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1 year ago

F(x)= x(x+3)(x+1)(x-4) has zeros at x=-3

Mathematics

1 answer:

anygoal [31]1 year ago

4 0

Answer: C) Sometimes positive; sometimes negative

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Explanation:

Pick a value between x = -1 and x = 0. Let's say we go for x = -0.5

Plug this into f(x)

f(x) = x(x+3)(x+1)(x-4)

f(-0.5) = -0.5(-0.5+3)(-0.5+1)(-0.5-4)

f(-0.5) = -0.5(2.5)(0.5)(-4.5)

f(-0.5) = 2.8125

We get a positive value.

This shows that f(x) is positive on the region of -1 < x < 0

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Now pick a value between x = 0 and x = 4. I'll use x = 1

f(x) = x(x+3)(x+1)(x-4)

f(1) = 1(1+3)(1+1)(1-4)

f(1) = 1(4)(2)(-3)

f(1) = -24

Therefore, f(x) is negative on the interval 0 < x < 4

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In short, f(x) is both positive and negative on the interval -1 < x < 4

It's positive when -1 < x < 0

And it's negative when 0 < x < 4

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Use the discriminant to determine how many real number solutions exist for the quadratic equations -4j+3j-28=0

aleksandr82 [10.1K]

<span>the correct equation is
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3 years ago

A manufacturing company produces steel housings for electrical equipment. The main component part of the housing is a steel trou

Mrrafil [7]

Answer:

Check the explanation

Step-by-step explanation:

(a)

H0: Population mean = 8.46

H1: Population mean is not equal to 8.46

The test statistic (t) is given by the following expression -

${t=\frac{\overline{x}-\mu_0}{s/\sqrt{n}}}$

We have calculated the sample mean (8.421) and sample standard deviation (0.0461). Therefore, the test statistic (t) will be -

t = (8.421 - 8.46)/(0.0461/sqrt(49)) = -5.92

The left-sided critical value from t-distribution (two-tailed) is -2.01. Thus the test statistic falls in the rejection region and we reject the null hypothesis at 95% confidence level and conclude that the population mean is different from 8.46 inches.

(b)

There are the following four assumptions for a single sample t-test.

The observations are in ratio scale.

The observations have been taken in such a manner that every single observation is independent and uncorrelated from the others.

There should not be any significant outliers in the sample observations.

The sample data should be approximately normal.

(c)

The first assumption is true as the data is in inches. The second assumption cannot be tested now. It will depend upon the situation and study design which we assume as correct in this case. The third assumption is checked by plotting a box plot and finding the outliers. The final assumption can be checked using the Anderson-Darling normality test.

Kindly check the graphical table in the attached images below.

(d)

The data is normal (the p-value of Anderson-Darling test in > 0.05). However, two points in the Box plot are outliers though not grossly different from the entire sample data set. So, we conclude that the assumptions are valid in this case for conducting the t-test.