1. An average of 50,000 people visit Riverside Park each day in the
summer. The park charges $21.00 for admission. Consultants predict
that for each $1.00 increase in the entrance price, the park would lose
an average of 3,500 customers per day.
(a) Express the daily revenue from ticket sales, R as a function of
the number of $1.00 price increases, x.
(b) What ticket price(rounded to nearest cent) maximizes the revenue
from ticket sales?
(c) What ticket price will lead to no revenue?

The explicit formula for the value of the truck after [tex]\(n\)[/tex] years is:

[tex]\[V(n) = -2,500n + 32,000\][/tex]

Where [tex]\(V(n)\)[/tex] represents the value of the truck after [tex]\(n\)[/tex] years.

To write an explicit formula for the value of the truck after n years with linear depreciation, we can use the formula for the equation of a straight line:

[tex]\[y = mx + b\][/tex]

Where:

- [tex]\(y\)[/tex] is the value of the truck after [tex]\(n\)[/tex] years.

- [tex]\(m\)[/tex] is the slope of the line, which represents the rate of depreciation per year.

- [tex]\(x\)[/tex] is the number of years.

-[tex]\(b\)[/tex] is the initial value of the truck.

Given:

- Initial value [tex](\(b\)): $32,000[/tex]

- Value after 3 years [tex](\(y\)): $24,500[/tex]

We can use these values to find the slope [tex](\(m\))[/tex]:

[tex]\[m = \frac{y_2 - y_1}{x_2 - x_1}\][/tex]

Substituting the given values:

[tex]\[m = \frac{24,500 - 32,000}{3 - 0}\]\[m = \frac{-7,500}{3}\]\[m = -2,500\][/tex]

Now, we have [tex]\(m = -2,500\)[/tex]. We can substitute this slope and the initial value into the equation:

[tex]\[y = -2,500x + 32,000\][/tex]

The explicit formula for the value of the truck after [tex]\(n\)[/tex] years is:

[tex]\[V(n) = -2,500n + 32,000\][/tex]

Where [tex]\(V(n)\)[/tex] represents the value of the truck after [tex]\(n\)[/tex] years.

The area of each strip of paper is 15/8 square inches.

The area of each strip of paper can be calculated by multiplying the length and width of the strip. In this case, the length is 2 1/2 inches and the width is 3/4 inches. To multiply fractions, multiply the numerators together to get the numerator of the product, and multiply the denominators together to get the denominator of the product. So, the area of each strip is:

(2 1/2 inches) x (3/4 inches) = (5/2 inches) x (3/4 inches) = 15/8 square inches.

The annual depreciation expense for the meat processor's refrigerator, using the straight-line method, is $2,600.

Straight-line Depreciation Calculation

The straight-line depreciation method spreads the cost of an asset evenly over its useful life. To calculate the annual depreciation expense for the refrigerator, we follow this formula: (Cost of the asset - Residual value) / Estimated useful life.

In this case, the numbers provided are a cost of $48,000, a residual value of $9,000, and an estimated useful life of 15 years. Using these values, the calculation is as follows:

(48,000 - 9,000) / 15 = 39,000 / 15 = $2,600 per year.

Therefore, the annual depreciation expense that the meat processor would record for this refrigerator is $2,600.

The probability that a couple will have a girl a boy a girl and a boy in this specific order is 1/16

Probability is the likelihood or chance that an event will occur.

Probability = Expected outcome/Total outcome

If the couple will give birth to a boy and a girl, hence the total outcome of the event is 2

Pr(having a boy) = 1/2

Pr(having a girl) = 1/2

Pr(a girl, a boy, a girl and a boy) = 1/2 * 1/2 *1/2 * 1/2

Pr(a girl, a boy, a girl, and a boy) = 1/16

Hence the probability that a couple will have a girl a boy a girl and a boy in this specific order is 1/16

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Amount earn as commission on real state building is $5,520

What information do we have?

Amount of real state house = $92,000

Percentage of Commission on sale = 6%

Amount earn as commission = Amount of real state house × Percentage of Commission on sale

Amount earn as commission = 92,000 × 6%

Amount earn as commission = 92,000 × 0.06

Amount earn as commission = $5,520

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Final answer:

After examining possible combinations of the given numbers (72, 21, 15, 36, 69, 81, 9, 27, 42, and 63), we find no subset that adds up to exactly 100. The closest sum we can get is 99, which is not sufficient to win the prize. Hence, we will not win the $100.

