40 divided by 346 whole number

Which number line correctly shows –0.8 + 0.6? A number line going from negative 1.6 to positive 1.6. An arrow goes from negative

Phoenix [80]

Answer:

Hi there, are you sure you didn't forget any parenthesis or signs? I can't find the right answer in the options, the one that would end up at -0.2 in the end I can edit my answer if you tell me some additional option or info about the expression I need to show on the number line

Step-by-step explanation:

which number line shows correctly -0.8 + 0.6?

well lets write it in words minus eight tenths plus six tenths\

so minus:

you go 0.8 units to the left from 0,

then plus:

you go 0.6 units right from -0.8, getting to -0.2

since there is no exact answer like that we can rearrange the addends 0.6-0.8, now:

we go from 0 6 tenths units to the right and get to 0.6, then we go from 0.6 eight tenths unit to the left to get to -0.2

4 0

2 years ago

Read 2 more answers

A square sheet of paper measures 42 centimeters on each side. What is the length of the diagonal of this paper?

Illusion [34]

59.4cm
Use Pythagorean theorem to find the diagonal (^ means exponent)
a^2 + b^2 = c^2
42^2 + 42^2= c^2
1764+1764=c^2
3528=c^2
square root both sides
c= 59.39696962
round and you get 59.4 cm

5 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: