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3 years ago

Solve: 1/3(−6q+4)=−1/12(4q+4)

Mathematics

2 answers:

STatiana [176]3 years ago

6 0

Answer:

Combine multiplied terms into a single fraction

13(−6+4)=−112(4+4)

13(−6q+4)=−112(4q+4)\frac{1}{3}(-6q+4)=\frac{-1}{12}(4q+4)31(−6q+4)=12−1(4q+4)

1(−6+4)3=−112(4+4)

1(−6q+4)3=−112(4q+4)\frac{1(-6q+4)}{3}=\frac{-1}{12}(4q+4)31(−6q+4)=12−1(4q+4)

Multiply by 1

Combine multiplied terms into a single fraction

Distribute

Multiply all terms by the same value to eliminate fraction denominators

Cancel multiplied terms that are in the denominator

Distribute

Subtract

161616

from both sides of the equation

Simplify

Add

4q4q4q

to both sides of the equation

Simplify

Divide both sides of the equation by the same term

Simplify

Show less

Solution

Step-by-step explanation:

Fittoniya [83]3 years ago

4 0

Answer:

q=1

Step 1: Simplify both sides of the equation.

Step 2: Add 1/3q to both sides.

Step 3: Subtract 4/3 from both sides.

Step 4: Multiply both sides by 3/(-5).

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In a sample of 70 stores of a certain company, 62 violated a scanner accuracy standard. It has been demonstrated that the condi

uysha [10]

Answer:

We need at least 243 stores.

Step-by-step explanation:

In a sample with a number n of people surveyed with a probability of a success of $\pi$ , and a confidence level of $1-\alpha$ , we have the following confidence interval of proportions.

$\pi \pm z\sqrt{\frac{\pi(1-\pi)}{n}}$

In which

z is the zscore that has a pvalue of $1 - \frac{\alpha}{2}$ .

The margin of error of the interval is:

$M = z\sqrt{\frac{\pi(1-\pi)}{n}}$

For this problem, we have that:

$n = 70, p = \frac{62}{70} = 0.886$

95% confidence level

So $\alpha = 0.05$ , z is the value of Z that has a pvalue of $1 - \frac{0.05}{2} = 0.975$ , so $Z = 1.96$ .

Determine the number of stores that must be sampled in order to estimate the true proportion to within 0.04 with 95% confidence using the large-sample method.

We need at least n stores.

n is found when M = 0.04. So

$M = z\sqrt{\frac{\pi(1-\pi)}{n}}$

$0.04 = 1.96\sqrt{\frac{0.886*0.114}}{n}}$