only be the BD if

x/4 = 5/7

x = 5 *4 / 7

x = 20/7 = 2.85 ≈ 3m

Thus if AD has length of 3m then it will enable the angle at B to be divided in half.

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Answer:

AD = 3m

Hope this helps!

Step-by-step explanation:

Best guess for the function is

[tex]\displaystyle f(x) = \sum_{n=1}^\infty \frac{x^n}{n^2}[/tex]

By the ratio test, the series converges for

[tex]\displaystyle \lim_{n\to\infty} \left|\frac{x^{n+1}}{(n+1)^2} \cdot \frac{n^2}{x^n}\right| = |x| \lim_{n\to\infty} \frac{n^2}{(n+1)^2} = |x| < 1[/tex]

When [tex]x=1[/tex], [tex]f(x)[/tex] is a convergent [tex]p[/tex]-series.

When [tex]x=-1[/tex], [tex]f(x)[/tex] is a convergent alternating series.

So, the interval of convergence for [tex]f(x)[/tex] is the closed interval [tex]\boxed{-1 \le x \le 1}[/tex].

The derivative of [tex]f[/tex] is the series

[tex]\displaystyle f'(x) = \sum_{n=1}^\infty \frac{nx^{n-1}}{n^2} = \frac1x \sum_{n=1}^\infty \frac{x^n}n[/tex]

which also converges for [tex]|x|<1[/tex] by the ratio test:

[tex]\displaystyle \lim_{n\to\infty} \left|\frac{x^{n+1}}{n+1} \cdot \frac n{x^n}\right| = |x| \lim_{n\to\infty} \frac{n}{n+1} = |x| < 1[/tex]

When [tex]x=1[/tex], [tex]f'(x)[/tex] becomes the divergent harmonic series.

When [tex]x=-1[/tex], [tex]f'(x)[/tex] is a convergent alternating series.

The interval of convergence for [tex]f'(x)[/tex] is then the closed-open interval [tex]\boxed{-1 \le x < 1}[/tex].

Differentiating [tex]f[/tex] once more gives the series

[tex]\displaystyle f''(x) = \sum_{n=1}^\infty \frac{n(n-1)x^{n-2}}{n^2} = \frac1{x^2} \sum_{n=1}^\infty \frac{(n-1)x^n}{n} = \frac1{x^2} \left(\sum_{n=1}^\infty x^n - \sum_{n=1}^\infty \frac{x^n}n\right)[/tex]

The first series is geometric and converges for [tex]|x|<1[/tex], endpoints not included.

The second series is [tex]f'(x)[/tex], which we know converges for [tex]-1\le x<1[/tex].

Putting these intervals together, we see that [tex]f''(x)[/tex] converges only on the open interval [tex]\boxed{-1 < x < 1}[/tex].

Answer:

Separating the 3x and -5 apart from the (x+1)

Step-by-step explanation:

It would turn out to be (3x-5)(x+1) !

Factor out x+1 from the expression

(x+1) x (3x-5)

The dipping sauce which has a stronger mustard flavor between burger barn and be sandwich town is burger barn

Burger bun:

Mustard : honey

= 2 : 4

= 2/4

= 1/2

= 0.5

Sandwich:

Mustard : honey

= 2 : 8

= 2/8

= 1/4

= 0.25

Therefore, burger barn has a more stronger mustard flavor of dipping sauce between burger barn and be sandwich town.

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Answer:

Step-by-step explanation:

The two dipping sauce have same taste.

Answer:

y = 1/5x -3

Step-by-step explanation:

Use y = mx +b as your model.  We plug in our slope for me and our y-intercept for b.

Answer:

[tex]\displaystyle x=\frac{5}{4},\;\;1\frac{1}{4}, \;\; or \;\; 1.25[/tex]

Step-by-step explanation:

    To solve for x, we need to isolate the x variable.

    Given:

[tex]\displaystyle x+\frac{1}{2} =\frac{7}{4}[/tex]

    Subtract [tex]\frac{1}{2}[/tex] from both sides of the equation:

[tex]\displaystyle (x+\frac{1}{2})-\frac{1}{2} =(\frac{7}{4})-\frac{1}{2}[/tex]

[tex]\displaystyle x=\frac{7}{4}-\frac{1}{2}[/tex]

    Now, we will create common denominators to simplify.

