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3 years ago

Find the product. If the result is negative, enter "-". If the result is positive, enter "+". -7(- a2 ) 2 ( -b3 ).

Mathematics

2 answers:

zhenek [66]3 years ago

8 0

Answer: +7a^4b^3

Step-by-step explanation:

m_a_m_a [10]3 years ago

5 0

Answer:

The product is positive, thus it is $\bold{+7a^4b^3}$

Step-by-step explanation:

The full question in proper notation is:

"Find the product. If the result is negative, enter "-". If the result is positive, enter "+".

$-7(-a^2)^2(-b^3)$ "

We have to work with it using Order of operations know as well as PEMDAS, thus expression inside parenthesis go first and exponents.

On this expression we have to work with exponents

$(-a^2)^2 = (-a^2)(-a^2) =a^4$

Thus we get

$-7(-a^2)^2(-b^3)=-7a^4(-b^3)$

Lastly we can work with multiplication and remembering that the multiplication of two negative signs becomes positive.

$-7(-a^2)^2(-b^3)=7a^4b^3$

So the final simplified expression is $\bold{7a^4b^3}$

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Which two values of x are roots of the polynomial below?

Julli [10]

Answer:

B. and C.

Step-by-step explanation:

1. Set the polynomial equal to zero:

x² + 5x + 7 = 0

2. Plug the given values of a, b, and c into the quadratic formula:

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

3. Solve the square root:

$x=\frac{-5+-\sqrt{25-4(1)(7)} }{2(1)}$

$x=\frac{-5+-\sqrt{25-28} }{2(1)}$

$x=\frac{-5+-\sqrt{-3} }{2(1)}$

4. Simplify the rest:

$x=\frac{-5+-\sqrt{-3} }{2}$

5. Separate the solutions:

$x=\frac{-5+\sqrt{-3} }{2} , x=\frac{-5-\sqrt{-3} }{2}$

Thus, options B and C are the answers.

hope this helps!

4 0

2 years ago

If x=4 and y=3 what is the value of 5x+2y-6 show how you got the answer.

Oksana_A [137]

Answer:

Given x = 4 and y = 3

Step-by-step explanation:

5x +2y -6 = 5*4 + 2*3 -6

= 20 + 6 - 6

7 0

3 years ago

Read 2 more answers

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likoan [24]

Answer:

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3 years ago

The library has 65,400 art books and 9,600 science book what part off all books are science books?

Mashcka [7]

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2 years ago

A personnel manager is concerned about absenteeism. She decides to sample employee records to determine if absenteeism is distri

nlexa [21]

Answer:

$df=categor-1=6-1=5$

The critical value can be founded with the following Excel formula:

=CHISQ.INV(1-0.05,5)

And we got $\chi^2_{critc}= 11.0705$

a. 11.070

And since our calculated value is lower than the critical we FAIL to reject the null hypothesis at 5% of significance

Step-by-step explanation:

Previous concepts

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".

Solution to the problem

For this case we want to test:

H0: Absenteeism is distributed evenly throughout the week

H1: Absenteeism is NOT distributed evenly throughout the week

We have the following data:

Monday Tuesday Wednesday Thursday Friday Saturday Total

12 9 11 10 9 9 60

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

The statistic to check the hypothesis is given by:

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

The table given represent the observed values, we just need to calculate the expected values with the following formula $E_i = \frac{60}{6}= 10$ and the expected value is the same for all the days since that's what we want to test.

now we can calculate the statistic:

$\chi^2 = \frac{(12-10)^2}{10}+\frac{(9-10)^2}{10}+\frac{(11-10)^2}{10}+\frac{(10-10)^2}{10}+\frac{(9-10)^2}{10}+\frac{(9-10)^2}{10}=0.8$

Now we can calculate the degrees of freedom (We know that we have 6 categories since we have information for 6 different days) for the statistic given by:

$df=categor-1=6-1=5$

The critical value can be founded with the following Excel formula:

=CHISQ.INV(1-0.05,5)

And we got $\chi^2_{critc}= 11.0705$

a. 11.070

And since our calculated value is lower than the critical we FAIL to reject the null hypothesis at 5% of significance

4 0

3 years ago