This question: 1 pt

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identify the type 1 error. the epa claims that fluoride in children's drinking water should be at a mean level of less than 1. 2 ppm, or parts per million, to reduce the number of dental cavities.

Answer:

14. x = 42, the angles are 84°

15. x = 22, the angles are 109°

16. x = 55, the angles are 212°

17. x = 25, the angles are 125°

Step-by-step explanation:

angles across from eachother are congruent :)

Answer:

C. 1

2

Step-by-step explanation:

I can give u the show ur work steps if u want

Option c is the correct one!!

Calculating the circumference of a circle based on the circle's diameter.

What kind of problems can a quantum computer solve?

10 Difficult Problems Quantum Computers can Solve Easily-

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The complete question is -

Which type of mathematical problem is too complex for a classical computer to solve efficiently?

a. converting an irregular fraction to an approximate decimal value

b. multiplying two numbers that both have a large number of digits

c. finding two prime factors that result in a specific value when multiplied

d. calculating the circumference of a circle based on the circle's diameter

The linearized equation around its equilibrium position is given as;f(x) = -4.8(x - 1.39)

The given equation is x² + 2x* + 2x² - 12x + 10 = 0.

It needs to be linearized around its equilibrium position.

Linearizing the given equation around its equilibrium position x0, we have;

f(x) = f(x0) + f'(x0)(x - x0)

Where f(x) = x² + 2x* + 2x² - 12x + 10

The equilibrium position is the point where f(x) = 0.

Hence, f(x0) = 0.

Thus, x² + 2x* + 2x² - 12x + 10 = 0⇒ 3x² - 6x = -10⇒ x² - 2x + (2.33) = 0(x-1)² = 0.77x - 0.77 or

x = 1 + (0.77)/(2) or x = 1.39

Hence, the equilibrium position x0 = 1.39.

Substitute x0 = 1.39 in the equation and simplify to get f'(x0):

f(x) = x² + 2x* + 2x² - 12x + 10

f(x) = 3.84 - 8.34 + 10

f'(x0) = f'(1.39) = -4.8

Therefore, the linearized equation around its equilibrium position is given as;f(x) = -4.8(x - 1.39)

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Considering the number of legs of each animal, it is found that there are 56 legs on the ground.

Hence the number of legs on the ground is given by:

N = 2 x 4 + 4 x 9 + 1 x 4 + 1 x 2 + 3 x 2 = 56.

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The equation of the line would be y = (-3/4)x + 5.

What is the slope-point form of the line?

For the line having slope "m" and the point (x1, y1) the equation of the line passing through the point (x1, y1) having slope 'm' would be

                      y - y1 = m(x - x1)

The given equation is \(y=-\frac{3}{4}x-17\)

The required line is parallel to the given line.

and we know that the slopes of the parallel lines are equal so the slope of the required line would be m = -3/4

And the required line passes through (8, -1)

so by using slope - point form of the line,

      y - (-1) = (-3/4)(x - 8)

      y + 1 = (-3/4)x - (-3/4)8

       y + 1 = (-3/4)x + 24/4

       y = (-3/4)x + (12/2 - 1)

       y = (-3/4)x + 5

Hence, the equation of the line would be y = (-3/4)x + 5.

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Answer:

echo

Step-by-step explanation:

True. Any solution of Problem 1 (max f(x)) is also a solution of Problem 2 (max f(x) subject to g(x) = c).

True. If Problem 1 does not have a solution, then Problem 2 does not have a solution.

True. Problem 2 (max f(x) subject to g(x) = c) is equivalent to min -f(x) subject to g(x) = c.

False. In Problem 2, the quasi-convexity of f is not a sufficient condition for a point satisfying the first-order conditions to be a global minimum.

The function f(x,y) = 5x - 17y is quasi-concave.

Any solution that maximizes f(x) will also satisfy the constraint g(x) = c. Therefore, any solution of Problem 1 is also a solution of Problem 2.

If Problem 1 does not have a solution, it means that there is no maximum value for f(x). In such a case, Problem 2 cannot have a solution since there is no maximum value to subject to the constraint g(x) = c.

Problem 2 can be reformulated as finding the minimum of -f(x) subject to the constraint g(x) = c. This is because maximizing f(x) is equivalent to minimizing -f(x) since the maximum of a function is the same as the minimum of its negative.

False. Quasi-convexity of f is not a sufficient condition for a point satisfying the first-order conditions to be a global minimum in Problem 2. Quasi-convexity guarantees that local minima are also global minima, but it does not ensure that the point satisfying the first-order conditions is a global minimum.

