4[3(10-7)+(11•2)]<br><br> Help please

To move backwards on the Y-axis, what operation should she use?

statuscvo [17]

It will either be -2 or -3

7 0

3 years ago

Can someone help me with this ?

Virty [35]

Mark drove for 5 hrs
pranav drove for 3 hrs

Distance = speed x time
Pranav =50kmh x 3hrs
= 150 km
* note that they drove the same distance

Speed of mark=distance / time
=150 km / 5hrs
= 30 kmh

3 0

3 years ago

Some scientists believe alcoholism is linked to social isolation. One measure of social isolation is marital status. A study of

alexandr402 [8]

Answer:

1) H0: There is significant evidence of independence between marital status and diagnostic

H1: There is significant evidence of an association between marital status and diagnostic

2) The statistic to check the hypothesis is given by:

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

3) $\chi^2 = \frac{(21-33.143)^2}{31.143}+\frac{(37-41.429)^2}{41.429}+\frac{(58-41.429)^2}{41.429}+\frac{(59-46.857)^2}{46.857}+\frac{(63-58.571)^2}{58.571}+\frac{(42-58.571)^2}{58.571} =19.72$

4) $p_v = P(\chi^2_{2} >19.72)=0.00005222$

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

"=1-CHISQ.DIST(19.72,2,TRUE)"

Since the p values is lower than the significance level we have enough evidence to reject the null hypothesis at 5% of significance, and we can say that we have associated between the two variables analyzed.

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:

Alcoholic Undiagnosed Alcoholic Not Alcoholic Total

Married 21 37 58 116

Not Married 59 63 42 164

Total 80 100 100 280

Part 1

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

H0: There is significant evidence of independence between marital status and diagnostic

H1: There is significant evidence of an association between marital status and diagnostic

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

Part 2

The statistic to check the hypothesis is given by:

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

The table given represent the observed values, we just need to calculate the expected values with the following formula $E_i = \frac{total col * total row}{grand total}$

And the calculations are given by:

$E_{1} =\frac{80*116}{280}=33.143$

$E_{2} =\frac{100*116}{280}=41.429$

$E_{3} =\frac{100*116}{280}=41.429$

$E_{4} =\frac{80*164}{280}=46.857$

$E_{5} =\frac{100*164}{280}=58.571$

$E_{6} =\frac{100*164}{280}=58.571$

And the expected values are given by:

Alcoholic Undiagnosed Alcoholic Not Alcoholic Total

Married 33.143 41.429 41.429 116

Not Married 46.857 58.571 58.571 164

Total 80 100 100 280

Part 3

And now we can calculate the statistic:

$\chi^2 = \frac{(21-33.143)^2}{31.143}+\frac{(37-41.429)^2}{41.429}+\frac{(58-41.429)^2}{41.429}+\frac{(59-46.857)^2}{46.857}+\frac{(63-58.571)^2}{58.571}+\frac{(42-58.571)^2}{58.571} =19.72$

Part 4

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

$df=(rows-1)(cols-1)=(2-1)(3-1)=2$

And we can calculate the p value given by:

$p_v = P(\chi^2_{2} >19.72)=0.00005222$

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

"=1-CHISQ.DIST(19.72,2,TRUE)"

Since the p values is lower than the significance level we have enough evidence to reject the null hypothesis at 5% of significance, and we can say that we have associated between the two variables analyzed.

8 0

3 years ago

There are 15 copies of a popular cd left to be sold in the store this is between 1% and 1.5% of the original number of copies of

Triss [41]

Answer: The original number of CDs is between 1000 and 1500.

Step-by-step explanation:

Since we have given that

Number of copies of a popular CD left to be sold in the store = 15

According to question, this is between 15 and 1.5% of the original number of copies of the CD in the store.

Let the original number of copies of the CD in the store be 'x'.

Now, 15 is equal to 1% of the original number of copies of the CD, it becomes

$1\%\ of\ x=15\\\\\frac{1}{100}\times x=15\\\\x=15\times 100\\\\x=1500$

Or if 15 is equal to 1.5% of the original number of copies of the CD, it becomes

$1.5\%\ of\ x=15\\\\\frac{1.5}{100}\times x=15\\\\\frac{15}{1000}x=15\\\\x=\frac{15\times 1000}{15}\\\\x=1000$

Hence, the original number of CDs is between 1000 and 1500.

3 0

3 years ago

Trying to find the initial value. help needed!!

m_a_m_a [10]

Answer: dkoakdoaskdoakodkaosdkaosdaksdpkadsokfpakofspokfsdp]fokpadsfkopadsfokpfodskfpaskofpd the answer is 2 sorry.

7 0

2 years ago

4[3(10-7)+(11•2)] Help please