A catering service offers 12 appetizers, 6 main courses, and 7 desserts. A banquet chairperson is to select 6 appetizers, 2 main courses, and 5 desserts for a banquet. In how many ways can this be done?

Answer:

Graph 1

Step-by-step explanation:

Here, the given equation is,

[tex]f(x)=x^2+3x+2-----(1)[/tex]

For x-intercept, f(x) = 0

[tex]x^2+3x+2=0[/tex]

[tex]x^2+2x+x+2=0[/tex]

[tex]x(x+2)+1(x+2)=0[/tex]

[tex](x+1)(x+2)=0[/tex]

[tex]\implies x=-1\text{ or } -2[/tex]

So, the x-intercept of the function are (-1,0) and (-2,0)

Since, the line must has at least one x-intercept.

⇒ Graph 2 and Graph 4 can not be the graph of the given function,

Also, for y-intercept,

Put x = 0 in equation (1),

We get, f(x) = 2,

Hence, the y-intercept of the given function is (0,2),

But in Graph 3 the y-intercept of the function = (0,1)

⇒ Graph 3 can not be the graph of the given function,

Therefore, Graph 1 is the correct graph of the given function.

C = 2* pi *r = 36 pi

 r = 36pi/2pi = 18

We know that A= r^2 * pi

A = 18^2 * pi = 324 pi

the area would be 324PI square units

Final answer:

The function t(x) with ordered pairs (0, 1), (1, 3), and (3, 27) most likely represents the exponential function because it shows rapidly increasing growth rates, which is a defining feature of exponential behavior.

Explanation:

The function that most likely represents the exponential function is the one whose growth rate increases significantly for larger values of x. Examining the ordered pairs, the function t(x), which passes through the points (0, 1), (1, 3), and (3, 27), clearly demonstrates this behavior as the increase between the y-values gets dramatically larger as x increases.

This is a classic characteristic of exponential growth. In contrast, the function p(x) that passes through (0, 2), (1, 3), (3, 11), (5, 27), and (6, 38) shows a more consistent increase in y-values as x increases, indicative of a quadratic function.

For two successes in seven binomial trials with a success probability of 0.06, use the binomial probability formula. Calculate (7 choose 2) * (0.06^2) * (0.94^5) to get the probability of 0.0554.

To calculate the probability of two successes in seven trials with a success probability of 0.06, you can use the binomial probability formula, which is P(X = k) = (n choose k) * p^k * q^(n-k), where 'n' is the number of trials (7), 'k' is the number of successes (2), 'p' is the probability of success (0.06), and 'q' is the probability of failure (q = 1 - p = 0.94).

First, calculate the binomial coefficient using 'n choose k', which is (7 choose 2). Then, raise the probability of success to the power of the number of successes (0.06^2) and the probability of failure to the power of the number of failures (0.94^5). Lastly, multiply these values together to get the probability.

The calculation is as follows:

(7 choose 2) * (0.06^2) * (0.94^5) = 21 * 0.0036 * 0.7339 = 0.0554

The dimensions of the rectangle that maximize the enclosed area are L = 80 yards and W = 80 yards. The maximum area is A = 80 * 80 = 6400 square yards.

To find the dimensions of the rectangle that maximize the enclosed area using 320 yards of fencing, we'll use the concept of optimization. Let's solve it step by step:

Let's assume the length of the rectangle is L and the width is W.

Perimeter constraint:

The perimeter of the rectangle is given as 2L + 2W, which must equal 320 yards:

2L + 2W = 320

Simplify the perimeter equation:

Divide both sides by 2 to get:

L + W = 160

Express one variable in terms of the other:

Solve the equation for L:

L = 160 - W

Area equation:

The area of the rectangle is given by A = L * W.

Substitute the value of L from the previous step into the area equation:

A = (160 - W) * W

A = 160W - W^2

Maximize the area:

To find the maximum area, we need to maximize the function A = 160W - W^2. This is achieved when the derivative is zero.

Take the derivative of A with respect to W:

dA/dW = 160 - 2W

Set dA/dW = 0 and solve for W:

160 - 2W = 0

2W = 160

W = 80

Substitute the value of W back into the perimeter equation to find the corresponding value of L:

L = 160 - W = 160 - 80 = 80

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The optimization problem involves finding the dimensions of a rectangle to maximize its area using a fixed amount of fencing. The dimensions that maximize the area are both 80 yards, making the maximum area 6400 square yards.

The subject of this question is Mathematics, specifically a problem about optimization in the field of Calculus. In the given problem, we wish to find a rectangular area that can be enclosed by 320 yards of fencing that maximizes the area.

Let's designate the rectangular area's width and length as x and y respectively. The problem can now be rephrased. With the total length of fencing equal to 320 yards, you can express this as 2x + 2y = 320. Simplifying this equation, we get x + y = 160, or y = 160 - x.