The function f(x,y) = 5x - 17y is quasi-concave. A function is quasi-concave if the upper contour sets, which are defined by f(x,y) ≥ k for some constant k, are convex. In this case, the upper contour sets of f(x,y) = 5x - 17y are convex, satisfying the definition of quasi-concavity.

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Answer:

52 inches

Step-by-step explanation:

Answer:

52 imches

Step-by-step explanation:

each foot is 12 inches

12x4=48

12/3=4

this is the 1/3 feet

add 48 and 4 and you get 52

To identify the vertical asymptotes, we have to factor the denominator. For horizontal asymptotes, we compare the degrees of the numerator and denominator.

For rational functions, there are vertical and horizontal asymptotes. To identify the vertical asymptotes, we first have to factor the denominator. After that, we should look for values that make the denominator zero. These values can be found by setting the denominator equal to zero and solving for x. The resulting x values would be the vertical asymptotes of the function.

The horizontal asymptote is the line that the function approaches as x goes towards infinity or negative infinity. For rational functions, the horizontal asymptote is found by comparing the degrees of the numerator and the denominator.

If the degree of the numerator is less than the degree of the denominator, the horizontal asymptote is y = 0. If the degree of the numerator is equal to the degree of the denominator, the horizontal asymptote is y = the ratio of the leading coefficients. If the degree of the numerator is greater than the degree of the denominator, there is no horizontal asymptote.

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This is the UML diagram for the development of the program, in which Shapes 2D is the superclass and Circle, Square, Triangle, Rectangle, Rhombus, and Parallelogram are subclasses.

​​​​​​​This is the program in C++ demonstrating the above classes.

#include<iostream>

using namespace std;

class shapes

{

public:

   string name;

   string color;

   virtual void getAttributes()=0;

};

class Circle: public shapes

{

public:

   float radius;

   Circle(string n,string c, float r)

   {

       name=n;

       color=c;

       radius=r;

   }

   float getPerimeter()

   {

       return(2*(3.142)*radius);

   }

   float getArea()

   {

       return((3.142)*(radius*radius));

   }

   void getAttributes()

   {

       cout<<"Name  :"<<name<<endl;

       cout<<"Color :"<<color<<endl;

   }

};

class Square:public shapes

{

public:

   float side;

   Square(string n,string c, float s)

   {

       name=n;

       color=c;

       side=s;

   }

   float getPerimeter()

   {

       return(4*side);

   }

   float getArea()

   {

       return(side*side);

   }

   void getAttributes()

   {

       cout<<"Name  :"<<name<<endl;

       cout<<"Color :"<<color<<endl;

   }

};

class Triangle:public shapes

{

public:

   float base;

   float height;

   float side1;

   float side2;

   float side3;

   Triangle(string n,string c)

   {

       name=n;

       color=c;

   }

   float getPerimeter()

   {

       cout<<"Enter side1\n";

       cin>>side1;

       cout<<"Enter side2\n";

       cin>>side2;

       cout<<"Enter side3\n";

       cin>>side3;

       return(side1+side2+side3);

   }

   float getArea()

   {

       cout<<"Enter base\n";

       cin>>base;

       cout<<"Enter height\n";

       cin>>height;

       return((0.5)*base*height);

   }

   void getAttributes()

   {

       cout<<"Name  :"<<name<<endl;

       cout<<"Color :"<<color<<endl;

   }

};

class Rectangle:public shapes

{

public:

   float length;

   float breadth;

   Rectangle(string n,string c, float l,float b)

   {

       name=n;

       color=c;

       length=l;

       breadth=b;

   }

   float getPerimeter()

   {

       return(2*(length+breadth));

   }

   float getArea()

   {

       return(length*breadth);

   }

   void getAttributes()

   {

       cout<<"Name  :"<<name<<endl;

       cout<<"Color :"<<color<<endl;

   }

};

class Rhombus:public shapes

{

public:

   float diagonal1;

   float diagonal2;

   float side;

   Rhombus(string n,string c)

   {

       name=n;

       color=c;

   }

   float getPerimeter()

   {

       cout<<"Enter Side\n";

       cin>>side;

       return(4*side);

   }

   float getArea()

   {

       cout<<"Enter diagonal 1\n";

       cin>>diagonal1;

       cout<<"Enter diagonal 2\n";

       cin>>diagonal2;

       return((0.5)*diagonal1*diagonal2);

