Nielson ratings are based on televisions in 5000 households. Nielsen estimates that 12,000 people live in these households. suppose Nielsen reports that American Idol had 75% of the TV audience. interpret this result with a 95% confidence interval, based on a sample of 12000 peoplewhat's the Margin of Error and the Confidence Interval in percentage

The volume of the cylinder is 461.28π cm³.The mass of the cylinder is 11404.84 g

From the question we are to calculate the mass of the cylinder.

First, we will determine the volume of the cylinder

The volume of a cylinder is given by the formula

V = πr²h

Where V is the volume of the cylinder

r is the radius

and h is the height

From the given information,

r = 6.2 cm

h = 12 cm

Thus,

V = π × (6.2)² × 12

V = 461.28π cm³

Now, to determine the mass of the cylinder

From the formula,

Density = Mass / Volume

Then, we can write that

Mass = Density × Volume

From the given information,

The density of iron is 7.87g/cm^3

Thus,

Mass of the cylinder = 461.28π cm³ × 7.87g/cm³

Mass of the cylinder = 11404.84 g

Hence, the mass is 11404.84 g

Here is most likely the complete question:

A cylinder has a radius of 6.2cm and a height of 12cm.You are told that the cylinder is made of iron. The density of iron is 7.87g/cm^3. Calculate the mass of the cylinder.

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Answer:

\(x = 4.8\)

Step-by-step explanation:

\( \frac{3}{4} = \frac{3.6}{x} \)

\(3x = 14.4\)

\(x = 4.8\)

Answer:

Kindly check explanation

Step-by-step explanation: Jullian's claim that the distance she drives to work is exactly 11.7miles is incorrect because, in other to record or get the exact result of a certain calculation such as Jullian's Distance, the value of the distance obtained will not be approximated or rounded. In this scenario, Distance was to the nearest tenth of a mile, thereby altering the true outcome of the calculation.

The word exact means that what is stated is very precise and does not fall below or above in any respect. However, a number whose accuracy is to the nearest tenth of a mile, violates this assertion.

Given z = 3 + i, right away we can find

(a) square

z ² = (3 + i )² = 3² + 6i + i ² = 9 + 6i - 1 = 8 + 6i

(b) modulus

|z| = √(3² + 1²) = √(9 + 1) = √10

(d) polar form

First find the argument:

arg(z) = arctan(1/3)

Then

z = |z| exp(i arg(z))

z = √10 exp(i arctan(1/3))

or

z = √10 (cos(arctan(1/3)) + i sin(arctan(1/3))

(c) square root

Any complex number has 2 square roots. Using the polar form from part (d), we have

√z = √(√10) exp(i arctan(1/3) / 2)

and

√z = √(√10) exp(i (arctan(1/3) + 2π) / 2)

Then in standard rectangular form, we have

\(\sqrt z = \sqrt[4]{10} \left(\cos\left(\dfrac12 \arctan\left(\dfrac13\right)\right) + i \sin\left(\dfrac12 \arctan\left(\dfrac13\right)\right)\right)\)

and

\(\sqrt z = \sqrt[4]{10} \left(\cos\left(\dfrac12 \arctan\left(\dfrac13\right) + \pi\right) + i \sin\left(\dfrac12 \arctan\left(\dfrac13\right) + \pi\right)\right)\)

We can simplify this further. We know that z lies in the first quadrant, so

0 < arg(z) = arctan(1/3) < π/2

which means

0 < 1/2 arctan(1/3) < π/4

Then both cos(1/2 arctan(1/3)) and sin(1/2 arctan(1/3)) are positive. Using the half-angle identity, we then have

\(\cos\left(\dfrac12 \arctan\left(\dfrac13\right)\right) = \sqrt{\dfrac{1+\cos\left(\arctan\left(\dfrac13\right)\right)}2}\)