[tex]\displaystyle x=\frac{7}{4}-\frac{2}{4}[/tex]

[tex]\displaystyle x=\frac{5}{4}[/tex]

The number which belongs to the set of rational numbers and the set of integers is -115 which is third option,-15 which is fourth option.

Given four options:

We are required to find the number which is included in the set of rational numbers and the set of integers.

Rational numbers are those numbers which can be written in the form of p/q in which q cannot be equal to zero because if q becomes zero then the fraction becomes infinity.

-5.5 is not a rational number,

-0.5 is also not a rational number.

-115 is a rational number and also an integer.

-15 is a rationalnumber and also an integer.

Hence the number which belongs to the set of rational numbers and the set of integers is -115 which is third option,-15 which is fourth option.

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Answer:

x = 5, y =2

Step-by-step explanation:

I guess the question is saying x+y = 7 and 3x-2y = 11?

then there are multiple ways but

I will multiply the first one by 2 so 2x+2y = 14

you add the equations to get 5x = 25 so x = 5 plug x into the first equation you get y = 2

if that isn't what the question means just comment and I'll change it

If T is the midpoint of PQ and PT = 3x+3, TQ = 7x-9, then PT = 12 units.

Determining the Value of PT

It is given that,

T is the midpoint of PQ ........ (1)

PT=3x+3 ......... (2)

TQ=7x-9 .......... (3)

From (1), the distance from P to T and the distance from T to Q will be equal.

⇒ PT = TQ [Since, a midpoint divides a line into two equal segments]

Hence, equating the equations of PT and TQ given in (2) and (3) respectively, equal, we get the following,

3x + 3 = 7x - 9

or 7x - 9 = 3x + 3

or 7x - 3x = 9 + 3

or 4x = 12

or x = 12/4

⇒ x = 3

Substitute this obtained value of x in equation (2)

PT = 3(3) + 3

PT = 9 + 3

PT = 12 units

Thus, if T is the midpoint of PQ, then the measure of PT and TQ is equal to 12 units.

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The answer is x < 1.

Bring the constant to the other side.

Divide by 4 on both sides.

[tex]\Large\texttt{Answer}[/tex]

[tex]\overline{\qquad\qquad\qquad\qquad\qquad\qquad\qquad\quad\space\space\qquad\qquad\qquad}}[/tex]

[tex]\Large\texttt{Process}[/tex]

[tex]\rm{4x-6 < -2}[/tex]

Do you remember that we need to get x by itself to find its value?

We should do this:

⇨ Add 6 to both sides

[tex]\rm{4x-6+6 < -2+6}[/tex]

On the left hand side (lhs), the 6s add up to zero; on the right hand side (rhs), the -2 and 6 result in 4. Hence

[tex]\rm{4x < 4}[/tex]

Now divide both sides by 4

[tex]\rm{\cfrac{4x}{4} < \cfrac{4}{4}}[/tex]

Simplifying fractions gives us

[tex]\rm{x < 1}[/tex]

* what this means is: numbers less than 1 will make the statement true

[tex]\Large\texttt{Verification}[/tex]

Substitute 1 into the original inequality [tex]\boxed{4x-6 < -2}[/tex]

[tex]\rm{4(1)-6 < -2}[/tex]

[tex]\rm{4-6 < -2}[/tex]

Do the arithmetic

[tex]\rm{-2 < -2}[/tex]

Hope that helped

Answer:

i'll give you answer.Dont worry. Since i came back from school

The mass of the sample should be reported as 3.54 g

The amount of matter in an object is expressed in terms of mass.

The most frequent way to determine mass is to weigh something.

The units of mass are grams, kilograms, tonnes (in metric units), or ounces and pounds (US units).

According to the question,

A label on an empty sample container reads 10.000 g

A sample of a compound is added and the mass of the sample container is found to be 13.54 g.

The mass of the sample should thus be reported as,

= 13.54 - 10.000

= 3.54 g

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We can comment that the maximum value of the given data is 0.2, maximum value is 42. The interquartile range is 15 to 37, first and third quartile values are 15 and 37 respectively. It can also be inferred from the box plot that there are no outliers. The median of the given data, as shown in the box plot, is 28.