   }

   void getAttributes()

   {

       cout<<"Name  :"<<name<<endl;

       cout<<"Color :"<<color<<endl;

   }

};

class Parallelogram:public shapes

{

public:

   float base;

   float height;

   Parallelogram(string n,string c, float b,float h)

   {

       name=n;

       color=c;

       base=b;

       height=h;

   }

   float getPerimeter()

   {

       return(2*(base+height));

   }

   float getArea()

   {

       return(base*height);

   }

   void getAttributes()

   {

       cout<<"Name  :"<<name<<endl;

       cout<<"Color :"<<color<<endl;

   }

};

int main()

{

   int choice;

   while(1)

   {

       cout<<"\n\nEnter your choice :";

       cout<<"\n1 for Circle\n";

       cout<<"2 for Square\n";

       cout<<"3 for Triangle\n";

       cout<<"4 for Rectangle\n";

       cout<<"5 for Rhombus\n";

       cout<<"6 for Parallelogram\n";

       cin>>choice;

       system("cls");

       switch(choice)

       {

       case 1:

           {

               float r;

               cout<<"Enter radius\n";

               cin>>r;

               Circle c("Circle","Yellow",r);

               c.getAttributes();

               cout<<"Perimeter : "<<c.getPerimeter()<<endl;

               cout<<"Area : "<<c.getArea()<<endl;

           }break;

       case 2:

           {

               float side;

               cout<<"Enter side\n";

               cin>>side;

               Square s("Square","Red",side);

               s.getAttributes();

               cout<<"Perimeter : "<<s.getPerimeter()<<endl;

               cout<<"Area : "<<s.getArea()<<endl;

           }break;

       case 3:

           {

               Triangle t("Triangle","Green");

               t.getAttributes();

               cout<<"Perimeter : "<<t.getPerimeter()<<endl;

               cout<<"Area : "<<t.getArea()<<endl;

           }break;

       case 4:

           {

               float l,b;

               cout<<"Enter Length and breadth\n";

               cin>>l>>b;

               Rectangle r("Rectangle","Blue",l,b);

               r.getAttributes();

               cout<<"Perimeter : "<<r.getPerimeter()<<endl;

               cout<<"Area : "<<r.getArea()<<endl;

           }break;

       case 5:

           {

               Rhombus r("Rhombus","Purple");

               r.getAttributes();

               cout<<"Perimeter : "<<r.getPerimeter()<<endl;

               cout<<"Area : "<<r.getArea()<<endl;

           }break;

       case 6:

           {

               float b,h;

               cout<<"Enter base\n";

               cin>>b;

               cout<<"Enter height\n";

               cin>>h;

               Parallelogram p("Parallelogram","Pink",b,h);

               p.getAttributes();

               cout<<"Perimeter : "<<p.getPerimeter()<<endl;

               cout<<"Area : "<<p.getArea()<<endl;

           }break;

       }

   }

}

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Answer:

=3p 3+17p+6

Step-by-step explanation:

The probability  that both bills are $1 is 0.76

Following the selection of one $1 bill, the remaining number of $1 bills in the bag is reduced to 19.

Based on statistical analysis, there is a high likelihood of selecting a $1 bill on the second attempt, with a confidence level of 95%.

The probability of drawing two $1 bills is mathematically deduced to be zero. The attainment of a numerical outcome of 76 is contingent upon the computation of the product derived from the probability of selecting a $1 bill during the initial endeavor, which stands at 4 out of 5, and the likelihood of selecting another $1 bill in a consequent endeavor, which amounts to 19 out of 20.

In an alternative expression, the probability of two bills being $1 bills simultaneously is 76%.

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The answer is ,  the Taylor series (centered at c=0) for the function f(x) = e^(4x) is given by:

\($$\large f(x) = \sum_{n=0}^{\infty} \frac{4^n}{n!}x^n$$\)

The Taylor series expansion is a way to represent a function as an infinite sum of terms that depend on the function's derivatives.

The Taylor series of a function f(x) centered at c is given by the formula:

\(\large f(x) = \sum_{n=0}^{\infty} \frac{f^{(n)}(c)}{n!}(x-c)^n\)

Using the definition of Taylor series to find the Taylor series (centered at c=0) for the function f(x) = e^(4x), we have:

\(\large e^{4x} = \sum_{n=0}^{\infty} \frac{e^{4(0)}}{n!}(x-0)^n\)

\(\large e^{4x} = \sum_{n=0}^{\infty} \frac{4^n}{n!}x^n\)

Therefore, the Taylor series (centered at c=0) for the function f(x) = e^(4x) is given by:

\($$\large f(x) = \sum_{n=0}^{\infty} \frac{4^n}{n!}x^n$$\)

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The Taylor series for f(x) = e^(4x) centered at c = 0 is:

f(x) = 1 + 4x + 8x^2 + 32x^3/3 + ...

