a bag contains 30 lottery balls numbered 1-30 a ball is selected replaced then another is drawn find each probability
p ( and even,then odd )
p ( 7, then a number greater than 16)
p ( a multiple of 5, then a prime number )
p ( two even number )

The correct next step for him to take in constructing a circumscribed circle for △JKL is to construct: the perpendicular bisector of JL

To construct a circumscribed circle for a triangle, the first thing we have to do is to place our compass at the center of one of the sides of the triangle.

This center point can be gotten by constructing a perpendicular bisector to a side of the triangle.

Therefore, the correct next step for him to take in constructing a circumscribed circle for △JKL is to construct: the perpendicular bisector of JL

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To translate the point P(x,y) , a units left and b units down, use P'(x−a,y−b) .

The rule for reflecting over the X axis is to negate the value of the y-coordinate of each point, but leave the x-value the same.

Given:

[tex]f(x)=x^2\\\\\text{Translating 2 units down, }\\\\g(x)=f(x)-2=x^2-2\\\\\text{On reflecting on x axis,}\\\\g(x)=x^2-2[/tex]

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To find when Rate 2 becomes cheaper, we set up an equation based on the rates given and solve for the number of miles. After solving the equation, we find that Rate 2 becomes cheaper after more than 800 miles.

The subject of your question is Mathematics, specifically, it's in the domain of linear equations. To find out the number of miles you should drive to pay less by taking rate 2, we need to determine when the total cost of rate 1 is more than rate 2.

Let's call M the number of miles you would drive. The cost for rate 1 would be $64 + $0.16 * M, and the cost for rate 2 would be $128 + $0.08 * M. To find when rate 2 is cheaper, we would set up the equation: 64 + 0.16 * M > 128 + 0.08 * M. By solving this equation for M, we can find the miles where rate 2 becomes cheaper. This equation simplifies to 0.08M > 64 and thus M > 800. So, for more than 800 miles, rate 2 would be the more cost-effective option.

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[tex] |\Omega|=6\\
|A|=3\\\\
P(A)=\dfrac{3}{6}=\dfrac{1}{2}=50\% [/tex]

Answer:  the number 82 belongs to the following subsets of the real numbers of Natural Numbers (N), Whole Numbers (W), Integers (I), Rational Numbers (Q), Real Numbers (R)

The number 82 belongs to several subsets of the real numbers as given in the following:

1. Natural Numbers (N): The natural numbers are the set of positive integers starting from 1, i.e.,(1, 2, 3,.......). Since 82 is a positive integer, it belongs to this subset.

2. Whole Numbers (W): The whole numbers are the set of non-negative integers, i.e., ( 0, 1, 2, 3,....). Since 82 is a non-negative integer, it belongs to this subset.

3. Integers (Z): The integers include all positive and negative whole numbers, including zero, i.e., ( -3, -2, -1, 0, 1, 2, 3,...). Since 82 is a whole number, it belongs to this subset.

4. Rational Numbers (Q): The rational numbers are numbers that can be expressed as the quotient of two integers [tex]\( \frac{a}{b} \)[/tex], where a and b are integers and [tex]\( b \neq 0 \)[/tex]. Since 82 can be written as [tex]\( \frac{82}{1} \)[/tex], it belongs to this subset.

5. Real Numbers (R): The real numbers include all the rational and irrational numbers. Since 82 is a rational number, it is also a real number.

The answer is D... :)

The third derivative of f(x) at x=0 is −150.

To find the third derivative of f(x) at x=0, we can utilize the Maclaurin series representation of f(x).

The Maclaurin series expansion of f(x) is given by:

[tex]f(x)=1-9 x+16 x^2-25 x^3+\ldots[/tex]

To find the third derivative, we first need to determine the general expression for the n-th term of the Maclaurin series. The general term of the series is given by:

[tex]f^{(n)}(0) \frac{x^n}{n !}[/tex]

where [tex]f^{(n)}(0)[/tex] represents the n-th derivative of f(x) evaluated at x=0.