What is a box plot?

A box and whisker plot, often known as a box plot, shows a data set's five-number summary. A box is drawn from the first quartile to the third quartile in a box plot. At the median, a vertical line passes through the box. The five-number summary of a box plot includes the following:

It also tells if there are any outliers in the data.

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The inequality in the box has to be written as

x² + 2x - 80 ≤ - 65

We have

(10 + x)1 * (16-2x) ≥ 130

Next we would have to open the bracket

160 + 16x - 20x - 2x² ≥ 130

Then we would have to arrange the equation

- 2x² - 4x + 160 ≥ 130

Divide the equation by two

- x² - 2x + 80 ≥ 65

This is arranged as

x² + 2x - 80 ≤ - 65

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Answer:

21

Step-by-step explanation:

In the diagram, the three tangents (segment touching a circle at one point) have equal length.

6y - 3 = 29 - 2y

8y = 32

y = 4

Since the lengths of segments AM and AN are equivalent, we can substitute the value of y into the expression, 6y - 3, to find AN.

6y - 3 = 6*4 - 3 = 24 - 3 = 21

Answer:

2

Step-by-step explanation:

given x=1 and y=7

now, given expression ,

x+y/4

by putting the values of the x and y ,we get

x+y/4

= 1+7/4

= 8/4

= 2 (Ans.)

Answer:

[tex]\dfrac{3}{2} \ln |x-4| - \dfrac{1}{2} \ln |x+2| + \text{C}[/tex]

Step-by-step explanation:

Fundamental Theorem of Calculus

[tex]\displaystyle \int \text{f}(x)\:\text{d}x=\text{F}(x)+\text{C} \iff \text{f}(x)=\dfrac{\text{d}}{\text{d}x}(\text{F}(x))[/tex]

If differentiating takes you from one function to another, then integrating the second function will take you back to the first with a constant of integration.

Given indefinite integral:

[tex]\displaystyle \int \dfrac{x+5}{(x-4)(x+2)}\:\:\text{d}x[/tex]

Take partial fractions of the given fraction by writing out the fraction as an identity:

[tex]\begin{aligned}\dfrac{x+5}{(x-4)(x+2)} & \equiv \dfrac{A}{x-4}+\dfrac{B}{x+2}\\\\\implies \dfrac{x+5}{(x-4)(x+2)} & \equiv \dfrac{A(x+2)}{(x-4)(x+2)}+\dfrac{B(x-4)}{(x-4)(x+2)}\\\\\implies x+5 & \equiv A(x+2)+B(x-4)\end{aligned}[/tex]

Calculate the values of A and B using substitution:

[tex]\textsf{when }x=4 \implies 9 = A(6)+B(0) \implies A=\dfrac{3}{2}[/tex]

[tex]\textsf{when }x=-2 \implies 3 = A(0)+B(-6) \implies B=-\dfrac{1}{2}[/tex]

Substitute the found values of A and B:

[tex]\displaystyle \int \dfrac{x+5}{(x-4)(x+2)}\:\:\text{d}x = \int \dfrac{3}{2(x-4)}-\dfrac{1}{2(x+2)}\:\:\text{d}x[/tex]

[tex]\boxed{\begin{minipage}{5 cm}\underline{Terms multiplied by constants}\\\\$\displaystyle \int ax^n\:\text{d}x=a \int x^n \:\text{d}x$\end{minipage}}[/tex]

If the terms are multiplied by constants, take them outside the integral:

[tex]\implies \displaystyle \dfrac{3}{2} \int \dfrac{1}{x-4}- \dfrac{1}{2} \int \dfrac{1}{x+2}\:\:\text{d}x[/tex]

[tex]\boxed{\begin{minipage}{5 cm}\underline{Integrating}\\\\$\displaystyle \int \dfrac{f'(x)}{f(x)}\:\text{d}x=\ln |f(x)| \:\:(+\text{C})$\end{minipage}}[/tex]

[tex]\implies \dfrac{3}{2} \ln |x-4| - \dfrac{1}{2} \ln |x+2| + \text{C}[/tex]

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For an alternative approach, expand and complete the square in the denominator to write