\(\sin\left(\dfrac12 \arctan\left(\dfrac13\right)\right) = \sqrt{\dfrac{1-\cos\left(\arctan\left(\dfrac13\right)\right)}2}\)

and since cos(x + π) = -cos(x) and sin(x + π) = -sin(x),

\(\cos\left(\dfrac12 \arctan\left(\dfrac13\right)+\pi\right) = -\sqrt{\dfrac{1+\cos\left(\arctan\left(\dfrac13\right)\right)}2}\)

\(\sin\left(\dfrac12 \arctan\left(\dfrac13\right)+\pi\right) = -\sqrt{\dfrac{1-\cos\left(\arctan\left(\dfrac13\right)\right)}2}\)

Now, arctan(1/3) is an angle y such that tan(y) = 1/3. In a right triangle satisfying this relation, we would see that cos(y) = 3/√10 and sin(y) = 1/√10. Then

\(\cos\left(\dfrac12 \arctan\left(\dfrac13\right)\right) = \sqrt{\dfrac{1+\dfrac3{\sqrt{10}}}2} = \sqrt{\dfrac{10+3\sqrt{10}}{20}}\)

\(\sin\left(\dfrac12 \arctan\left(\dfrac13\right)\right) = \sqrt{\dfrac{1-\dfrac3{\sqrt{10}}}2} = \sqrt{\dfrac{10-3\sqrt{10}}{20}}\)

\(\cos\left(\dfrac12 \arctan\left(\dfrac13\right)+\pi\right) = -\sqrt{\dfrac{10-3\sqrt{10}}{20}}\)

\(\sin\left(\dfrac12 \arctan\left(\dfrac13\right)+\pi\right) = -\sqrt{\dfrac{10-3\sqrt{10}}{20}}\)

So the two square roots of z are

\(\boxed{\sqrt z = \sqrt[4]{10} \left(\sqrt{\dfrac{10+3\sqrt{10}}{20}} + i \sqrt{\dfrac{10-3\sqrt{10}}{20}}\right)}\)

and

\(\boxed{\sqrt z = -\sqrt[4]{10} \left(\sqrt{\dfrac{10+3\sqrt{10}}{20}} + i \sqrt{\dfrac{10-3\sqrt{10}}{20}}\right)}\)

Answer:

\(\displaystyle \text{a. }8+6i\\\\\text{b. }\sqrt{10}\\\\\text{c. }\\\sqrt{\sqrt{\frac{5}{2}}+\frac{3}{2}}+i\sqrt{\frac{\sqrt{10}-3}{2}},\\-\sqrt{\sqrt{\frac{5}{2}}+\frac{3}{2}}-i\sqrt{\frac{\sqrt{10}-3}{2}}\\\\\\\text{d. }\\\text{Exact: }z=\sqrt{10}\left(\cos\left(\arctan\left(\frac{1}{3}\right)\right), i\sin\left(\arctan\left(\frac{1}{3}\right)\right)\right),\\\text{Approximated: }z=3.16(\cos(18.4^{\circ}),i\sin(18.4^{\circ}))\)

Step-by-step explanation:

Recall that \(i=\sqrt{-1}\)

Part A:

We are just squaring a binomial, so the FOIL method works great. Also, recall that \((a+b)^2=a^2+2ab+b^2\).

\(z^2=(3+i)^2,\\z^2=3^2+2(3i)+i^2,\\z^2=9+6i-1,\\z^2=\boxed{8+6i}\)

Part B:

The magnitude, or modulus, of some complex number \(a+bi\) is given by \(\sqrt{a^2+b^2}\).