Now, let's identify the pattern in the given series:

[tex]f(x)=1-9 x+16 x^2-25 x^3+\ldots[/tex]

The coefficients of the terms seem to be perfect squares of consecutive odd numbers. So, the n-th term of the series can be expressed as [tex](-1)^n \cdot(2 n+1)^2[/tex].

Now, let's find the third derivative of f(x) at x=0:

[tex]\begin{aligned}& f(x)=1-9 x+16 x^2-25 x^3+\ldots \\& f^{\prime}(x)=-9+32 x-75 x^2+\ldots \\& f^{\prime \prime}(x)=32-150 x+\ldots \\& f^{\prime \prime \prime}(x)=-150+\ldots\end{aligned}[/tex]

Now, evaluating f′′′(x) at x=0, we get:

[tex]f^{\prime \prime \prime}(0)=-150[/tex]

So, the third derivative of f(x) at x=0 is −150.

Complete Question:

Find the third derivative of f at x = 0 if the Maclaurin series for:

[tex]f(x) = 1 - 9x + 16x^2 - 25x^3 +.....[/tex]

Tire size affects the number of rotations, with larger tires making fewer rotations for a given distance. However, tire size doesn't directly influence when you need to replace your tires, which depends more on usage and wear factors. Changing tire size can impact vehicle performance.

The tire size directly affects the number of rotations a wheel makes in traversing a particular distance. A wheel's circumference, which is related to its size, determines how far it will travel in one rotation. A larger wheel will have a greater circumference, meaning it will cover a greater distance in one rotation. Therefore, for a given distance, larger tires will complete fewer rotations than smaller ones.

Switching to a different tire size wouldn't necessarily cause you to have to switch your tires sooner. Tire wear is primarily affected by factors like driving habits, road conditions, tire material, and maintenance, not rotational frequency. However, changing your tire size can affect vehicle mechanics and performance, including speedometer and odometer accuracy, so it's important to consult a professional before making such changes.

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I took a test?

1.5 w/out a negative.

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Also saw a lot of struggle to answer the other question,... so in case you need it.

From the graph in the figure attached to the question, The best estimate of the residual value when x = 1 on the graph is 1.5

A residual value is an indication of how far a regression line fails to reach a data point vertically. From the graph in the figure attached to the question;

We can see how the regression line lies at the midpoint between 3 to 4 on the vertical axis at x = 1. However, the data point is located at 5, so we have the distance of the residual value to be from the midpoint of 3 to 5 which is equal to 1.5

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Answer:

x + 64 = 110

x = 46

Step-by-step explanation:

You need to isolate the x, so you need to subtract 64 from 110 ( which is 46 ). By you doing that, you need to now subtract 64 from 64 itself.

Would look like this:  x + 64   = 110

                                      -  64   - 64

Which would leave you with x = 46

Periodic data is useful when analyzing trends or patterns that repeat at regular intervals over time.

Examples include analyzing sales data by month, tracking website traffic by day of the week, or monitoring seasonal fluctuations in temperature.

Example 1: Monthly Sales Data

To illustrate the usefulness of periodic data, let's consider a retail business that tracks its monthly sales over the course of a year. Suppose the sales data for January to December are as follows:

January: $50,000

February: $55,000

March: $60,000

April: $65,000

May: $70,000

June: $80,000

July: $85,000

August: $90,000

September: $95,000

October: $100,000

November: $110,000

December: $120,000

To analyze this periodic data, we can calculate the average monthly sales:

Total Sales = $50,000 + $55,000 + ... + $120,000

           = $780,000

Number of Months = 12

Average Monthly Sales = Total Sales / Number of Months

                     = $780,000 / 12

                     = $65,000

So, the average monthly sales for this retail business is $65,000.

Periodic data, such as monthly sales figures, allows businesses to identify seasonal trends, peak periods, and areas for improvement. By analyzing this data, businesses can make informed decisions about inventory management, marketing strategies, and resource allocation. In this example, calculating the average monthly sales helps the business understand its typical revenue stream and plan accordingly. Additionally, periodic data analysis enables businesses to compare performance across different time periods and track progress towards goals. Therefore, periodic data is essential for strategic planning and optimizing business operations.

Complete question:

When is periodic data useful? Give examples to support your answer.