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zzz [600]
1 year ago
15

2(x + 3) = x - 4 Solve this equation

Mathematics
2 answers:
a_sh-v [17]1 year ago
8 0

Hii!

\leadsto\parallel\boldsymbol{Answer.}\parallel\gets

__________________________________________________________

x=-10

__________________________________________________________

\multimap\parallel\boldsymbol{Explanation.}\parallel\gets

Let's distribute 2 first. \sf 2(x+3)== > 2x+6}

\sf 2x+6=x-4}.

Now subtract 6 from both sides.

\sf 2x=x-4-6} \\ 2x=x-10

Now we subtract x from both sides.

\sf 2x-x=-10}\\x=-10}

Hope that this helped! Best Wishes.

\textsl{Reach far. Aim high. Dream big.}

\boldsymbol{-Greetings!-}

____________________________________________________________

marissa [1.9K]1 year ago
7 0

Answer:

Its x=-10

Step-by-step explanation:

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In order to estimate the difference between the average hourly wages of employees of two branches of a department store, the fol
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Answer:

(9-8) -2.02 \sqrt{\frac{2^2}{25} +\frac{1^2}{20}}= 0.0743

(9-8) +2.02 \sqrt{\frac{2^2}{25} +\frac{1^2}{20}}= 1.926

And we are 9% confidence that the true mean for the difference of the population means is given by:

0.0743 \leq \mu_1 -\mu_2 \leq 1.926

Step-by-step explanation:

For this problem we have the following data given:

\bar X_1 = 9 represent the sample mean for one of the departments

\bar X_2 = 8 represent the sample mean for the other department

n_1 = 25 represent the sample size for the first group

n_2 = 20 represent the sample size for the second group

s_1 = 2 represent the deviation for the first group

s_2 =1 represent the deviation for the second group

Confidence interval

The confidence interval for the difference in the true means is given by:

(\bar X_1 -\bar X_2) \pm t_{\alpha/2} \sqrt{\frac{s^2_1}{n_1}+\frac{s^2_2}{n_2}}

The confidence given is 95% or 9.5, then the significance level is \alpha=0.05 and \alpha/2 =0.025. The degrees of freedom are given by:

df=n_1 +n_2 -2= 20+25-2= 43

And the critical value for this case is t_{\alpha/2}=2.02

And replacing we got:

(9-8) -2.02 \sqrt{\frac{2^2}{25} +\frac{1^2}{20}}= 0.0743

(9-8) +2.02 \sqrt{\frac{2^2}{25} +\frac{1^2}{20}}= 1.926

And we are 9% confidence that the true mean for the difference of the population means is given by:

0.0743 \leq \mu_1 -\mu_2 \leq 1.926

4 0
3 years ago
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