The number of members f(x) in Joe's health club increased by 25% every year over a period of x years. The function below shows t
he relationship between f(x) and x: f(x) = 15(1.25)x Which of the following graphs best represents the function?
2 answers:
Solution:
It is established that the number of members f(x) in Joe's health club grows by 25% annually over a span of x years.
The relationship connecting f(x) and x is expressed as:
By substituting f(x) with y in the equation above
y=
To create the graph without utilizing Desmos, determine various y values corresponding to different x values. For instance, when x=1
This yields, y= = 18.75
For, x=2
y= =15 × 1.5625=23.4375
For, x=3
y= = 29.296875
Thus, for varying x values, we derive different y outcomes. Plot these points on the plane and connect them. This will illustrate the graph of the function:⇒ f(x)=
The graph is shown below.
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Answer:
Step-by-step explanation:
Considering the differential equation x^4(dy/dx) + x^3y = -sec(xy). We will solve it employing the method of separation of variables;
By substituting v and dv/dx into the previous equation, we acquire;
We then separate the variables:
The end expression provides the solution to the differential equation.
Answer:
The likelihood that Albert's sample of 64 will have a mean waiting time between 13.5 and 16.5 minutes is 0.9973.
Step-by-step explanation:
Prior concepts
A normal distribution is characterized as a "probability distribution that is symmetric around the mean, indicating that data close to the mean are more frequent than those further away".
The Z-score refers to "a statistical measurement that reflects the relationship of a value to the mean of a group, measured in standard deviations".
Let X denote the random variable of interest, and we identify its distribution:
Also, let signify the sample mean, whose distribution is:
In this case,
Solution to the problem
We seek this probability
Applying the Z-score formula to the probability results in:
To determine these probabilities, we can refer to normal distribution tables, use Excel, or a calculator.
The likelihood that Albert's sample of 64 will have a mean waiting time between 13.5 and 16.5 minutes is 0.9973.