To find the Taylor series for the function f(x) = e^(4x) centered at c = 0, we can use the definition of the Taylor series. The general formula for the Taylor series expansion of a function f(x) centered at c is given by:

f(x) = f(c) + f'(c)(x - c) + f''(c)(x - c)^2/2! + f'''(c)(x - c)^3/3! + ...

First, let's find the derivatives of f(x) = e^(4x):

f'(x) = d/dx(e^(4x)) = 4e^(4x)

f''(x) = d^2/dx^2(e^(4x)) = 16e^(4x)

f'''(x) = d^3/dx^3(e^(4x)) = 64e^(4x)

Now, let's evaluate these derivatives at x = c = 0:

f(0) = e^(4*0) = e^0 = 1

f'(0) = 4e^(4*0) = 4e^0 = 4

f''(0) = 16e^(4*0) = 16e^0 = 16

f'''(0) = 64e^(4*0) = 64e^0 = 64

Now we can write the Taylor series expansion:

f(x) = f(0) + f'(0)(x - 0) + f''(0)(x - 0)^2/2! + f'''(0)(x - 0)^3/3! + ...

Substituting the values we found:

f(x) = 1 + 4x + 16x^2/2! + 64x^3/3! + ...

Simplifying the terms:

f(x) = 1 + 4x + 8x^2 + 32x^3/3 + ...

Therefore, the Taylor series for f(x) = e^(4x) centered at c = 0 is:

f(x) = 1 + 4x + 8x^2 + 32x^3/3 + ...

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The probability that the sample proportion is between 0.2 and 0.42 can be calculated using the standard normal distribution.

To calculate the probability, we need to assume that the sample proportion follows a normal distribution. This assumption holds true when the sample size is sufficiently large and the conditions for the central limit theorem are met.

First, we need to calculate the standard error of the sample proportion. The standard error is the standard deviation of the sampling distribution of the sample proportion and is given by the formula sqrt(p(1-p)/n), where p is the estimated proportion and n is the sample size.

Next, we convert the sample proportion range into z-scores using the formula z = (x - p) / SE, where x is the given proportion and SE is the standard error. In this case, we use z-scores of 0.2 and 0.42.

Once we have the z-scores, we can use a standard normal distribution table or a statistical software to find the corresponding probabilities. The probability of the sample proportion falling between 0.2 and 0.42 is equal to the difference between the two calculated probabilities.

Alternatively, we can use the z-table to find the individual probabilities and subtract them. The z-table provides the cumulative probabilities up to a certain z-score. By subtracting the lower probability from the higher probability, we can find the desired probability.

In conclusion, the probability that the sample proportion is between 0.2 and 0.42 can be calculated using the standard normal distribution and z-scores. This probability represents the likelihood of observing a sample proportion within the specified range.

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The cosine of angle J is given as follows:

\(\cos{J} = \frac{14\sqrt{2}}{49}\)

The three trigonometric ratios are the sine, the cosine and the tangent of an angle, and they are obtained according to the rules presented as follows:

For the angle J in this problem, we have that:

Hence the cosine of angle J is given as follows:

\(\cos{J} = \frac{4}{\sqrt{98}} \times \frac{\sqrt{98}}{\sqrt{98}}\)

\(\cos{J} = \frac{4\sqrt{98}}{98}\)

\(\cos{J} = \frac{2\sqrt{98}}{49}\)

As 98 = 2 x 49, we have that \(\sqrt{98} = \sqrt{49 \times 2} = 7\sqrt{2}\), hence:

\(\cos{J} = \frac{14\sqrt{2}}{49}\)

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The dollar gains depreciated relative to the pound when the price of pounds in dollars increases, for instance, from $1 = 1 pound to $2 = 1 pound.

Devaluation of a currency can take place in both absolute and relative terms. When the value of one currency declines in relation to the values of other currencies, this is referred to as a relative devaluation. For instance, the British pound sterling may be worth more today than it did yesterday in terms of US dollars.

