Student identification codes at a high school are 4-digit randomly generated codes beginning with 1 letter and ending with 3 numbers. there are 26,000 possible codes. what is the probability that you will be assigned the code a123?

Answer:

(-0.55,0),(4.55,0)

To solve for the number of different possible pizzas with 7 toppings out of 11 and the arrangement of these toppings is not important, therefore we use the combination formula.

The formula for combination is:

n C r = n! / r! (n – r)!

where,

n = is the total number of toppings = 11

r = the number of toppings in a pizza = 7

Substituting the values into the equation:

11 C 7 = 11! / 7! (11 – 7)!

11 C 7 = 11! / 7! * 4!

11 C 7 = 330

Therefore there are a total different 330 pizzas with 7 different combinations toppings.

Final answer:

330 different combinations.

Explanation:

The student's question is about combinations in Mathematics, specifically how many 7-topping pizzas can be made from a choice of 11 toppings. This is a combinatorics problem, and the solution involves the use of the combination formula, which is given as:

C(n, k) = n! / (k!(n-k)!)

where n is the total number of items to choose from, k is the number of items to choose, n! denotes the factorial of n, and C(n, k) represents the number of combinations.

In this case, n is 11 (the total number of toppings) and k is 7 (the number of toppings on the pizza).

Therefore, the number of different 7-topping pizzas possible is:

C(11, 7) = 11! / (7!(11-7)!) = 11! / (7!4!) = (11x10x9x8)/(4x3x2x1) = 330

Hence, 330 different 7-topping pizzas are possible.

Ordered pairs that are solutions to the equation x-4y=4 is (4, 0) and (0, -1).

The given equation is x-4y=4.

We need to find an ordered pair that is a solution to the equation.

The solutions of linear equations are the points at which the lines or planes representing the linear equations intersect or meet each other. A solution set of a system of linear equations is the set of values to the variables of all possible solutions.

From the graph, we can observe that (4, 0) and (0, -1) are solutions.

Verification of the solution (4, 0):

4-4y=4

⇒-4y=0

⇒y=0

Verification of the solution (0, -1):

0-4y=4

⇒-4y=4

⇒y=-1

Therefore, ordered pairs that are solutions to the equation x-4y=4 is (4, 0) and (0, -1).

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The maximum value of the function y = 3 cos x will be 3.

The minimum value of the function y = 3 cos x will be - 3.

Mathematical expression is defined as the collection of the numbers variables and functions by using operations like addition, subtraction, multiplication, and division.

Given that;

The function is,

⇒ y = 3 cos x

In the interval [-2π, 2π].

Now,

Since, The function is,

⇒ y = 3 cos x

Hence, We get;

The maximum value of the function y = 3 cos x is,

⇒ y = 3 cos2π

⇒ y = 3 × 1

⇒ y = 3

And, The minimum value of the function y = 3 cos x is,

⇒ y = 3 cos(-2π)

⇒ y = 3 × - 1

⇒ y = - 3

Thus, The maximum value of the function = 3.

The minimum value of the function = - 3.

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x^2+9x+8=0

 (x+1)(x+8)=0

x+1=0

x=-1

x+8 = 0

x=-8

 solutions are -1 and -8

The domain for the function f(x) = 2x² + 5√(x + 2) is all real numbers greater than or equal to -2.

In interval notation, we can express this as:

Domain: x ∈ [-2, ∞)

Given is a function f(x) = 2x² + 5√2, we need to determine the domain of the function,

To complete the statement, we need to determine the domain for the function f(x) = 2x² + 5√(x + 2).

The domain of a function represents all the possible values of x for which the function is defined.

In this case, we need to consider two factors that can restrict the domain:

For a real number to be a valid input for the square root (√), the expression inside the square root (x + 2) must be greater than or equal to 0.

Otherwise, we would encounter the issue of taking the square root of a negative number, which is not defined in the real number system.

There are no other restrictions in the function that would cause it to be undefined for certain values of x.

Since x² is defined for all real numbers, there are no issues there.

Let's solve the inequality for the square root expression:

x + 2 ≥ 0

Subtract 2 from both sides:

x ≥ -2

So, the domain for the function f(x) = 2x² + 5√(x + 2) is all real numbers greater than or equal to -2.

