$\frac{2/3x+20}{2} =\frac{2}{3} x+3\\\frac{2/3x+20}{2} - 3 = \frac{2}{3}x\\\frac{2/3x+20}{2} - \frac{6}{2} = \frac{2}{3}x\\\frac{2/3x+20-6}{2} = \frac{2}{3}x\\\frac{2/3x+14}{2} = \frac{2}{3}x\\2(\frac{2/3x+14}{2}) = 2(\frac{2}{3}x)\\2/3x+14=\frac{4}{3}x\\14=\frac{4}{3}x-\frac{2}{3}x\\14=\frac{2}{3}x\\\frac{14}{1}/\frac{2}{3}=x\\\frac{14}{1}*\frac{3}{2}=x\\7*3=x\\x=21$

The number is 21.

4 0

3 years ago

Good night, logging out !

musickatia [10]

Good night :). Sleep tight!

5 0

2 years ago

The owner of a movie theater wants to survey theater customers about their favorite types of movies.

Nutka1998 [239]

C) 50 randomly selected people leaving the movie teather

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2 years ago

Read 2 more answers

A certain change in a process for manufacture of component parts is being considered. Samples are taken using both the existing

Katyanochek1 [597]

Answer:

a) The confidence interval is $0.00\leq\pi_1-\pi_2\leq0.02$ .

Step-by-step explanation:

We have to calculate a confidence interval (CI) of a difference of proportions.

For the existing procedure, the proportion is:

$p_1=75/1500=0.05$

For the new procedure, the proportion is:

$p_2=80/2000=0.04$

To calculate the CI, we need to estimate the standard deviation

$s=\sqrt{\frac{p_1(1-p_1)}{n_1} +\frac{p_2(1-p_2)}{n_2} }\\\\s=\sqrt{\frac{0.05(1-0.05)}{1500} +\frac{0.04(1-0.04)}{2000} }=\sqrt{ 0.000032 + 0.000019 }=\sqrt{ 0.000051 }\\\\s=0.007$

For a 90% CI, the z-value is 1.64.

Then, the CI is:

$(p_1-p_2)-z*\sigma \leq\pi_1-\pi_2\leq(p_1-p_2)+z*\sigma \\\\(0.05-0.04)-1.64*0.007\leq\pi_1-\pi_2\leq(0.05-0.04)+1.64*0.007\\\\0.01-0.01\leq\pi_1-\pi_2\leq0.01+0.01\\\\0.00\leq\pi_1-\pi_2\leq0.02$

6 0

2 years ago