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zlopas [31]
3 years ago
10

Copy the diagram and calculate the sizes of x°, yº and zº. What is the sum of the angles of the

Mathematics
2 answers:
Dominik [7]3 years ago
5 0

Answer:

180

Step-by-step explanation:

to find angle :

x =180 - 150=30

y =180-80=100

z = 180-130=50

so, 30+50+100=180

artcher [175]3 years ago
4 0
Angle x = 30 degrees
Angle y= 100 degrees
Angle z = 50 degrees
Sum of the angles = 180 degrees
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A sequence is defined recursively by the formula f(n + 1) = –2f(n). The first term of the sequence is –1.5. What is the next ter
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3 years ago
Nola was selling tickets at the high school dance. At the end of the evening she picked up the cash box and noticed a dollar lyi
Blizzard [7]

Answer:

The $1 belongs to the cash box

Step-by-step explanation:

Given

See attachment for complete question

Required

Determine if the $1 belongs to the cash box or not

Represent singles with s and couples with c.

From the attachment, we have:

s + c = 47 --- total attendance

5s + 4c =200 --- ticket sold

Solve for s and c.

Make s the subject in (1)

s = 47-c

Substitute 47 - c for s in (2)

5(47 - c) + 4c = 200

Open bracket

235 - 5c + 4c = 200

235 - c = 200

c = 235 - 200

c = 35

This means that the total individual which makes up the couples are 35. This is not possible because couples are in 2's and the total should be an even number.

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4 0
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

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