A currency's value declines when compared to other currencies, which is known as currency depreciation. Political unrest, interest rate differences, weak economic fundamentals, and investor risk aversion are a few examples of the causes of currency devaluation.

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Finding the six trigonometric functions of θ: Since (-3,3) is a point on the terminal side of θ, we can use the coordinates of this point to determine the values of the trigonometric functions.

Let's label the legs of the right triangle formed as opposite = 3 and adjacent = -3, and use the Pythagorean theorem to find the hypotenuse.

Using Pythagorean theorem: hypotenuse² = opposite² + adjacent²

hypotenuse² = 3² + (-3)²

hypotenuse² = 9 + 9

hypotenuse² = 18

hypotenuse = √18 = 3√2

Now we can calculate the trigonometric functions:

sin θ = opposite/hypotenuse = 3/3√2 = √2/2

cos θ = adjacent/hypotenuse = -3/3√2 = -√2/2

tan θ = opposite/adjacent = 3/-3 = -1

csc θ = 1/sin θ = 2/√2 = √2

sec θ = 1/cos θ = -2/√2 = -√2

cot θ = 1/tan θ = -1/1 = -1

Therefore, the exact values of the six trigonometric functions of θ are:

sin θ = √2/2, cos θ = -√2/2, tan θ = -1, csc θ = √2, sec θ = -√2, cot θ = -1.

Part 2: Finding the remaining trigonometric functions given tan θ and sin θ:

Given that tan θ = and sin θ < 0, we can deduce that θ lies in the third quadrant of the unit circle where both the tangent and sine are negative. In this quadrant, the cosine is positive, while the cosecant, secant, and cotangent can be determined by taking the reciprocals of the corresponding functions in the first quadrant.

Since tan θ = opposite/adjacent = sin θ/cos θ, we have:

sin θ = -1 and cos θ =

Using the Pythagorean identity sin² θ + cos² θ = 1, we can find cos θ:

(-1)² + cos² θ = 1

1 + cos² θ = 1

cos² θ = 0

cos θ = 0

Now we can calculate the remaining trigonometric functions:

csc θ = 1/sin θ = 1/-1 = -1

sec θ = 1/cos θ = 1/0 = undefined

cot θ = 1/tan θ = 1/-1 = -1

Therefore, the exact values of the remaining five trigonometric functions of θ are:

sin θ = -1, cos θ = 0, tan θ = -1, csc θ = -1, sec θ = undefined, cot θ = -1.

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Answer:

179,000-94000

Step-by-step explanation:

85000 He made $85000

Answer:

the answer is $85,000

Step-by-step explanation:

so basically you just subtract because less stands for subtraction

179000

-94000

________

$85,000

hope i helped and sorry if i was late :)

Answer: I hope it helps :)

Step-by-step explanation:

1.

\(Hypotenuse =x\\Opposite =y \\Adjacent =6\\\alpha = 6\\Let's\: find\: the \:hypotenuse\: first\\Using SOHCAHTOA\\Cos \alpha = \frac{adj}{hyp} \\Cos 60 = \frac{6}{x} \\\frac{1}{2} =\frac{6}{x} \\Cross\:Multiply\\x = 12\\Let's\: find\: y\\Hyp^2=opp^2+adj^2\\12^2=y^2+6^2\\144=y^2+36\\144-36=y^2\\108=y^2\\\sqrt{108} =\sqrt{y^2} \\y=6\sqrt{3}\)

2.

\(Opposite =x\\Hypotenuse = 46\\Adjacent =y \\\alpha =60\\Using \: SOHCAHTOA\\Sin \alpha =\frac{opp}{adj} \\Sin 60=\frac{x}{46}\\\\\frac{\sqrt{3} }{2} =\frac{x}{46} \\2x=46\sqrt{3} \\x = \frac{46\sqrt{3} }{2} \\x =23\sqrt{3} \\\\Hyp^2=opp^2+adj^2\\46^2=(23\sqrt{3} )^2+y^2\\2116=1587+y^2\\2116-1587=y^2\\529=y^2\\\sqrt{529} =\sqrt{y^2} \\y = 23\)

3.

\(Hypotenuse = u\\Opposite =6\sqrt{3} \\Adjacent = v\\\alpha =60\\Sin\: 60 = \frac{6\sqrt{3} }{u} \\\frac{\sqrt{3} }{2} =\frac{6\sqrt{3} }{u} \\12\sqrt{3} =u\sqrt{3} \\\\\frac{12\sqrt{3} }{\sqrt{3} } =\frac{u\sqrt{3} }{\sqrt{3} } \\u = 12\\Hyp^2=opp^2+adj^2\\12^2= (6\sqrt{3} )^2+v^2\\144=108+v^2\\144-108=v^2\\36 = v^2\\\sqrt{36} =\sqrt{v^2} \\\\v =6\)

4.