In \(3+i\), assign values:

\(|z|=\sqrt{3^2+1^2},\\|z|=\sqrt{9+1},\\|z|=\sqrt{10}\)

Part C:

In Part A, notice that when we square a complex number in the form \(a+bi\), our answer is still a complex number in the form

We have:

\((c+di)^2=a+bi\)

Expanding, we get:

\(c^2+2cdi+(di)^2=a+bi,\\c^2+2cdi+d^2(-1)=a+bi,\\c^2-d^2+2cdi=a+bi\)

This is still in the exact same form as \(a+bi\) where:

Thus, we have the following system of equations:

\(\begin{cases}c^2-d^2=3,\\2cd=1\end{cases}\)

Divide the second equation by \(2d\) to isolate \(c\):

\(2cd=1,\\\frac{2cd}{2d}=\frac{1}{2d},\\c=\frac{1}{2d}\)

Substitute this into the first equation:

\(\left(\frac{1}{2d}\right)^2-d^2=3,\\\frac{1}{4d^2}-d^2=3,\\1-4d^4=12d^2,\\-4d^4-12d^2+1=0\)

This is a quadratic disguise, let \(u=d^2\) and solve like a normal quadratic.

Solving yields:

\(d=\pm i \sqrt{\frac{3+\sqrt{10}}{2}},\\d=\pm \sqrt{\frac{{\sqrt{10}-3}}{2}}\)

We stipulate \(d\in \mathbb{R}\) and therefore \(d=\pm i \sqrt{\frac{3+\sqrt{10}}{2}}\) is extraneous.

Thus, we have the following cases:

\(\begin{cases}c^2-\left(\sqrt{\frac{\sqrt{10}-3}{2}}\right)^2=3\\c^2-\left(-\sqrt{\frac{\sqrt{10}-3}{2}}\right)^2=3\end{cases}\\\)

Notice that \(\left(\sqrt{\frac{\sqrt{10}-3}{2}}\right)^2=\left(-\sqrt{\frac{\sqrt{10}-3}{2}}\right)^2\). However, since \(2cd=1\), two solutions will be extraneous and we will have only two roots.

Solving, we have:

\(\begin{cases}c^2-\left(\sqrt{\frac{\sqrt{10}-3}{2}}\right)^2=3 \\c^2-\left(-\sqrt{\frac{\sqrt{10}-3}{2}}\right)^2=3\end{cases}\\\\c^2-\sqrt{\frac{5}{2}}+\frac{3}{2}=3,\\c=\pm \sqrt{\sqrt{\frac{5}{2}}+\frac{3}{2}\)

Given the conditions \(c\in \mathbb{R}, d\in \mathbb{R}, 2cd=1\), the solutions to this system of equations are:

\(\left(\sqrt{\sqrt{\frac{5}{2}}+\frac{3}{2}}, \sqrt{\frac{\sqrt{10}-3}{2}}\right),\\\left(-\sqrt{\sqrt{\frac{5}{2}}+\frac{3}{2}},- \frac{\sqrt{10}-3}{2}}\right)\)

Therefore, the square roots of \(z=3+i\) are:

\(\sqrt{z}=\boxed{\sqrt{\sqrt{\frac{5}{2}}+\frac{3}{2}}+i\sqrt{\frac{\sqrt{10}-3}{2}} },\\\sqrt{z}=\boxed{-\sqrt{\sqrt{\frac{5}{2}}+\frac{3}{2}}-i\sqrt{\frac{\sqrt{10}-3}{2}}}\)

Part D:

The polar form of some complex number \(a+bi\) is given by \(z=r(\cos \theta+\sin \theta)i\), where \(r\) is the modulus of the complex number (as we found in Part B), and \(\theta=\arctan(\frac{b}{a})\) (derive from right triangle in a complex plane).

We already found the value of the modulus/magnitude in Part B to be \(r=\sqrt{10}\).

The angular polar coordinate \(\theta\) is given by \(\theta=\arctan(\frac{b}{a})\) and thus is:

\(\theta=\arctan(\frac{1}{3}),\\\theta=18.43494882\approx 18.4^{\circ}\)

Therefore, the polar form of \(z\) is:

\(\displaystyle \text{Exact: }z=\sqrt{10}\left(\cos\left(\arctan\left(\frac{1}{3}\right)\right), i\sin\left(\arctan\left(\frac{1}{3}\right)\right)\right),\\\text{Approximated: }z=3.16(\cos(18.4^{\circ}),i\sin(18.4^{\circ}))\)

The expression given is 3c + 5d and the expression when c = 24 and d =25 will be 197.