Explanation:

To determine whether we will win the $100 by finding a combination of the given numbers that add up to 100, we should examine possible combinations. The provided numbers are 72, 21, 15, 36, 69, 81, 9, 27, 42, and 63.

Looking at these numbers, we can see if any subsets add to 100:

It becomes evident that combinations of the given numbers will either fall short of or exceed 100. The closest combinations we can make are 99, just one short of the required total, meaning we cannot win the $100 with the provided numbers on the card. Therefore, the answer is no; we will not win $100 as we cannot make a sum of exactly 100 using the numbers on the card.

Step 1

Find the equation of the line perpendicular to [tex] y = 4x + 5 [/tex] that pas through the origin

we know that

the slope of the equation [tex] y = 4x + 5 [/tex] is

[tex] m1=4 [/tex]

if two lines are perpendicular

then

the product of their slopes is equal to [tex] -1 [/tex]

[tex] m1*m2=-1 [/tex]

the slope of the line perpendicular is equal to

[tex] m2=-\frac{1}{m1} \\\\ m2=-\frac{1}{4} [/tex]

with m2 and the origin find the equation of the line

[tex] y-y1=m(x-x1)\\ y-0=(-\frac{1}{4} )*(x-0)\\ y=-\frac{1}{4} x [/tex]

Step 2

Solve the system

[tex] y = 4x + 5 [/tex]---> equation [tex] 1 [/tex]

[tex] y=-\frac{1}{4} x [/tex]-----> equation [tex] 2 [/tex]

Multiply equation [tex] 1 [/tex] by [tex] -1 [/tex]

Adds equation [tex] 1 [/tex] and equation [tex] 2 [/tex]

[tex] -y = -4x - 5 [/tex]

[tex] y=-\frac{1}{4} x \\ ----- [/tex]

[tex] 0=-4x-\frac{1}{4} x-5\\ \\ \frac{17}{4} x=-5\\ \\ x=-\frac{20}{17} [/tex]

[tex] y = 4x + 5 [/tex]

[tex] y = 4*(-\frac{20}{17}) + 5 \\ \\ y=\frac{5}{17} [/tex]

the solution is the point [tex] (-\frac{20}{17} ,\frac{5}{17} ) [/tex]

[tex] (-\frac{20}{17} ,\frac{5}{17} ) [/tex]=[tex] (-1.176 ,0.294 ) [/tex]

therefore

the answer is

the point is [tex] (-\frac{20}{17} ,\frac{5}{17} ) [/tex]

see the attached figure

The point [tex]\boxed{(-1.176,0.294)}[/tex] on the line [tex]y=4x+5[/tex] is the closest point to the origin.  

Further explanation:

The general form of linear function is as follows:

[tex]\boxed{y=mx+c}[/tex]

A linear function has one independent variable and one dependent variable. The independent variable is [tex]x[/tex] and the dependent variable is [tex]y[/tex].

Here, [tex]c[/tex] is the constant term and [tex]m[/tex] is the slope and gives the rate of change of dependent variable.  

The point slope form of a line is given as follows:

[tex]\boxed{y-y_{1}=m(x-x_{1})}[/tex]  

where [tex](x_{1},y_{1})[/tex] is the point on the line and [tex]m[/tex] is the slope of the line.

It is given that the equation of the line is as follows:

[tex]y=4x+5[/tex]

where [tex]4[/tex] is the slope of the line.

Consider the slope of the line [tex]y=4x+5[/tex] as [tex]m_{1}=4[/tex].

We first find the line perpendicular to the line [tex]y=4x+5[/tex] that passes through the origin as shown in Figure 1 in the attachment below.

If two lines are perpendicular then the product of their slopes is [tex]-1[/tex]  that is [tex]m_{1}\tiimes m_{2}=-1[/tex].

And [tex]m_{2}[/tex] is the slope of the perpendicular line.

Calculated the value of [tex]m_{2}[/tex] as follows:

[tex]\begin{aligned}m_{1}\times m_{2}&=-1\\m_{2}&=-\dfrac{-1}{m_{1}}\\&=\dfrac{-1}{4}\\&=-0.25\end{aligned}[/tex]

Therefore, the slope of perpendicular line is [tex]-0.25[/tex].

