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Lisa [10]
3 years ago
10

What is the inverse function of f(x)=4x-7

Mathematics
2 answers:
evablogger [386]3 years ago
5 0

To find the inverse, turn the equation into x=4y-7 and solve for y. The inverse would be y=1/4x+7/4, or f(x)=1/4x+7/4. I hope this helps!

Serjik [45]3 years ago
3 0
<h3>Hey there! </h3><h3>f(x) ("f" of "x") is usually the "y" variable of the mathematical problem, same thing with g(x) ("g" of "x"), they are both part of the "y-intercepts" </h3><h3>So, basically , the inverse of f(x) = 4x-7 would most likely be: y =4x -7</h3><h3>Good luck on your assignment and enjoy your day! </h3><h3>~LoveYourselfFirst:)</h3>
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2 years ago
The heights of a random sample of 50 college students showed a mean of 174.5 centimeters and a standard deviation of 6.9 centime
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Answer:

Step-by-step explanation:

Hello!

For me, the first step to any statistics exercise is to determine what is the variable of interest and it's distribution.

In this example the variable is:

X: height of a college student. (cm)

There is no information about the variable distribution. To estimate the population mean you need a variable with at least a normal distribution since the mean is a parameter of it.

The option you have is to apply the Central Limit Theorem.

The central limit theorem states that if you have a population with probability function f(X;μ,δ²) from which a random sample of size n is selected. Then the distribution of the sample mean tends to the normal distribution with mean μ and variance δ²/n when the sample size tends to infinity.

As a rule, a sample of size greater than or equal to 30 is considered sufficient to apply the theorem and use the approximation.

The sample size in this exercise is n=50 so we can apply the theorem and approximate the distribution of the sample mean to normal:

X[bar]~~N(μ;σ2/n)

Thanks to this approximation you can use an approximation of the standard normal to calculate the confidence interval:

98% CI

1 - α: 0.98

⇒α: 0.02

α/2: 0.01

Z_{1-\alpha /2}= Z_{1-0.01}= Z_{0.99} =2.334

X[bar] ± Z_{1-\alpha /2} * \frac{S}{\sqrt{n} }

174.5 ± 2.334* \frac{6.9}{\sqrt{50} }

[172.22; 176.78]

With a confidence level of 98%, you'd expect that the true average height of college students will be contained in the interval [172.22; 176.78].

I hope it helps!

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