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x= child

2x = adult

6.20x +9.80(2x) =593.40

6.20x +19.6x=593.40

25.80x =593.40

x=593.40/25.80 = 23

 23 childrens tickets were sold

The functions for House 1 and House 2 are [tex]\(f_1(x) = 5079.6x + 248900.4\)[/tex] and [tex]\(f_2(x) = 4666.\overline{66}x + 256000\)[/tex] respectively.

After 45 years, House 1 will be valued at $477,555, and House 2 will be valued at $465,000.

Here's the step-by-step solution with complete calculations for the given problem:

Part A: Identifying the Function Type

- House 1:

 - Year 1 to Year 2: [tex]\(259,059.60 - 253,980 = 5,079.60\)[/tex]

 - Year 2 to Year 3: [tex]\(264,240.79 - 259,059.60 = 5,181.19\)[/tex]

- House 2:

 - Year 1 to Year 2: [tex]\(263,000 - 256,000 = 7,000\)[/tex]

 - Year 2 to Year 3: [tex]\(270,000 - 263,000 = 7,000\)[/tex]

The functions are linear because the changes are constant for House 2 and almost constant for House 1.

Part B: Formulating the Equations

- House 1 Equation:

 - Slope (m): [tex]\(5079.6\)[/tex]

 - Y-intercept (c): [tex]\(248900.4\)[/tex]

 - Equation: [tex]\(f_1(x) = 5079.6x + 248900.4\)[/tex]

- House 2 Equation:

 - Slope (m): [tex]\(4666.\overline{66}\)[/tex]

 - Y-intercept (c): [tex]\(256000\)[/tex]

 - Equation: [tex]\(f_2(x) = 4666.\overline{66}x + 256000\)[/tex]

Part C: Calculating Future Values

- House 1 Value at Year 45:

[tex]- \(f_1(45) = 5079.6 \times 45 + 248900.4 = 477555\)[/tex]

- House 2 Value at Year 45:

[tex]- \(f_2(45) = 4666.\overline{66} \times 45 + 256000 = 465000\)[/tex]

These equations predict the values of the houses after 45 years based on the given data. House 1 will be valued at $477,555, and House 2 will be valued at $465,000.

total = 2372* (1+0.045)^20=

2372*1.045^20=

2372 * 2.411714025= 5720.585

she will have 5720.59 in 20 years

Answer:

Hence, the difference in Mean of two teams is:

0.52

Step-by-step explanation:

Five members of the soccer team and five members of the track team ran the 100-meter dash.

Their time is listed as:

Soccer          Track

   12.3              12.3

   13.2               11.2

   12.5               11.7

   11.3                12.2

    14.4               13.7

The mean of the soccer team is given by:

[tex]Mean_1=\dfrac{12.3+13.2+12.5+11.3+14.4}{5}\\\\Mean_1=12.74[/tex]

The mean of track team is given by:

[tex]Mean_2=\dfrac{12.3+11.2+11.7+12.2+13.7}{5}\\\\Mean_2=\dfrac{61.1}{5}\\\\Mean_2=12.22[/tex]

Hence, the Difference in Mean is:

[tex]Mean_1-Mean_2\\\\=12.74-12.22\\\\=0.52[/tex]

Hence, the difference in Mean of two teams is:

0.52

The sequence 3/4,-3/2, 3, -6 is the arithmetic sequence.

The difference between two successive terms of an arithmetic progression is known as a common difference.

(a)

We will find the common difference between each term of the given sequences by subtracting a term and its previous term.

[tex](\frac{6}{11}- \frac{-7}{11}) \neq (\frac{-5}{11} -\frac{6}{11} )\neq (\frac{4}{11} -\frac{-5}{11})[/tex]

[tex]\frac{13}{11}\neq \frac{-11}{11}\neq \frac{9}{11}[/tex]

So, option (a) is incorrect.

(b)

We will take the common difference between the terms.

[tex](\frac{-3}{5} -\frac{-3}{4} )\neq (\frac{-3}{6} -\frac{-3}{5} )\neq (\frac{-3}{7} -\frac{-3}{6} )\\[/tex]

[tex]\frac{3}{20} \neq \frac{3}{30}\neq \frac{3}{42}[/tex]

So, option (b) is also incorrect.

