What is the distributive property to express 16 48?

Mathematics

1 answer:

Norma-Jean [14]3 years ago

5 0

So, I will un-distribute the GCF of the two numbers. 16(1 + 3) Distributing we get: 16(1) + 16(3) 16 + 48

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

And replacing we got:

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And the confidence interval for the difference of means would be given by:

$0.2903 \leq \mu_1 -\mu_2 \leq 0.5097$

Step-by-step explanation:

Previous concepts

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".

We have the following data given:

$\bar X_1 = 9.2 , \bar X_2 = 8.8, \sigma_1= 0.3, n_1 = 27, \sigma_2 = 0.1, n_2 = 30$

In order to find the critical value we need to take in count that we are finding the interval for a proportion, so on this case we need to use the z distribution. Since our interval is at 93% of confidence, our significance level would be given by $\alpha=1-0.93=0.07$ and $\alpha/2 =0.035$ . And the critical value would be given by: