25 divided 5 + 7 - [ 4 x 3 ] =

Determine the next step for solving the quadratic equation by completing the square

GaryK [48]

Answer:

The two solutions are: $x=\frac{1}{2}+/-\frac{\sqrt{7} }{2}$

The next and every step are below.

Explanation:

1. $-3=-2x^2+2x$ : Given (addition property / add - 3 to both sides)

2. $-3 = - 2(x^2-x)$ : Given (commom factor - 2)

3. $-3-1/2=-2(x^2-x+1/4)$

To obtain the perfect square it was added the square of half of the coefficient of x: (1/2)² = 1/4, inside the parenthesis.

Since, the terms inside the parentthesis are multiplied by - 2, you have to add - 2 (1/4) = - 1/2 to the left side of the equation.

4. Now, you have that the trinomial x² - x + 1/4 is a square perfect trinomial which is factored as (x - 1/2)² and get the expression:

$-7/2=-2(x-1/2)^2$

5. Divide both sides by - 2 to get the next expression:

$-7/4=(x-1/2)^2$

6. The last step is to extract squere root from both sides of the equality:

$(x-1/2)=+/-\sqrt{\frac{7}{4}}\\ \\ x-1/2=+/-\frac{\sqrt{7} }{2}\\ \\ x=\frac{1}{2}+/-\frac{\sqrt{7} }{2}$

8 0

2 years ago

ASAP PLS!!!!!! LOGISTICS MODEL STUDY GUIDE!!!! 40 POINTS

Monica [59]

This all depends on the content you are studying, and what learning style works for you. One thing that works for most people is repetition. Although it sucks, it works because that’s how our brains are wired. Tell yourself, “I will not stop studying until I understand this concept or calculation,” for example. Then take a break once you are comfortable with it. Repeat again as needed. Start studying early and only study what you don't know. Try to be as consistent as possible. Good luck!

3 0

2 years ago

A study of long-distance phone calls made from General Electric's corporate headquarters in Fairfield, Connecticut, revealed the

Jet001 [13]

Answer:

a) 0.4332 = 43.32% of the calls last between 3.6 and 4.2 minutes

b) 0.0668 = 6.68% of the calls last more than 4.2 minutes

c) 0.0666 = 6.66% of the calls last between 4.2 and 5 minutes

d) 0.9330 = 93.30% of the calls last between 3 and 5 minutes

e) They last at least 4.3 minutes

Step-by-step explanation:

Problems of normally distributed samples are 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.

In this problem, we have that:

$\mu = 3.6, \sigma = 0.4$

(a) What fraction of the calls last between 3.6 and 4.2 minutes?

This is the pvalue of Z when X = 4.2 subtracted by the pvalue of Z when X = 3.6.

X = 4.2

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

$Z = \frac{4.2 - 3.6}{0.4}$

$Z = 1.5$

$Z = 1.5$ has a pvalue of 0.9332

X = 3.6

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

$Z = \frac{3.6 - 3.6}{0.4}$

$Z = 0$

$Z = 0$ has a pvalue of 0.5

0.9332 - 0.5 = 0.4332

0.4332 = 43.32% of the calls last between 3.6 and 4.2 minutes

(b) What fraction of the calls last more than 4.2 minutes?

This is 1 subtracted by the pvalue of Z when X = 4.2. So

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

$Z = \frac{4.2 - 3.6}{0.4}$

$Z = 1.5$

$Z = 1.5$ has a pvalue of 0.9332

1 - 0.9332 = 0.0668

0.0668 = 6.68% of the calls last more than 4.2 minutes

(c) What fraction of the calls last between 4.2 and 5 minutes?

This is the pvalue of Z when X = 5 subtracted by the pvalue of Z when X = 4.2. So

X = 5

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

$Z = \frac{5 - 3.6}{0.4}$

$Z = 3.5$

$Z = 3.5$ has a pvalue of 0.9998

X = 4.2

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

$Z = \frac{4.2 - 3.6}{0.4}$

$Z = 1.5$

$Z = 1.5$ has a pvalue of 0.9332

0.9998 - 0.9332 = 0.0666

0.0666 = 6.66% of the calls last between 4.2 and 5 minutes

(d) What fraction of the calls last between 3 and 5 minutes?

This is the pvalue of Z when X = 5 subtracted by the pvalue of Z when X = 3.

X = 5

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

$Z = \frac{5 - 3.6}{0.4}$

$Z = 3.5$

$Z = 3.5$ has a pvalue of 0.9998

X = 3

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

$Z = \frac{3 - 3.6}{0.4}$

$Z = -1.5$

$Z = -1.5$ has a pvalue of 0.0668

0.9998 - 0.0668 = 0.9330

0.9330 = 93.30% of the calls last between 3 and 5 minutes

(e) As part of her report to the president, the director of communications would like to report the length of the longest (in duration) 4% of the calls. What is this time?

At least X minutes

X is the 100-4 = 96th percentile, which is found when Z has a pvalue of 0.96. So X when Z = 1.75.

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

$1.75 = \frac{X - 3.6}{0.4}$