You deposit $3000 each year into an account earning 7% interest compounded annually. How much will you have in the account in 20 years?

Hundreds digit as ones digit. The value of my tens digit is 50. The value of my ones digit is 4,then the number is 454.

The system which deals with numbers and decimals is called number system.

The digit at ones place=4

The digit at tens place=50

hundreds digit as ones digit=400

Therefore the number is 400+50+4

=454

Therefore if same hundreds digit as ones digit. The value of my tens digit is 50. The value of my ones digit is 4. Then 454 is the Number.

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Answer:

its devon

Step-by-step explanation:

The probability that at least 2 people have the same birthday is 29.37%

Probability is the likelihood of an event occurring. Probability is the number of ways of achieving success. Probability is also the total number of possible outcomes.

Suppose there are 30 people at a party. Do you think any two share the same birthday?

Let's use the random-number table to simulate the birthdays of the 30 people at the party, ignoring leap year.

Let's assume that the year has 365 days. number the days, with 1 representing January 1, 2 representing January 2, and so forth, with 365 representing December 31.

Draw a random sample of 30 days (with replacement). These days represent the birthdays of the people at the party. Would you expect any two of the birthdays to be the same?

[tex]1^{st} people = \frac{365}{365} \\ 2^{nd} people = \frac{364}{365} \\ 3^{rd} people = \frac{363}{365}[/tex]

For 30 people

365! = 365*364*363*...336

So

[tex]= \frac{365*364*363*...336}{(365^{30})}  =  \frac{365!}{(365^{30})}[/tex]

[tex]\frac{365!}{(365^{30})}[/tex] [tex]= \frac{365!/(365-30)!}{365^{30}}[/tex]

[tex]= \frac{365!/335!}{365^{30}} \\ = 0.2937 = 29.37%[/tex]

The probability that at least 2 people have the same birthday is 29.37%

Grade:  9

Subject:  mathematics

Chapter:  probability

Keywords: the same birthday, probability, random sample, party, simulate

Exponential notation is used to express very large and very small numbers as a product of two numbers. We can cube the digit term in the usual way and multiply the exponent of the exponential term by 3 to write expressions in exponential form using the digit 3 and 5.

Exponential notation is used to express very large and very small numbers as a product of two numbers. In the case of writing expressions in exponential form using the digit 3 and 5, we can cube the digit term in the usual way and multiply the exponent of the exponential term by 3. For example, if we want to write 3 to the power of 5 in exponential form, it can be written as 3³ x 5³ = 243 x 125. Therefore, there are multiple different ways to use the digits 3 and 5 to write expressions in exponential form.

Final answer:

To write expressions in exponential form using the digits 3 and 5, follow the rules of exponential arithmetic.

Explanation:

To write expressions in exponential form using the digits 3 and 5, we can follow the rules of exponential arithmetic. In exponential form, a number is written as the digit term multiplied by 10 raised to the power of the exponential term.

For example, if we want to write 35 in exponential form, we can write it as 3×105. If we want to write 53 in exponential form, it would be 5×103.

By using these rules, you can create different expressions by combining the digits 3 and 5 with different exponential terms.

Answer:

I'm pretty sure the answer for this is ±[tex]\frac{2\sqrt{2} }{3}[/tex]

Step-by-step explanation:

I did the assignment.

35 pounds of $0.20 coffee beans and 85 pounds of $0.68 coffee beans create a 120-pound mixture valued at $0.54 per pound when mixed. Solving the system of linear equations confirms these amounts as reasonable and correct.

To solve the problem of determining how much of each type of coffee bean is used in a mixture, we can use a system of equations to model the scenario. Let's define x as the number of pounds of the $0.20 coffee beans and y as the number of pounds of the $0.68 coffee beans. Given the total weight of the mixture is 120 pounds, we can formulate our first equation:

x + y = 120

The total value of the coffee mixture is $0.54 per pound, hence the second equation representing the total value of the mixture is:

0.20x + 0.68y = 0.54(120)

By solving this system of equations, we can find the precise quantities of each type of bean used in the mixture.

Step 1: Multiply the second equation by 100 to clear the decimals for easier calculation:

20x + 68y = 54 times 120

Step 2: Solve for one of the variables, let's say x:

x = 120 - y

Step 3: Substitute x in the second equation:

20(120 - y) + 68y = 54 times 120

This simplifies to:

2400 - 20y + 68y = 6480

48y = 4080

Step 4: Solve for y:

y = 4080 / 48

y = 85

Step 5: Substitute y back into the first equation to find x:

x = 120 - 85

x = 35

If 35 pounds of the $0.20 beans and 85 pounds of the $0.68 beans are mixed together, we achieve the desired mixture weighing 120 pounds valued at $0.54 per pound. This solution is reasonable because the sum of the weights equals the total weight of the mixture, and the calculated total value per pound matches the desired value.

