X-1/x-1=1 what is the answer????

Consider the following pair of equations:

miv72 [106K]

$3x + 3 = x - 1 \\ 3x - x = - 1 - 3 \\ 2x = - 4 \\ x = \frac{ - 4}{2 } = - 2$
Take X=-2 into y
$y = x - 1 \\ y = - 2 - 1 \\ y = - 3$
(x, y) =(-2,-3)

6 0

3 years ago

3.47472 rounded to the nearest 10th

Neporo4naja [7]

Answer:

It would be 0 its nowhere near a ten

Step-by-step explanation:

hope i helped

8 0

3 years ago

Read 2 more answers

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

What is 2x^2-x-15 in vertex form

In-s [12.5K]

Answer:

(1/4,-121/8)

Step-by-step explanation:

5 0

2 years ago

Read 2 more answers

How do you solve these?

Vinil7 [7]

For the first one, the answer is -1 1/5

for the second one, the answer is 2.75

for the third one and fourth one,
To divide fractions take the reciprocal (invert the fraction) of the divisor and multiply the dividend. This is the quickest technique for dividing fractions. The top and bottom are being multiplied by the same number and, since that number is the reciprocal of the bottom part, the bottom becomes one

6 0

2 years ago