Reema bought pencils for school in August. She gave

Eric works at the deli on weekends to earn extra money. He makes $10 per hour making sandwiches, and $14 per hour delivering ord

artcher [175]

Answer:

1/2(10s+14d)

he saves $53 this weekend

Step-by-step explanation:

8 0

3 years ago

Solve for A, find the exact value of each expression below.

nataly862011 [7]

$\begin{gathered} \theta=\frac{7\pi}{6} \\ 2\sin (\frac{7\pi}{6})=2(\frac{-1}{2})=-1 \\ 2\sin (\frac{7\pi}{6})=-1 \\ \\ \sin ^2(\frac{7\pi}{6})=(\frac{-1}{2})^2=\frac{1}{4} \\ \sin ^2(\frac{7\pi}{6})=\frac{1}{4} \\ \\ \sin (-\frac{7\pi}{6})=-\sin (\frac{7\pi}{6})=-(\frac{-1}{2})=\frac{1}{2} \\ \sin (-\frac{7\pi}{6})=\frac{1}{2} \end{gathered}$

4 0

1 year ago

Select the correct answer from the drop-down menu. Find the missing term. ) + (7 - 31) + (5 + 91) + 131 = 10 - 51.

pantera1 [17]

Answer:

1st option is correct

Step-by-step explanation:

-2 - 24i is the asnwer because ,to find the missing term we will take other all terms to other side and we will simplify this equation

7 0

3 years ago

The author drilled a hole in a die and filled it with a lead weight, then proceeded to roll it 200 times. Here are the observed

algol [13]

Answer:

$\chi^2 = \frac{(27-33.33)^2}{33.33}+\frac{(31-33.33)^2}{33.33}+\frac{(42-33.33)^2}{33.33}+\frac{(40-33.33)^2}{33.33}+\frac{(28-33.33)^2}{33.33}+\frac{(32-33.33)^2}{33.33} =5.860$

$p_v = P(\chi^2_{5} >5.860)=0.32$

Since the p value is higher than the significance level assumed 0.05 we FAIL to reject the null hypothesis at 5% of significance, and we can conclude that we can assume that we have equally like results.

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:

Number: 1, 2 , 3 , 4 , 5 ,6

Frequency: 27, 31, 42, 40, 28, 32

We need to conduct a chi square test in order to check the following hypothesis:

H0: The outcomes are equally likely.

H1: The outcomes are not equally likely.

The level of significance assumed for this case is $\alpha=0.05$

The statistic to check the hypothesis is given by:

$\chi^2 =\sum_{i=1}^n \frac{(O_i -E_i)^2}{E_i}$

The observed values are given:

$O_{1}=27$ $O_{2}=31$

$O_{3}=42$ $O_{4}=40$

$O_{5}=28$ $O_{6}=32$

The expected values are given by:

$E_{1} =\frac{1}{6}*200=33.33$ $E_{2} =\frac{1}{6}*200=33.33$

$E_{3} =\frac{1}{6}*200=33.33$ $E_{4} =\frac{1}{6}*200=33.33$

$E_{5} =\frac{1}{6}*200=33.33$ $E_{6} =\frac{1}{6}*200=33.33$

And now we can calculate the statistic:

$\chi^2 = \frac{(27-33.33)^2}{33.33}+\frac{(31-33.33)^2}{33.33}+\frac{(42-33.33)^2}{33.33}+\frac{(40-33.33)^2}{33.33}+\frac{(28-33.33)^2}{33.33}+\frac{(32-33.33)^2}{33.33} =5.860$

Now we can calculate the degrees of freedom for the statistic given by:

$df=Categories-1=6-1=5$

And we can calculate the p value given by:

$p_v = P(\chi^2_{5} >5.860)=0.32$

And we can find the p value using the following excel code:

"=1-CHISQ.DIST(5.860,5,TRUE)"

Since the p value is higher than the significance level assumed 0.05 we FAIL to reject the null hypothesis at 5% of significance, and we can conclude that we can assume that we have equally like results.

4 0

3 years ago

Three of thiry computers are put of service whar percent of the computers are working

Schach [20]

Answer: 10%

Step-by-step explanation:

30/3 = 10

3 0

3 years ago