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3 years ago

8x= 3x + 22 What's the solution?

2 answers:

8 0

Step 1: Subtract 3x from both sides.8x−3x=3x+22−3x5x=22Step 2: Divide both sides by 5.5x5=225x=225Answer:x=225

Diano4ka-milaya [45]3 years ago

5 0

Well you subtract 3 from both sides then you get 5x and then divide it by 22 and then your answer is a decimal 4.4

You might be interested in

The scatter plot shows the number of years of experience x and the hourly pay rate y for each of 25 cashiers in Ohio.

astra-53 [7]

Answer:

a) $y= 0.95*7 +7.95 =14.6$

b) $y= 0.95*0 +7.95 =7.95$

c) For this case we can use the slope from the linear model to answer this question, the 0.95 on this case represent the increase in the hourly pay rate per each year of experience, for example if we have one year of experience the pay rate increase 0.95. So then the answer for this case is 0.95.

Step-by-step explanation:

Assuming the following question: "The scatter plot shows the number of years of experience, x, and the hourly pay rate, y, for each of 25 cashiers in Ohio. x represent the number of years of experience and y the hourly pay rate. The line adusted is: y=0.95x+7.95, to answer the questions below. Give exact answers, not rounded approximations. "

(a) What is the predicted hourly pay rate for a cashier with 7 years of experience?

For this case we can use the lineal model given:

$y = 0.95 x +7.95$

And we can replace x =7 years and we got:

$y= 0.95*7 +7.95 =14.6$

(b) What is the predicted hourly pay rate for a cashier who doesn't have any experience?

For this case we just need to replace in the linear model x=0 since that means no experience and we got:

$y= 0.95*0 +7.95 =7.95$

(c) For an increase of one year of experience, what is the predicted increase in the hourly pay rate?"

For this case we can use the slope from the linear model to answer this question, the 0.95 on this case represent the increase in the hourly pay rate per each year of experience, for example if we have one year of experience the pay rate increase 0.95. So then the answer for this case is 0.95.

6 0

3 years ago

The data set represents a progression of hourly temperature measurements.Use the regression equation to predict the temperature

zimovet [89]

I will attach google sheet that I used to find regression equation.
We can see that linear fit does work, but the polynomial fit is much better.
We can see that R squared for polynomial fit is higher than R squared for the linear fit. This tells us that polynomials fit approximates our dataset better.
This is the polynomial fit equation:
$T(h)=20.2-3.6h-0.875h^2$
I used h to denote hours. Our prediction of temperature for the sixth hour would be:
$T(6)=20.2-3.6(6)-0.875(6)^2=-32.9$
Here is a link to the spreadsheet (https://docs.google.com/spreadsheets/d/17awPz5U8Kr-ZnAAtastV-bnvoKG5zZyL3rRFC9JqVjM/edit?usp=sharing)

4 0

3 years ago

Read 2 more answers

01:59:16

Nezavi [6.7K]

Answer: E

Step-by-step explanation:

8 0

3 years ago

Read 2 more answers

How far away is a star with a parallax of 2 arc sec in parsecs? In AU?

Vesna [10]

Answer:

ear

Step-by-step explanation:

8 0

3 years ago

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

frez [133]

Answer:

1) H0: There is independence between the marital status and the diagnostic of alcoholic

H1: There is association between the marital status and the diagnostic of alcoholic

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

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

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

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

And we can calculate the p value given by:

$p_v = P(\chi^2_{2} >19.72)=5.22x10^{-5}$

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

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

Since the p value is lower than the significance level so then we can reject the null hypothesis at 5% of significance, and we can conclude that we have association 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:

Diag. 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 independence between the marital status and the diagnostic of alcoholic

H1: There is association between the marital status and the diagnostic of alcoholic

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

Part 2

The statistic to check the hypothesis is given by:

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

Part 3

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:

Diag. 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

And now we can calculate the statistic:

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

Part 4

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

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

And we can calculate the p value given by:

$p_v = P(\chi^2_{2} >19.72)=5.22x10^{-5}$

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

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

7 0

4 years ago