X2 - 4x + 4 = 0 solve following quadratic equations graphically and a table

1 answer:

6 0

Let's create a table, and use arbitrary values of x between -3 and 3 (if the solution is not in this interval, we can use another one later). If we call:

$f(x)=x^2-4x+4$

We want to find the x's for with f(x) = 0

This is the table:

And now we can evaluate f(x) in each value of x to complete the table:

$\begin{gathered} f(-3)=(-3)^2-4\cdot(-3)+4=9+12+4=25 \\ f(-2)=(-2)^2-4\cdot(-2)+4=4+8+4=16 \\ f(-1)=(-1)^2-4\cdot(-1)+4=1+4+4=9 \\ f(0)=0^2-4\cdot0+4=4 \\ f(1)=1^2-4\cdot1+4=1-4+4=1 \\ f(2)=2^2-4\cdot2+4=4-8+4=0 \\ f(3)=3^2-4\cdot3+4=9-12+4=1 \end{gathered}$

The table is:

If we use this values and plot them in the cartesian plane.

We get:

And now if we plot a line that connects the points, we get the graph of the quadratic equation:

Since the problem ask us to find the value of x for which f(x) = 0, we can see both in the table and in the graph that this value is x = 2

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It appears that people who are mildly obese are less active than leaner people. One study looked at the average number of minute

NemiM [27]

Answer:

a) 10.93% probability that the mean number of minutes of daily activity of the 5 mildly obese people exceeds 420 minutes.

b) 99.22% probability that the mean number of minutes of daily activity of the 5 lean people exceeds 420 minutes.

Step-by-step explanation:

The Central Limit Theorem estabilishes that, for a random variable X, with mean $\mu$ and standard deviation $\sigma$ , a large sample size can be approximated to a normal distribution with mean $\mu$ and standard deviation $\frac{\sigma}{\sqrt{n}}$ .

Normal probability distribution

Problems of normally distributed samples can be solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$ , the zscore of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

Mildly obese

Normally distributed with mean 373 minutes and standard deviation 67 minutes. So $\mu = 373, \sigma = 67$

A) What is the probability that the mean number of minutes of daily activity of the 5 mildly obese people exceeds 420 minutes?

So $n = 5, s = \frac{67}{\sqrt{5}} = 29.96$

This probability is 1 subtracted by the pvalue of Z when X = 410.

$Z = \frac{X - \mu}{s}$

$Z = \frac{410 - 373}{29.96}$

$Z = 1.23$

$Z = 1.23$ has a pvalue of 0.8907.

So there is a 1-0.8907 = 0.1093 = 10.93% probability that the mean number of minutes of daily activity of the 5 mildly obese people exceeds 420 minutes.

Lean

Normally distributed with mean 526 minutes and standard deviation 107 minutes. So $\mu = 526, \sigma = 107$