100 Points and Brainly.

Rationalize the denominator of sqrt -9 / (4-7i) - (6-6i) ...?

Colt1911 [192]

√(- 9 ) / (( 4 - 7 i ) - ( 6 - 6 i )) = 3 i / ( 4 - 7 i - 6 + 6 i ) =
= 3 i / ( - 2 - i ) = - 3 i / ( 2 + i ) =
$= \frac{-3i}{2 + i}* \frac{2 - i}{2 - i}= \\ = \frac{-6i+3i ^{2} }{2- i^{2} }= \\ = \frac{-3-6i}{2+1}= \frac{-3-6i}{3}=$
= - 1 - 2 i

5 0

3 years ago

Y=x+2<br> 3x+3y=6<br> solving systems by substitution

motikmotik

Answer:6

Step-by-step explanation:Reorder

2

and

−

x

-

x

.

y

=

−

x

+

2

y

=

-

x

+

2

3

x

+

3

y

=

6

3

x

+

3

y

=

6

Replace all occurrences of

y

in

3

x

+

3

y

=

6

3

x

+

3

y

=

6

with

−

x

+

2

-

x

+

2

.

y

=

−

x

+

2

y

=

-

x

+

2

3

x

+

3

(

−

x

+

2

)

=

6

3

x

+

3

(

-

x

+

2

)

=

6

Simplify

3

x

+

3

(

−

x

+

2

)

3

x

+

3

(

-

x

+

2

)

.

Tap for more steps...

y

=

−

x

+

2

y

=

-

x

+

2

6

=

6

=

6

Since

6

=

6

=

6

, the equation will always be true.

y

=

−

x

+

2

y

=

-

x

+

2

Always true

Remove any equations from the system that are always true.

y

=

−

x

+

2

6 0

3 years ago

Estimate the difference by rounding each mixed number to the nearest half or whole number. 8 4/10 - 2 1/7

Brilliant_brown [7]

The answer would be 6 9/35 which would round to 6.5

7 0

2 years ago

Read 2 more answers

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)"

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.

7 0

3 years ago

You survey your class and record each students first name, last name, and age. You then make ordered pairs of (first name, age).

ExtremeBDS [4]

Answer: Yes

Step-by-step explanation:

It is a relation because each person will most likely have a different name and you can have plenty of the same outputs just as long as the input doesn't repeat at all.

5 0

3 years ago