It should be noted that an expression is simply used to illustrate the relationship between the variables.

In this case, the expression given is 3c + 5d.

Therefore, the expression given is 3c + 5d and the expression when c = 24 and d =25 will be:

= 3c + 5d

= 3(24) + 5(25)

= 72 + 125

= 197

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Complete question:

The expression given is 3c + 5d. Evaluate the expression when c = 24 and d =25.

Answer:

\(2\sin(\theta)\)

Step-by-step explanation:

\(\csc(\theta)=\dfrac{1}{\sin(\theta)}\)

\(\sin^2(\theta)+\cos^2(\theta)=1\implies \sin^2(\theta)=1-\cos^2(\theta)\)

\(2\csc(\theta)- 2cos^2(\theta)\times csc(\theta)\)

\(=\dfrac{2}{\sin(\theta)}-\dfrac{ 2cos^2(\theta)}{\sin(\theta)}\)

\(=\dfrac{ 2[1-cos^2(\theta)]}{\sin(\theta)}\)

\(=\dfrac{ 2sin^2(\theta)}{\sin(\theta)}\)

\(=2\sin(\theta)\)

\(\\ \rm\Rrightarrow 2cscA-2cos^2A(cscA)\)

\(\\ \rm\Rrightarrow 2(1/sinA)-2cos^2A(1/sinA)\)

\(\\ \rm\Rrightarrow \dfrac{2}{sinA}-\dfrac{2cos^2A}{sinA}\)

\(\\ \rm\Rrightarrow \dfrac{2-2cos^2A}{sinA}\)

\(\\ \rm\Rrightarrow \dfrac{2(1-cos^2A)}{sinA}\)

\(\\ \rm\Rrightarrow \dfrac{2sin^2A}{sinA}\)

\(\\ \rm\Rrightarrow sinA\)

Answer:

five kids .each 6 sweaters and 9 trousers

Step-by-step explanation:

Answer:

{x = -1, 3}

Step-by-step explanation:

See picture below :)

Answer:

Step-by-step explanation:

Multiples of 5 5 , 10 , 15 , 20 , 25 , 30 , 35 , 40 , 45 , 50 , 55 , 60

Multiples of 6 6, 12, 18 , 24 , 30, 36 , 42 , 48, 54, 60

Common Multiples of 5 and 6 30 and 60

Answer:

Step-by-step explanation:

\(\dfrac{5*2}{6*2}= \dfrac{10}{12}\\\\ \dfrac{5*3}{6*3}= \dfrac{15}{18}\\\\\\ \dfrac{5*4}{6*4}= \dfrac{20}{24}\)

The number that produces an irrational number when multiplied by 0.4 is 3π. The correct option is A. \(3\pi\)

To determine which number produces an irrational number when multiplied by 0.4

First and foremost, 0.4 is a rational number.

A rational number is a number that is of the form \(\frac{p}{q}\) where p and q are integers and q is not equal to 0.

0.4 can be written as \(\frac{4}{10}\) or \(\frac{2}{5}\), hence, it is a rational number.

Now,

Any irrational number when multiplied by a rational number other than zero will also be irrational.

Therefore, we are to determine which of the given options is an irrational number.

π is an irrational number, therefore 3π is also an irrational number

0.444... can be expressed as 4/9, therefore it is a rational number

\(\sqrt{9}\) = 3, which can be expressed as 3/1, therefore it is a rational number

\(\frac{2}{7}\) is a rational number

The only irrational number among the given options is 3π

Hence, the number that produces an irrational number when multiplied by 0.4 is 3π. The correct option is A. \(3\pi\)

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Answer:

1 pound 3.4 ounces

Step-by-step explanation:

Answer:

The scale factor is now 3

Step-by-step explanation:

6/2=3

12/4=3

The smaller square was increased by the scale factor of 3

Answer:

Step-by-step explanation:

Answer:

G(x)=1/x+15

Step-by-step explanation:

.

