Name the property of 5x•(4y+3x)=5x•(3x+4y)

156 rounded to the nearest hundred is?

aliina [53]

200 because 156 is closer to 200 than 100

6 0

3 years ago

40,000 is 10 times as much as

Dimas [21]

40,000 is 10 times as much as 4,000

3 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

A study of a company's practice regarding the payment of invoices revealed that an invoice was paid an average of 20 days after

morpeh [17]

Answer:

=0.1587 or 15.87%

So option A is correct answer so 15.87% of the invoices were paid within 15 days of receipt.

Step-by-step explanation:

In order to find the percent of the invoices paid within 5 days of receipt we have to find the value of Z first.

$Z=\frac{X-u}{S}$

where:

X is the random varable which in our case is 15 days

u is the mean or average value which is 20 days

S is the standard deviation which is 5 days

$Z=\frac{15-20}{5}$

Z=-1.0

We have to find Probability at Z less than -1

P(Z<-1.0) which can be written as:

=1-P(Z>1.0)

From Cumulative distribution table:

=1-(0.3413+0.5)

=0.1587 or 15.87%

So option A is correct answer so 15.87% of the invoices were paid within 15 days of receipt.

7 0

3 years ago

I need help ASAP!!!!!!

topjm [15]

Answer:

1. YES 2. NO

Step-by-step explanation:

1. Since x=0.121212... and the question is asking for 100x, we need to multiply each side by 100. 0.121212... multiplied by 100 is 12.1212... So YES, 100x is still a repeating decimal.

2. 12.121212... minus 0.121212... would just be 12. This is because if we know that all the numbers behind the decimal point is 0.121212 and so on, 12.121212... minus 0.121212... the numbers behind the decimal would cancel out. So we are left with 12. So NO, the answer for 99x is NOT a repeating decimal.

6 0

3 years ago