(c)

We will take the common difference between the terms.

[tex](2-\frac{1}{2} )= ( \frac{7}{2}-2 )= (5 - \frac{7}{2} )[/tex]

[tex]\frac{3}{2}= \frac{3}{2} = \frac{3}{2}[/tex]

Since the difference between the terms is common.

Thus, option (c) is correct.

(d)

We will take the common difference between the terms.

[tex](\frac{-3}{2}- \frac{3}{4})\neq (3-\frac{3}{2})\neq (-6-3)[/tex]

[tex]\frac{-18}{2}\neq \frac{9}{2}\neq (-9)[/tex]

So, option (d) is incorrect.

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Answer:

Option (d) is correct.

The area of triangle is 38.5 square units

Step-by-step explanation:

 Given: An obtuse triangle.

We have to find the area of this obtuse angle.

Consider the given obtuse triangle

Area of triangle = [tex]\frac{1}{2}\cdot b \cdot h[/tex]

where b = Base

h = height

Given : base = 11  units

and height = 7 units

Thus, Area of triangle = [tex]\frac{1}{2}\cdot 11 \cdot 7[/tex]

Simplify, we have,

Area of triangle = 38.5

Thus, The area of triangle is 38.5 square units

 {(3, 5), (–4, 5), (–5, 0), (1, 1), (4, 0)}

As long as there are the same x-value does not have multiple y-value results, it will be a function. This data array doesn't contain any recurring x-values. Therefore, this is a function (in simple terms of speaking).

A= apple pie

B = blueberry pie

a+b=38

a=38-b

11a + 13b =460

11(38-b) + 13b = 460

418-11b +13b = 460

2b=42

b=42/2 =21

they sold 21 blueberry pies and 17 apple pies

Answer:  the correct option is (B) 29.

Step-by-step explanation:  We are given to find the length of the hypotenuse x, if (20, 21, x) is a Pythagorean triple.

We know that

in a right-angled triangle, the lengths of the sides (hypotenuse, perpendicular, base) is a Pythagorean triple, where

[tex]Hypotenuse^2=Perpendicular^2+base^2.[/tex]

So,  for the given Pythagorean triple, we have

[tex]x^2=20^2+21^2\\\\\Rightarrow x^2=400+441\\\\\Rightarrow x^2=841\\\\\Rightarrow x=\sqrt{841}~~~~~~~~~~~~~~~~~[\textup{taking square root on both sides}]\\\\\Rightarrow x=\pm29.[/tex]

Since the length of the hypotenuse cannot be negative, so x = 29.

Thus, the length of the hypotenuse, x = 29.

Option (B) is CORRECT.

Answer:

42.06

Step-by-step explanation:

Two points (22,27) and (2,-10)

using distance formula:

[tex]d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}[/tex]

So, distance between (22,27) and (2,-10) is D

[tex]D=\sqrt{(22-2)^2+(27-(-10))^2}[/tex]

[tex]D=\sqrt{(20)^2+(37)^2}[/tex]

[tex]D=\sqrt{400+1369}[/tex]

[tex]D=\sqrt{1769}[/tex]

[tex]D=42.059[/tex]

Round off two decimal place.

[tex]D=42.06[/tex]

Hence, The distance between the points is 42.06

Answer: A cosine function would be a simpler model for the situation.

The minimum depth (low tide) occurs at

t = 0. A reflection of the cosine curve also has a minimum at t = 0.

A sine model would require a phase shift, while a cosine model does not.

Step-by-step explanation:

Using a cosine function is simpler to model the tide changes because it starts at the maximum value, aligning with the high tide occurrence.

The depth of the water at the end of a pier changes periodically due to tides. Given that low tides occur at 12:00 am and 12:30 pm with a depth of 2.5 m, and high tides occur at 6:15 am and 6:45 pm with a depth of 5.5 m, we need to model this situation with a periodic function.

Let's align this with a cosine function for simplicity. In general, the cosine function can be modeled as y = A cos(B(t - C)) + D, where:

Therefore, the function becomes: y(t) = 1.5 cos((2π/747.5)(t - 375)) + 4. Using a cosine function is simpler because it starts at the maximum value, which corresponds to the high tide.