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3 years ago

8 times the difference between a number and 6 ls equal to 4 times the number.

Mathematics

2 answers:

In-s [12.5K]3 years ago

6 0

Answer:

A. 12

Step-by-step explanation:

Let the number be A

8 x (A - 6) = 4 x A

Expand the left side

8 x A - 8 x 6 = 4A

8A - 48 = 4A

Add 48 to both sides

8A - 48 + 48 = 4A + 48

8A = 4A + 48

Subtract 4A from both sides

8A - 4A = 4A - 4A + 48

4A = 48

Divide both sides by 4

A = 48/4

A = 12

Sveta_85 [38]3 years ago

4 0

Answer:

the answer is 12 because 8 × 6 = 48 ÷ 4 = 12

Step-by-step explanation:

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Answer:

The 95% confidence interval would be given (0.0109;0.1651).

We are confident at 95% that the difference between the two proportions is between $0.0109 \leq p_{elementary} -p_{High school} \leq 0.1651$

Step-by-step explanation:

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".

The margin of error is the range of values below and above the sample statistic in a confidence interval.

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

$p_A$ represent the real population proportion for elementary school

$\hat p_A =\frac{226}{397}=0.569$ represent the estimated proportion for elementary school

$n_A=397$ is the sample size required for Brand A

$p_B$ represent the real population proportion for high school teachers

$\hat p_B =\frac{129}{268}=0.481$ represent the estimated proportion for high school teachers

$n_B=268$ is the sample size required for Brand B

$z$ represent the critical value for the margin of error

The population proportion have the following distribution

$p \sim N(p,\sqrt{\frac{p(1-p)}{n}})$

The confidence interval for the difference of two proportions would be given by this formula

$(\hat p_A -\hat p_B) \pm z_{\alpha/2} \sqrt{\frac{\hat p_A(1-\hat p_A)}{n_A} +\frac{\hat p_B (1-\hat p_B)}{n_B}}$

For the 95% confidence interval the value of $\alpha=1-0.95=0.05$ and $\alpha/2=0.025$ , with that value we can find the quantile required for the interval in the normal standard distribution.