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Yakvenalex [24]
2 years ago
5

22/39 of 3/11 plz help asap :)

Mathematics
2 answers:
Natalija [7]2 years ago
7 0

Answer:

2/13

Step-by-step explanation:

To find a fraction of a fraction, you just multiply the two.

So 22/39 x 3/11 is 22x39 over 3/11

or 66/429.

Simplified would be 22/39.

Anton [14]2 years ago
4 0

Answer:

2/13

Step-by-step explanation:

22/39x3/11

2/13

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6x – 4y = -5<br>4x + 5y = -11​
kipiarov [429]

Answer:

x= -1.5, y= -1

Step-by-step explanation:

6x= -5+4y

x= (-5+4y) / 6

4((-5+4y) / 6) + 5y = -11

4(-5+4y) + 5*6y = -11*6

-20+16y+30y= -66

16y + 30y = -66+20 = -46

y(16 + 30) = -46

y= -46 / (16 + 30)

y= -46 / 46

y= -1

x= (-5+4(-1)) / 6

x= (-5-4) / 6

x= -9 /6

x= -1.5

3 0
3 years ago
Can u help with these two
Artemon [7]
The volume of a rectangular prism can be given by the form
V= b*h*w

you know Volume, base, and height but you dont know width

so divide b and h to both sides
\frac{V}{b*h} = w
and you cna solve for width

now for 6.
dividing by a number is the same as multiplying by its reciprocal

n \div 6  = n * \frac{1}{6}
8 0
3 years ago
Read 2 more answers
Solve this query plzzz​
Olin [163]

Step-by-step explanation:

\frac{\sin \:4A}{\cos \:2A}  \times \frac{1 - \cos \:2A}{1 - \cos\:4A}   = \tan \:A \\  \\ LHS =  \frac{\sin \:4A}{\cos \:2A}  \times \frac{1 - \cos \:2A}{1 - \cos\:4A}   \\  \\  =  \frac{2\sin \:2A.\cos \:2A}{\cos \:2A}   \times  \frac{1 - (2 { \cos}^{2}A - 1) }{1 - (2 { \cos}^{2}2A - 1) } \\  \\  = 2\sin \:2A   \times  \frac{1 - 2 { \cos}^{2}A  +  1}{1 - 2 { \cos}^{2}2A  + 1 } \\  \\  = 2\sin \:2A   \times  \frac{2- 2 { \cos}^{2}A  }{2 - 2 { \cos}^{2}2A   } \\  \\   = 2\sin \:2A   \times  \frac{2(1 - { \cos}^{2}A)  }{2 (1-  { \cos}^{2}2A)   } \\  \\   = 2\sin \:2A   \times  \frac{1 - { \cos}^{2}A}{1-  { \cos}^{2}2A   } \\  \\     = 2\sin \:2A   \times  \frac{ { \sin}^{2}A}{{ \sin}^{2}2A   } \\  \\    = 2  \times  \frac{ { \sin}^{2}A}{{ \sin}2A   } \\  \\   = 2  \times  \frac{ { \sin}^{2}A}{{ 2\sin}A. \cos \:   A } \\  \\   = \frac{ { \sin}A}{ \cos \:   A }  \\  \\  = tan \: A \\  \\  = RHS \\

7 0
3 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
Use the Distributive Property to write -3x-4x in factored form
Alchen [17]
-x(3+4) 

If you distribute it to all the numbers the simplest form would be -3x-4x 
5 0
3 years ago
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