To solve the equation 67 times a number minus 58 is equal to 36 less than the number, we can use algebraic manipulation to isolate the variable x. The solution is x = 1/3.

To solve the equation, we can set up the equation as follows:

67x - 58 = x - 36

Next, we can simplify the equation by combining like terms:

67x - x = 58 - 36

66x = 22

Finally, we can solve for x by dividing both sides of the equation by 66:

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Answer:  The required inequality is [tex]35+0.65x\leq 45[/tex] and teh maximum number of times that Chelsea can play the game is 15.

Step-by-step explanation:  Given that Chelsea had $45 to spend at the fair. she spends $20 on admissible and $15 on snacks. She wants to play a game that costs $0.65 per game.

We are to write an inequality  to find the maximum number of times, x, Chelsea can play games.

Also, to find the maximum number of times she can play he game.

According to the given information, the inequality can be written as follows :

[tex]20+15+0.65\times x\leq 45\\\\\Rightarrow 35+0.65x\leq 45.[/tex]

And, the solution of the above inequality is as follows :

[tex]35+0.65x\leq 45\\\\\Rightarrow 0.65x\leq 45-35\\\\\Rightarrow 0.65x\leq 10\\\\\Rightarrow x\leq \dfrac{10}{0.65}\\\\\Rightarrow x\leq 15.38.[/tex]

Thus, the maximum number of times Chelsea can play the game is 15.

Answer:

10.25 inches in the sketch

Step-by-step explanation:

Since we are talking about scales drawing, we can solve this problem using proportions, the ratio John will use is 25:1. We will write the proportion using this information and then we will solve for x.

[tex]\frac{25}{1} =\frac{256.25}{x} \\x=256.25/25\\x=10.25[/tex]

Therefore, the building will be 10.25 inches in the sketch.

The work done by the weightlifter to lift a barbell and a chain with a total weight of 2,200 N through a vertical distance of 1 meter is 2,200 joules.

The question involves calculating the work done when a weightlifter lifts a barbell from one height to a higher height. To find the work done, we can use the formula W = F  imes d  imes  ext{cos}(\theta), where W is the work, F is the force, d is the distance through which the force acts, and \theta is the angle between the force and the direction of motion. Since the weightlifter is lifting vertically, the angle \theta is 0 degrees, and cos(0) is 1, simplifying the formula to W = F  imes d.

The total weight of the barbell and the chain is the sum of both weights, which is 1,700 N + 500 N = 2,200 N. The barbell is initially 1 meter above the ground and needs to be lifted to a height of 2 meters, so the distance d is 2 m - 1 m = 1 m. Therefore, the work done to lift the barbell and chain is W = 2,200 N  imes 1 m = 2,200 J.

The average rate of change for the function is 6 on the interval [10, 60].

Lagrange mean value theorem states that, if a function f is continuous over the closed interval [a, b] and differentiable over the open interval (a, b), then there must be at least one point c in the interval (a, b) where the slope of the tangent at the point c is equal to the slope of the tangent through the curve's endpoints, resulting in the expression f'(c) = {F(b) -F(a)}/(b-a)

According to the attached function graph,

At point x = 10,

F(10) = 400

At point x = 60,

F(60) = 700

Since the formula for the average rate of change of the function between x = a and x = b is,

The average rate of change = {F(b) -F(a)}/(b-a)

Here a = 10, b = 60 and F(10) = 400, F(60) = 700

Substitute the values in the formula,

So the average rate of change = (700 - 400)/(60 - 10)

The average rate of change = 300/50.

The average rate of change = 6

Hence, the average rate of change for the function is 6 on the interval [10, 60].

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Your question is incomplete, probably the missing graph is:

To find the average rate of change of f on the interval [10, 60], calculate the difference between the values of f at these two points and divide it by the difference between the two points.

To find the average rate of change of f on the interval [10, 60], we need to calculate the difference between the values of f at these two points and divide it by the difference between the two points. Let's assume that f(10) = a and f(60) = b. The average rate of change of f on the interval [10, 60] is then given by (b-a)/(60-10). To round the answer to the nearest integer, we can use the nearest neighbor rounding method.