The area of a rectangle is computed as width times length, or in this case, x(160 - x). This is a quadratic function, and its maximum value happens at the vertex of the parabola defined by this function. For a quadratic in standard form like y = ax^2 + bx + c, the x-coordinate of the vertex is at -b/2a. In this case, the maximum area happens when x = 160/2 = 80.

Substituting this value back into the equation for the rectangle's dimensions gives y = 160 - 80 = 80. So, the dimensions that maximize the area for a rectangle with a parameter of 320 yards are both 80 yards. Therefore, the maximum area possible is 80*80 = 6400 square yards.

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A good rule of thumb is one minute per year of your child's age. So Karen’s child is 4 years old, she would get four minutes of time-out. If you find that the shorter time-outs aren't having the wanted result, increase the length by half the time (so your 4-year-old would get an extra two minutes, for a total of six minutes).

In adults, the arm span is approximately 5 cm greater than the height in adult males and 1.2 cm in adult females. To calculate the arm span for the heights given, we add 5cm to their height. The following are the results:

Height                   Arm Span Length (in cm)

4’10                        152.32

4’11                        154.86

5’0                          157.4

5’4                          167.56

5’5                          170.10

5’7                          175.18

5’8                          177.72

5’9                          180.26

5’10                        182.80

5’11                        185.34

6’0                          187.88

To add, the total measurement of the length from the furthermost part of an individual's arms to the other end when raised equidistant to the ground at shoulder height at a 90º angle is called the arm span or wingspan.

Answer is 4

1. Multiple 29 by both side to get N by itself

2. solve 29/7.25 on a calculator, because it's easier

3. answer is 4.

4. Check your answer.  Plug in 4 to see if you get the same answer on both sides.  You should get .138 aprox. on both side. 

After 9 years many years are the total costs for the cars the same.

Expression in maths is defined as the collection of the numbers variables and functions by using signs like addition, subtraction, multiplication and division

Given that:-

We can form two-equation for the total cost of the two cars:-

A   =   17655  +   1230y,

B   =   15900  +   1425y

We wish to know when the cost will become the same for both the cars.

A     =     B

15900    +   1425y  =    17655  +    1230y  

Now subtract 15900 from both sides

1425y     =    1230y   +    1755  

Now subtract 1230y from both sides

195y     =    1755  

Now divide both sides by 195

y   =   9 years

Therefore After 9 years, many years are the total costs for the cars the same.

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Answer: C:6

Step-by-step explanation:

I just got it right on the quiz

Answer:

5 cm is the correct answer it would be half of DG

Step-by-step explanation:

use AE and CE to find the angle

AE = 20, CE = 6

 so the angle FOR CAE = tan^-1(6/20) = 16.7 degrees

 to Find DF

a^2 = b^2 +c^2-2abcos(A)

a^2 = 10^2 + 14^2 -2(14)(10)cos(16.7)

a^2 = 30

sqrt((30)=5.47 rounded to 5.5

EF =

a^2 = b^2 +c^2-2abcos(A)

a^2 = 20^2 + 14^2 -2(14)(20)cos(16.7)

a^2 = 64

sqrt(64)=8

Answer:

population in 2003 is 234 million.

Step-by-step explanation:

A country's population in 1991 was 231 million

In 1999 it was 233 million.

We have to calculate the population in 2003.

Since population growth is always represented by exponential function.

It is represented by [tex]P(t)=P_{0}e^{kt}[/tex]

Here t is time in years, k is the growth constant, and  is initial population.

For year 1991 ⇒

233 = [tex]P_{0}e^{8k}[/tex] = 231 [tex]e^{8k}[/tex]

[tex]\frac{231}{233}= e^{8k}[/tex]

Taking ln on both the sides ⇒

[tex]ln(\frac{233}{231})=lne^{8k}[/tex]

ln 233 - ln 231 = 8k  [since ln e = 1 ]

5.451 - 5.4424 = 8k

k = [tex]\frac{0.0086}{8}=0.001075[/tex]

For year 2003 ⇒

[tex]P(t)=P_{0}e^{kt}[/tex]

P (t) = 231 × [tex]e^{(0.001075)(12)}[/tex]

     = 231 × [tex]e^{0.0129}[/tex]

     = 231 × 1.0129

     = 233.9 ≈ 234 million

Therefore, population in 2003 is 234 million.

Using the concept of proportion, Rita would make $156 for 12 hours of work at the same rate.

We have,

We can use proportions to solve this problem.

Let x be the amount of money Rita would make for 12 hours of work.

We know that Rita made $221 for 17 hours of work,

which can be written as:

221/17 = x/12

To solve for x, we can cross-multiply and simplify:

221(12) = 17x

2652 = 17x

x = 156

Therefore,

Rita would make $156 for 12 hours of work at the same rate.

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