[tex](x-4)(x+2) = x^2 - 2x - 8 = (x - 1)^2 - 9[/tex]

In the integral, substitute [tex]x - 1 = 3 \sin(u)[/tex] and [tex]dx=3\cos(u)\,du[/tex] to transform it to

[tex]\displaystyle \int \frac{x+5}{(x - 1)^2 - 9} \, dx = \int \frac{3\sin(u) + 6}{9 \sin^2(u) - 9} 3\cos(u) \, du \\\\ ~~~~~~~~~~~~ = - \int \frac{\sin(u) + 2}{\cos(u)} \, du \\\\ ~~~~~~~~~~~~ = - \int (\tan(u) + 2 \sec(u)) \, du[/tex]

Using the known antiderivatives

[tex]\displaystyle \int \tan(x) \, dx = - \ln|\cos(x)| + C[/tex]

[tex]\displaystyle \int \sec(x) \, dx = \ln|\sec(x) + \tan(x)| + C[/tex]

we get

[tex]\displaystyle \int \frac{x+5}{(x - 1)^2 - 9} \, dx = \ln|\cos(u)| - 2 \ln|\sec(u) + \tan(u)| + C \\\\ ~~~~~~~~~~~~ = - \ln\left|\frac{(\sec(u) + \tan(u))^2}{\cos(u)}\right|[/tex]

Now, for [tex]n\in\Bbb Z[/tex],

[tex]\sin(u) = \dfrac{x-1}3 \implies u = \sin^{-1}\left(\dfrac{x-1}3\right) + 2n\pi[/tex]

so that

[tex]\cos(u) = \sqrt{1 - \dfrac{(x-1)^2}9} = \dfrac{\sqrt{-(x-4)(x+2)}}3 \implies \sec(u) = \dfrac3{\sqrt{-(x-4)(x+2)}}[/tex]

and

[tex]\tan(u) = \dfrac{\sin(u)}{\cos(u)} = -\dfrac{x-1}{\sqrt{-(x-4)(x+2)}}[/tex]

Then the antiderivative we found is equivalent to

[tex]\displaystyle - \int \frac{x+5}{(x - 1)^2 - 9} \, dx = - \ln\left|-\frac{3(x+2)}{(x-4) \sqrt{-(x-4)(x+2)}}\right| + C[/tex]

and can be expanded as

[tex]\displaystyle - \int \frac{x+5}{(x - 1)^2 - 9} \, dx = -\ln\left| \frac{3(x+2)^{1/2}}{(x-4)^{3/2}}\right| + C \\\\ ~~~~~~~~~~~~ = - \ln\left|(x+2)^{1/2}\right| + \ln\left|(x-4)^{3/2}\right| + C \\\\ ~~~~~~~~~~~~ = \boxed{\frac32 \ln|x-4| - \frac12 \ln|x+2| + C}[/tex]

We need at least 7 terms of the Maclaurin series for ln(1 + x)  to estimate ln 1.4 to within 0.0001

For given question,

We have been given a function f(x) = ln(1 + x)

We need to find  the estimate of In(1.4) within 0.001 by applying the function of the Maclaurin series for f(x) = In (1 + x)

The expansion of ln(1 + x) about zero is:

[tex]ln(1+x)=x-\frac{x^2}{2} + \frac{x^3}{3} -\frac{x^4}{4} +\frac{x^5}{5} -\frac{x^6}{6} +.~.~.[/tex]

where -1 ≤ x ≤ 1

To estimate the value of In(1.4), let's replace x with 0.4

[tex]\Rightarrow ln(1+0.4)=0.4-\frac{0.4^2}{2} + \frac{0.4^3}{3} -\frac{0.4^4}{4} +\frac{0.4^5}{5} -\frac{0.4^6}{6} +.~.~.[/tex]

From the above calculations, we will realize that the value of  [tex]\frac{0.4^5}{5}=0.002048[/tex] and [tex]\frac{0.4^6}{6}=0.000683[/tex]  which are approximately equal to 0.001

Hence, the estimate of In(1.4) to the term [tex]\frac{0.4^6}{6}[/tex]  is enough to justify our claim.