\(Hypotenuse = a\\Opposite =18 \\Adjacent = b\\\alpha =45\\Tan \alpha = opp/adj\\Tan \:45 =18/b\\1=\frac{18}{b}\\ b = 18\\\\Hyp^2=Opp^2+Adj^2\\a^2 = 18^2+18^2\\a^2=324+324\\a^2=648\\\sqrt{hyp^2} =\sqrt{648}\\ \\a =18\sqrt{2}\)

5.

\(Hypotenuse = 13\sqrt{2}\\ Opposite =x\\Adjacent = y\\\alpha =45\\Sin\:\alpha = opp/hyp\\Sin 45=x/13\sqrt{2}\\ \\\frac{\sqrt{2} }{2} =\frac{x}{13\sqrt{2} } \\2x=26\\2x/2=26/2\\\\x = 13\\\\Hyp^2=opp^2+adj^2\\(13\sqrt{2})^2=13^2+y^2\\ 338=169+y^2\\338-169=y^2\\169=y^2\\\sqrt{169} =\sqrt{y^2} \\13 = y\)

Answer: 24,500 ways

Step-by-step explanation:

The number of ways to select 4 appetizers out of 8 is given by the combination C(8,4) = 70. Similarly, the number of ways to select 3 main courses out of 5 is given by C(5,3) = 10, and the number of ways to select 3 desserts out of 7 is given by C(7,3) = 35.

To find the total number of ways to select the full meal consisting of 4 appetizers, 3 main courses, and 3 desserts, we can use the multiplication principle of counting, which states that the total number of ways to perform a sequence of independent tasks is equal to the product of the number of ways to perform each individual task.

Therefore, the total number of ways to select the full meal is given:

70 x 10 x 35 = 24,500

Hence, there are 24,500 ways to select 4 appetizers, 3 main courses, and 3 desserts from the given options.

Answer:

Step-by-step explanation:

The number of ways to select 4 appetizers out of 8 is given by the combination C(8,4) = 70.
Similarly, the number of ways to select 3 main courses out of 5 is given by C(5,3) = 10, and the number of ways to select 3 desserts out of 7 is given by C(7,3) = 35.

To find the total number of ways to select the full meal consisting of 4 appetizers, 3 main courses, and 3 desserts, we can use the multiplication principle of counting, which states that the total number of ways to perform a sequence of independent tasks is equal to the product of the number of ways to perform each individual task.

Therefore, the total number of ways to select the full meal is given:70 x 10 x 35 = 24,500.

Hence, there are 24,500 ways to select 4 appetizers, 3 main courses, and 3 desserts from the given options.

It will take Rob and Tricia 75 minutes to paint 2/3 of the room together. Rob can paint 1/3 of the room in 2 hours, and Tricia can paint 1/2 of the room in 5 hours.

To determine how quickly they can paint the room together, we'll first find their individual painting rates.

Rob's rate: (1/3 room) / (2 hours) = 1/6 room/hour
Tricia's rate: (1/2 room) / (5 hours) = 1/10 room/hour

Now, we'll add their rates together to find their combined rate:
(1/6 + 1/10) room/hour = (5/30 + 3/30) room/hour = 8/30 room/hour

Next, we need to find how long it takes for them to paint 2/3 of the room together:
(2/3 room) / (8/30 room/hour) = (2/3) * (30/8) hours = 5/4 hours

Finally, convert this time to minutes:
(5/4 hours) * (60 minutes/hour) = 75 minutes

So, it will take Rob and Tricia 75 minutes to paint 2/3 of the room together.

Learn more about Tricia here:

https://brainly.com/question/17046336

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Answer:

20

Step-by-step explanation:

1 hour is 60 minutes 1/3x60=20

Answer:

1200 seconds

Step-by-step explanation:

1/3 of an hour is 20 minutes. There are 60 seconds in a minute. 60 times 20 = 1200.

Answer:

4+4=8

Step-by-step explanation:

Answer:

median, 60

Step-by-step explanation:

Answer:

it is 60 my friend.

Step-by-step explanation:

Answer:

$1,680

Step-by-step explanation:

i think it is right because 12,000 at 2 percent=280 then 280×7= 1,680