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3 years ago

PLEASE HELP me!! will give brainliest!! (:

Mathematics

2 answers:

Cloud [144]3 years ago

5 0

Answer:

The answer to your question is: Yes, x - 5 is a factor of the polynomial.

Step-by-step explanation:

To answer this question you need to divide the polynomial by the factor and if there is nothing left, they are factor.

x³ -4x² -7x +10

x -5

x² + x -2

x-5 x³ -4x² -7x +10

-x³ + 5x²

x² - 7x

-x² + 5x

-2x + 10

+2x -10

0 0 These zeros tell us that the linear

expression is a factor of the polynomial

rjkz [21]3 years ago

3 0

Hey!

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Solution:

$f(x)= \frac{x^3-4x^2-7x+10}{x-5}\\f(x)= x^2 + x - 2\\f(x)= (x^2 + x - 2)(x-5)\\f(x)= (x+2) (x-1) (x-5)$

-------------------------------------------------

Answer:

$f(x) = (x+2) (x-1) (x-5)$

x - 5 is a factor of this polynomial.

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Hope This Helped! Good Luck!

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Help please and thank you

Sidana [21]

Answer:

It's discrete and non-linear

Step-by-step explanation:

it is not continuous and linear because it does not form a line going in one direction the entire time

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3 years ago

There is one error in one of five blocks of a program. To find the error, we test three randomly selected blocks. Let X be the n

Molodets [167]

The probability, or expectation value for X in a block, equals the number of occurrences of X in all the blocks divided by the total number of blocks

well E(X)=µ and

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7 0

3 years ago

Some scientists believe alcoholism is linked to social isolation. One measure of social isolation is marital status. A study of

frez [133]

Answer:

1) H0: There is independence between the marital status and the diagnostic of alcoholic

H1: There is association between the marital status and the diagnostic of alcoholic

2) The statistic to check the hypothesis is given by:

$\sum_{i=1}^n \frac{(O_i -E_i)^2}{E_i}$

3) $\chi^2 = \frac{(21-33.143)^2}{33.143}+\frac{(37-41.429)^2}{41.429}+\frac{(58-41.429)^2}{41.429}+\frac{(59-46.857)^2}{46.857}+\frac{(63-58.571)^2}{58.571}+\frac{(42-58.571)^2}{58.571} =19.72$

4) $df=(rows-1)(cols-1)=(3-1)(2-1)=2$

And we can calculate the p value given by:

$p_v = P(\chi^2_{2} >19.72)=5.22x10^{-5}$

And we can find the p value using the following excel code:

"=1-CHISQ.DIST(19.72,2,TRUE)"

Since the p value is lower than the significance level so then we can reject the null hypothesis at 5% of significance, and we can conclude that we have association between the two variables analyzed.

Step-by-step explanation:

A chi-square goodness of fit test "determines if a sample data matches a population".

A chi-square test for independence "compares two variables in a contingency table to see if they are related. In a more general sense, it tests to see whether distributions of categorical variables differ from each another".

Assume the following dataset:

Diag. Alcoholic Undiagnosed Alcoholic Not alcoholic Total

Married 21 37 58 116

Not Married 59 63 42 164

Total 80 100 100 280

Part 1

We need to conduct a chi square test in order to check the following hypothesis:

H0: There is independence between the marital status and the diagnostic of alcoholic

H1: There is association between the marital status and the diagnostic of alcoholic

The level os significance assumed for this case is $\alpha=0.05$

Part 2

The statistic to check the hypothesis is given by:

$\sum_{i=1}^n \frac{(O_i -E_i)^2}{E_i}$

Part 3

The table given represent the observed values, we just need to calculate the expected values with the following formula $E_i = \frac{total col * total row}{grand total}$

And the calculations are given by:

$E_{1} =\frac{80*116}{280}=33.143$

$E_{2} =\frac{100*116}{280}=41.429$

$E_{3} =\frac{100*116}{280}=41.429$

$E_{4} =\frac{80*164}{280}=46.857$

$E_{5} =\frac{100*164}{280}=58.571$

$E_{6} =\frac{100*164}{280}=58.571$

And the expected values are given by:

Diag. Alcoholic Undiagnosed Alcoholic Not alcoholic Total

Married 33.143 41.429 41.429 116

Not Married 46.857 58.571 58.571 164

Total 80 100 100 280

And now we can calculate the statistic:

$\chi^2 = \frac{(21-33.143)^2}{33.143}+\frac{(37-41.429)^2}{41.429}+\frac{(58-41.429)^2}{41.429}+\frac{(59-46.857)^2}{46.857}+\frac{(63-58.571)^2}{58.571}+\frac{(42-58.571)^2}{58.571} =19.72$

Part 4

Now we can calculate the degrees of freedom for the statistic given by:

$df=(rows-1)(cols-1)=(3-1)(2-1)=2$

And we can calculate the p value given by:

$p_v = P(\chi^2_{2} >19.72)=5.22x10^{-5}$

And we can find the p value using the following excel code:

"=1-CHISQ.DIST(19.72,2,TRUE)"

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