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Tresset [83]
2 years ago
7

Write a polynomial function of least degree with integral coefficients that has the given zeros. –2, –3,3 – 6i

Mathematics
1 answer:
Anuta_ua [19.1K]2 years ago
5 0

(x-(-2))(x-(-3))(x-(3-6i)(x-(3+6i))=\\(x+2)(x+3)(x-3+6i)(x-3-6i)=\\(x^2+3x+2x+6)((x-3)^2+36)=\\(x^2+5x+6)(x^2-6x+9+36)=\\(x^2+5x+6)(x^2-6x+45)=\\x^4-6x^3+45x^2+5x^3-30x^2+225x+6x^2-36x+270=\\x^4-x^3+21x^2+189x+270

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This question is incomplete, the complete question is;

Research indicates that many Americans do not save enough for retirement, on average. A group of economists at the Federal Reserve Bank of St. Louis are interested in conducting a study to examine whether Americans between the ages of 55 and 65 have saved too little. The analysts obtain a random sample of Americans in this age group and proceed to test if there is evidence of insufficient savings from these data. The variable considered is the total amount of savings for each individual, reported in thousands of US dollars (i.e. if value 3 appears in the data set, it stands for $3,000.)

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Assume that financial specialists suggest that the minimum level of retirement savings should be 1.5 million US dollars. Use the JMP output to report the value of the test statistic used to gather evidence against the null hypothesis. Report your answer as a number ( no symbols ) and round up to two decimal places.

Test mean

Hypothesized value : 1500

Actual Estimate        : 1613.07

DF                             : 251

Std Dev                    : 782.626

t Test

prob > |t|  0.0226*

Answer:

the value of the test statistic used to gather evidence against the null hypothesis is 2.29

Step-by-step explanation:

Given the data in the question,

we determine Test statistics of using the formula

t = (x" - μ) / (s/√n)

where;

μ is the theoretical mean ( 1500 )

x" is the observed mean ( 1613.07)

s is the standard deviation ( 782.626 )

n is the sample size ( DF = n - 1, n = DF + 1 , n = 251 + 1 = 252 )

so we substitute

t = (1613.07 - 1500 ) / (782.626 / √252)

t = 113.07 / 49.3008

t = 2.29

Therefore, the value of the test statistic used to gather evidence against the null hypothesis is 2.29

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Answer:

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