Step-by-step explanation:

2x + ay + 2 = 0

the gradient or slope of the line is the factor m of x in the form

y = mx + b

so let's transform the given equation into that form :

ay = -2x - 2

y = (-2/a)x - 2/a

we know that m = 1.

so,

-2/a = 1

-2 = a

Plan A lasts for 0.5 hours, and Plan B lasts for 0.75 hours based on the given information and the solution to the System of equations.

To determine the duration of each workout plan, let's assume that both Plan A and Plan B have a fixed duration. Let's denote the duration of Plan A as "x" hours and the duration of Plan B as "y" hours.

From the given information, we can form the following equations:

On Wednesday:

5 clients did Plan A, so the total duration for Plan A is 5x hours.

6 clients did Plan B, so the total duration for Plan B is 6y hours.

The total training duration on Wednesday is 7 hours, so we have the equation: 5x + 6y = 7   ... Equation (1)

On Thursday:

3 clients did Plan A, so the total duration for Plan A is 3x hours.

2 clients did Plan B, so the total duration for Plan B is 2y hours.

The total training duration on Thursday is 3 hours, so we have the equation: 3x + 2y = 3   ... Equation (2)

We now have a system of equations (Equation 1 and Equation 2) that we can solve to find the values of x and y.

Multiplying Equation (1) by 2 and Equation (2) by 5 to eliminate the y variable, we get:

10x + 12y = 14   ... Equation (3)

15x + 10y = 15   ... Equation (4)

Multiplying Equation (3) by 3 and Equation (4) by 2 to create a new system of equations:

30x + 36y = 42   ... Equation (5)

30x + 20y = 30   ... Equation (6)

Subtracting Equation (6) from Equation (5), we can eliminate the x variable:

(30x + 36y) - (30x + 20y) = 42 - 30

16y = 12

y = 12/16

y = 0.75

Substituting the value of y into Equation (2), we can solve for x:

3x + 2(0.75) = 3

3x + 1.5 = 3

3x = 3 - 1.5

3x = 1.5

x = 1.5/3

x = 0.5

Therefore, the duration of Plan A is 0.5 hours, and the duration of Plan B is 0.75 hours.

Plan A lasts for 0.5 hours, and Plan B lasts for 0.75 hours based on the given information and the solution to the system of equations.

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The given information does not provide any clear indication for determining the position that should be taken on 3/01 and 3/02. Without additional information, it is not possible to make a decision. The table only displays the closing prices of the July GBP Futures Contract on different days, and it is unclear what trading strategy or what scenario is being considered. Additional information about the goals and objectives, the market conditions, and other relevant factors would be necessary to make a decision about trading positions.

The range of value for d is d ≥ 1

Inequality, In mathematics is a statement of an order relationship. Greater than, greater than or equal to, less than, or less than or equal to—between two numbers or algebraic expressions.

Greater than is >

Greater then or equal to is ≥

less than <

less then or equal to ≤

d + 5≥ 6

add the additive inverse of 5 to both sides

d+5+(-5) ≥ 6+(-5)

d ≥ 6-5

d ≥ 1

therefore the range of value of d is d ≥ 1

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\(\sf\sin^2 150^\circ + \sin^2 60^\circ = 1 \\\)

Step 1: Convert degrees to radians:

\(\sf\sin^2 \left(\frac{150\pi}{180}\right) + \sin^2 \left(\frac{60\pi}{180}\right) = 1 \\\)

Step 2: Simplify the expressions using the trigonometric identity:

\(\sf\sin^2 \left(\frac{\pi}{6}\right) + \sin^2 \left(\frac{\pi}{3}\right) = 1 \\\)

Step 3: Recall the values of sine for angles \(\sf \frac{\pi}{6} \\\) and \(\sf \frac{\pi}{3} \\\):

\(\sf\left(\frac{1}{2}\right)^2 + \left(\frac{\sqrt{3}}{2}\right)^2 = 1 \\\)

Step 4: Evaluate the squares and simplify further:

\(\sf\frac{1}{4} + \frac{3}{4} = 1 \\\)

Step 5: Combine the fractions:

\(\sf\frac{4}{4} = 1 \\\)

Step 6: Simplify the fraction:

\(\sf1 = 1 \\\)

Thus, the equation \(\sf \sin^2 150^\circ + \sin^2 60^\circ = 1 \\\) is verified and true.