Therefore,  we need at least 7 terms of the Maclaurin series for function ln(1 + x)  to estimate ln 1.4 to within 0.0001

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Answer:

5,250 km

Step-by-step explanation:

[tex]\frac{hours}{miles}[/tex] = [tex]\frac{hours}{miles}[/tex]

[tex]\frac{3}{3500}[/tex] = [tex]\frac{4.5}{x}[/tex]  Cross Multiply

3x = (4.5)(3500)

3x = 15750 Divide both sides of the equation by 3

x = 5,250

The x value or observed value corresponding to  z-score, z = 1.50 is 130.

According to the question.

For a population with µ = 100 and σ = 20.

Since,  we know that

The z-score is a statistical evaluation of a value's correlation to the mean of a collection of values, expressed in terms of standard deviation.

And it is given by

z = (x - μ) / σ

Where,

x is the observed value.

μ is the mean.

and, σ is the standard deviation.

Therefore, the x value or observed value corresponding to z = 1.50 is given by

[tex]1.50 = \frac{x -100}{20}[/tex]

⇒ 1.50 × 20 = x - 100

⇒ 30 = x - 100

⇒ x = 30 + 100

⇒ x = 130

Hence, the x value or observed value corresponding to  z-score, z = 1.50 is 130.

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Answer:

mass of a clay pot = 4.95 kg

Kindly award branliest

Step-by-step explanation:

Let the mass of a clay pot be x

Let the mass of a metal pot be y

Thus; 2x + 2y = 13.2

And ;

x = 3 times y

x = 3y

2x + 2y = 13.2

2(3y) + 2y = 13.2

6y + 2y = 13.2

8y = 13.2

y = 13.2/8 = 1.65

x = 3y = 3(1.65) = 4.95

mass of a clay pot = 4.95 kg

The value of the function g(-3) from the given piecewise function is 1

A piecewise-defined function is a function defined by multiple sub-functions, where each sub-function applies to a different interval in the domain.

From the given piecewise function, we are to find the value of the function when x is -3 that is g(-3)

In order to determine the equivalent function, we need to determine the function where x = -3

The equivalent function is g(x) = x+ 4

Substitute x = -3 into the resulting function

g(x) = x + 4

g(-3) = -3 + 4

g(-3) = 1

Hence the value of the function g(-3) from the given piecewise function is 1

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The z-score corresponding to a sample mean of m = 69 is -0.167

In this problem, we have been given :

population mean (μ) = 70, standard deviation (σ) = 12,  sample size (n) = 4, sample mean (m) = 69

We know that, the Z-score measures how many standard deviations the measure is from the mean.

Also, the formula when calculating the z-score of a sample with known population standard deviation is:

[tex]Z=\frac{m-\mu}{\frac{\sigma}{\sqrt{n} } }[/tex]

where z = standard score

μ = population mean

σ = population standard deviation

m = the sample mean

and [tex]\frac{\sigma}{\sqrt{n} }[/tex] is the Standard Error of the Mean for a Population

First we find the Standard Error of the Mean for a Population

σ /√n

= 12 / √4

= 12 / 2

= 6

So, the z-score would be,

⇒ [tex]Z=\frac{m-\mu}{\frac{\sigma}{\sqrt{n} } }[/tex]

⇒ [tex]Z=\frac{69-70}{6 }[/tex]

⇒ Z = -1/6

⇒ Z = -0.167

Therefore, the z-score corresponding to a sample mean of m = 69 is -0.167

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Split up the interval [0, 3] into 3 equally spaced subintervals of length [tex]\Delta x = \frac{3-0}3 = 1[/tex]. So we have the partition

[0, 1] U [1, 2] U [2, 3]

The left endpoint of the [tex]i[/tex]-th subinterval is

[tex]\ell_i = i - 1[/tex]

where [tex]i\in\{1,2,3\}[/tex].

Then the area is given by the definite integral and approximated by the left-hand Riemann sum

[tex]\displaystyle \int_0^3 f(x) \, dx \approx \sum_{i=1}^3 f(\ell_i) \Delta x \\\\ ~~~~~~~~~~ = \sum_{i=1}^3 (i-1)^3 \\\\ ~~~~~~~~~~ = \sum_{i=0}^2 i^3 \\\\ ~~~~~~~~~~ = 0^3 + 1^3 + 2^3 = \boxed{9}[/tex]