The equation T^2 = A^3 shows the relationship between a planet’s orbital period, T, and the planet’s mean distance from the sun,

Answer:

Approximately 59 stacked cups.

Step-by-step explanation:

We have the following measurements:

Height of one cup = 12.5 cm,

Height of two cups stacked = 14 cm,

Height of three cups stacked = 15.5 cm,

...and so on.

This situation can be described by an arithmetic sequence,

12.5, 14, 15.5,....

The first term is defined as a = 12.5,

with a common difference of d = 1.5 cm.

Thus, the height of x stacked cups is given by

As per the problem,

h(x) = 200

⇒ 1.5x + 11 = 200

⇒ 1.5x = 189

⇒ x = 59.3333333333 ≈ 59.

Therefore, you will need approximately 59 stacked cups.

Here are 3 questions with their respective answers.

1) Find




Answer: 4.

Explanation:

This expression indicates the limit as the function f(x) approaches 2 from the right side.

You should apply the function (the line) from the right side of 2 and get as close to x = 2 as you can.

That line has an open circle at y = 4, which is the limit we are looking for.

2) Analyze the graph to see if the limit exists.

Answer:





To find each limit, utilize the function approaching from the direction of x.

It's important to note that since the two limits differ, it is concluded that the limit of the function as x approaches 2 does not exist.

3)
Answer: -1




To determine the limit as the function approaches 3 from the left, follow the line ending with an open circle at (3, -1).

Hence, the limit is -1.

We consider all workers as either full-time or part-time. 36 = 24 + 12 If there are 24 or fewer full-time workers, there must be at least 12 part-time workers. (This conclusion is based on the understanding of sums.) You can formulate the inequality in two steps. First, present and resolve an equation for full-time workers in relation to part-time workers. Then, apply the specified limit on full-time workers. This results in an inequality that can be solved for part-time workers. Let f and p represent full-time and part-time positions, respectively. f + p = 36... given f = 36 - p... subtract p to express f in terms of p. f ≤ 24......... given (36 - p) ≤ 24.... substitute for f. This gives your inequality in terms of p. 36 - 24 ≤ p.... rearranging gives p ≥ 12........ this is the solution to the inequality

Answer:

Step-by-step explanation:

<pGreetings!

a. The variable X represents the height of a Goomba, which follows a normal distribution with a mean of μ= 12 inches and a standard deviation of δ= 6 inches.

To find the probability that a Goomba picked at random has a height between 13 and 15 inches, you express it as:

P(13≤X≤15)

Considering that standard normal probability tables provide cumulative values, you can express this range as the cumulative probability up to 15 minus the cumulative probability up to 13. You'll first need to standardize these variable heights to obtain corresponding Z values:

P(X≤15) - P(X≤13)

P(Z≤(15-12)/6) - P(Z≤(13-12)/6)

P(Z≤0.33) - P(Z≤0.17)= 0.62930 - 0.56749= 0.06181

b. Now we have Y as the variable indicating the height of a Koopa Troopa. This variable also follows a normal distribution, with a mean μ= 15 inches and a standard deviation δ=3 inches.

The query concerns the probability that a Koopa Troopa stands taller than 75% of Goombas.

First step:

You need to determine the height of a randomly chosen Koopa Troopa that exceeds 75% of the Goomba population.

This entails determining the value of X corresponding to the limit below which 75% of the population falls, denoted by:

P(X ≤ b)= 0.75

Step 2:

Search the standard normal distribution for the Z value that has 0.75 beneath it:

Next, you will reverse the standardization to solve for "b"

Z= (b - μ)/δ

b= (Z*δ)+μ

b= (0.674*6)+12

b= 16.044 inches

Step 3:

With the height that identifies a Koopa Troopa taller than 75% of the Goomba population determined, compute the probability of selecting that Koopa Troopa:

P(Y≤16.044)

This time, utilize the Koopa’s average height and standard deviation to find the probability:

P(Z≤(16.044-15)/3)

P(Z≤0.348)= 0.636

The likelihood of randomly selecting a Koopa Troopa that is taller than 75% of Goombas is 63.6%

I hope this information is useful!

Thus, the algebraic expression representing the weight of coffee B is established. Given that x denotes the pounds of coffee A, the total weight for coffee A and coffee B adds up to 100 pounds, leading to an expression for coffee B's quantity.