Answer:

144 ml

Step-by-step explanation:

0.008 x 180 = 144

The annual percentage yield (APY) of Rocio is 3.11 %

What is Annual Percentage Yield?

The annual percentage yield (APY) is the real rate of return earned on an investment, taking into account the effect of compounding interest.

Given data ,

Deposit amount of Russ = $ 3200

Interest rate R = 3.1 %

And it is given that the interest rate is compounded daily , so

Interest rate R = 3.1 % / 365

                       = 0.031 / 365

                       = 0.000084

Now , The annual percentage yield (APY) is calculated as

The interest amount = 3200 x ( 1 + 0.000084 )³⁶⁵

                                 = 3200 x ( 1.000084 )³⁶⁵

                                 = 3200 x 1.031133

                                 = 3299.6272

                                 ≈ 3299.627

Therefore , the interest will be

= 3299.627 - 3200

= $ 99.627

Now , the annual percentage yield (APY) is given by

= Interest / Deposit

= 99.627 / 3200

= 0.03113

≈ 3.11 %

Hence , annual percentage yield (APY) of Rocio is 3.11 %

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Answer:

As per the given information there are 4 different price intervals.

Let x be the age and y be the admission price, then we have following equations;

ANSWER

784 cheetahs

EXPLANATION

The population of cheetahs is given by an exponential decrease,

Where r is the decrease rate and p₀ is the initial population.

In this problem, the initial population is 1250 cheetahs, the decrease rate is 0.11 and we have to find the population of cheetahs after t = 4 years,

Hence, there will be 784 cheetahs after 4 years.

For the linear function built from the scatter plot, it is found that:

The slope-intercept representation of a linear function is the rule presented as follows:

y = mx + b

The coefficients of the function and their meaning are listed as follows:

From the graph, we have that when x = 0, y = 500, hence the intercept of the line is:

b = 500.

When the input increases by 200, the output increases by 1000, hence the slope is:

m = 1000/2 = 5.

Thus the equation is:

y = 5x + 500.

The graph is given by the image at the end of the answer.

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Given:

v = 150d

v represents company's production for the coming year

d represents the number of days

150 is the daily production

The rate of change of the function representing the number of vehicles manufactured for the coming year is CONSTANT (150) , and its graph is a STRAIGHT LINE . So, the function is a LINEAR function.

I hope this helps!

Answer:

v = 150d

v represents company's production for the coming year

d represents the number of days

150 is the daily production

The system has an infinite number of solutions, but the only solution is (a, b) = (0, 0).

The given system of equations can be solved using the substitution method. We can begin by solving the first equation,\(a^2 + b^2\), for either a or b. Let's solve for a:

\(a^2 + b^2 = 0\)

\(a^2 + b^2 = 0\)

\(a^2 = -b^2\)

\(a = \pm\sqrt(-b^2)\)

We can substitute this expression for a into the second equation, \(3a^2 - 2ab - b^2 = 0\), and simplify:

\(3(\pm\sqrt(-b^2))^2 - 2(\pm\sqrt(-b^2))b - b^2 = 0\)

\(3b^2 - 2b^2 - b^2 = 0\)

0 = 0

Since 0 = 0, this means that the system of equations has an infinite number of solutions. In other words, any values of a and b that satisfy the equation \(a^2 + b^2 = 0\) will also satisfy the equation \(3a^2 - 2ab - b^2 = 0\)

However, the equation \(a^2 + b^2 = 0\) only has a single solution, which is a = b = 0. Therefore, the solution to the system of equations is (a, b) = (0, 0).

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