We have the point [tex](0,0)[/tex] on the line and the slope of the line.

Thus, the equation of line is,

[tex]\begin{aligned}y-y_{1}&=m_{2}(x-x_{1})\\y-0&=(-0.25)(x-0)\\y&=-0.25x\end{aligned}[/tex]  

The equation of the perpendicular line is as follows:

[tex]\boxed{y=-0.25x}[/tex]        .......(2)

Substitute [tex]y=-0.25x[/tex] in equation (1).

[tex]\begin{aligned}-\dfrac{x}{4}&=4x+5\\-\dfrac{x}{4}-4x&=5\\-\dfrac{17x}{4}&=5\\-17x&=5\times 4\\x&=-\dfrac{20}{17}\\x&\approx-1.176\end{aligned}[/tex]  

Now, put the value of [tex]x[/tex] in equation (2) to get the value of [tex]y[/tex] as,

[tex]\begin{aligned}y&=-0.25\times (-1.176)\\&=0.294\end{aligned}[/tex]  

Therefore, the closest point is [tex]\boxed{(-1.176,0.294)}[/tex].

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Answer details:

Grade: High school

Subject: Mathematics

Chapter: Linear equations

Keywords: Linear equations, slope of a line, equation of the line, function, real numbers, ordinates, abscissa, interval, open interval, closed intervals, semi-closed intervals, semi-open intervals, sets, range domain, codomain.

Answer:

6, 6, 2

Step-by-step explanation:

Factor 72:

72 = 2³ × 3²

Look for groups of 3 numbers whose products are 72 and their sums:

2, 4, 9: sum 15

8, 9, 1: sum 18

8, 3, 3: sum 14

6, 6, 2: sum 14

6, 4, 3: sum 13

12, 6, 1: sum 19

12, 3, 2: sum 17

18, 4, 1: sum 23

18, 2, 2: sum 22

24, 3, 1: sum 28

36, 2, 1: sum 39

72, 1, 1: sum 74

Jim knows the sum, but he can't guess the ages. that means the sum must be 14 for which there are two sets of ages which have a product of 72:

8, 3, 3 and 6, 6, 2

Then when he says "my youngest", that means one age is less than the other two. The only choice is 6, 6, 2.

Using the concept of combinations, the person can order a two course meal in 63 ways

To determine the number of ways a person can order a two-course meal from the given options of appetizers and main courses, we can use the concept of combinations.

In this case, we need to choose one appetizer and one main course. Since these choices are independent, we can use the principle of multiplication.

The number of ways to choose one appetizer from 7 options is 7, and the number of ways to choose one main course from 9 options is 9.

Therefore, the total number of ways to order a two-course meal is given by the product of these two choices:

Number of ways = 7 * 9 = 63.

Hence, a person can order a two-course meal in 63 different ways from the given options of appetizers and main courses.

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The trivial lower-bound classes for the problems provided include O(n) for finding the largest element in an array, O(n^2) for graph completeness, O(2^n) for subset generation, and O(n log n) for distinctness checking. Some of these bounds are tight, such as O(n) and O(2^n), while others may not be.

For each of the following problems, we can establish a trivial lower-bound class and discuss whether it is tight:

Where the above figure is given as the attendant derivatives F', the maximum value attained by F is 400.

It is to be noted that:

Max occurs at x=50 where F'(x) = 0.

We need to find value for F(50) using fundamental theorem of calculus:

[tex]\int_{50}^{50} F' dx = F(50) - F(20)[/tex]

= F (50) -100
F(50) = 100 + A

Where A is the estimated Area under the graph.

If we assume the shaded region is close to a right triangle then the Area is

(1/2) x 30 x 20

= 300

Thus,

F (50) = 100 + 300

= 400

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The mean thickness of the washers is 1.55 mm and the standard deviation is approximately 0.296 mm.

To find the mean of the thicknesses, we add up all the values and divide it by the total number of values. In this case, the sum of the thicknesses is 24.81. Dividing this by the total number of values, which is 16, gives us a mean of 1.55 mm.

To find the standard deviation, we need to calculate the variance first. The formula for variance is sum of (value - mean)^2 divided by n - 1, where n is the number of values. After calculating the variance, we take its square root to find the standard deviation. In this case, the variance is 0.0877, and the standard deviation is approximately 0.296 mm.