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guapka [62]
2 years ago
3

The equation x3 + 3x2 + 5x + 15 = 0 has ______ real and _______ imaginary solution(s).

Mathematics
2 answers:
egoroff_w [7]2 years ago
7 0
The equation x3 + 3x2 + 5x + 15 = 0 has ⇒ 1 real and ⇒ 2 <span>imaginary solution(s). </span>
arlik [135]2 years ago
5 0

Answer:

Given Equation has 1 real Solution and 2 Imaginary Solution.  

Real Solution of given Equation is -3 and Imaginary solution are \sqrt{5}i\:\:and\:\:-\sqrt{5}i.

Step-by-step explanation:

Given Equation : x³ + 3x² + 5x + 15 = 0

let p(x) =  x³ + 3x² + 5x + 15

From polynomial factor theorem,

put x = -3

p(-3) = (-3)³ + 3(-3)² + 5(-3) + 15

        = -27 + 27 - 15 + 15

       = 0

So, -3 is one of the zero of the polynomial and solution of the equation.

Also, x + 3 is factor of p(x)

We get another factor by dividing p(x) by x + 3

that is x² + 5

so, to get other solution of given equation we put this factor equal to 0

x² + 5 = 0

x² = -5

x = ± √-5

x = + √5 × √-1 and   x = - √5 ×√-1

x=\sqrt{5}i\:\:and\:\:x=-\sqrt{5}i

Therefore, Given Equation has 1 real Solution and 2 Imaginary Solution.  

Real Solution of given Equation is -3 and Imaginary solution are \sqrt{5}i\:\:and\:\:-\sqrt{5}i.

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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 -

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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.

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2 years ago
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zhuklara [117]

Answer:

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Step-by-step explanation:


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rjkz [21]

Answer:

option B

Step-by-step explanation:

The symmetric distribution have close or approximate values of median and mean. In the given scenario mean is 48.5 and median is 33. We can see that the both values significantly differs. Also, given mean is greater than median(48.5>33) . Thus, the distribution is positively skewed and not symmetrical.

Hence, the distribution is skewed because mean and median significantly differ from each other.

3 0
2 years ago
If a1 = 6 and an = 3 + 2(an-1), then a2 equals<br><br> no links
Salsk061 [2.6K]

Answer:

15

Step-by-step explanation:

Given that a1 = 6

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