Which statement is true?

First, let me do the Mathematical part of that, and then I shall explain the theory behind it.

Mathematical part:
We are going to multiply 513 with 46. So the two partial products that we are going to choose are 40 and 6.

Multiply 513 with 6 first.

  513
x46
--------------------------
  18 (as 6*3 = 18)
  60 (as 6*10 = 60; In 513, the digit at tenths place is 1, so 1*10=10)
  3000 (as 6*500 = 3000; In 513, 5 is at hundredth place, so 5*100=500)
  120 (as 40*3 = 120; since 4 is at the tenth place, so 4*10=40)
  400 (as 40*10 = 400)
20000 (as 40*500 = 20000)
--------------------------
23598 (Add all of them)

Theory:
As you can see above that we have chosen the two partial products individually which are 6 and 40. Since 4 in 46 is in tenth place, we have to consider it 40 (since 4*10 = 40). One by one, we first multiply 6 with 513. Then we move to the tenth place, and multiply 513 with 40. At the end, we have added all the results we found after multiplication.

Check: If we check the multiplication result by using the calculator, we would get the same result (23598).

Another Method (instant):
513 * (40+6) = (513*40) + (513*6) = 23598.

3 0

4 years ago

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According to the University of Nevada Center for Logistics Management, 6% of all mer-

Fudgin [204]

Answer:

a) The point estimate of the proportion of items returned for the population of

sales transactions at the Houston store = 12/80 = 0.15

b) The 95% confidence interval for the proportion of returns at the Houston store = [0.0718 < p < 0.2282].

c) Yes.

We set an hypothesis and construct a test statistics. The test statistics result gives us:

Z calculated = 2.2545, and this gives us the p-value = 0.0121. We assumed 95% confident interval. Hence, the level of significance (α) = 5%. Conclusively, since the p-value ==> 0.0121 is less than (α) = 5%, the test is significant. Hence, the proportion of returns at the Houston store is significantly different from the returns for the nation as a whole.

Step-by-step explanation:

a) Point estimate of the proportion = number of returned items/ total items sold = 12/80 = 0.15.

b) By formula of confident interval:

CI(95%) = p ± Z* $\sqrt{\frac{p*(1-p)}{n} } = 0.15 \pm 1.96 *\sqrt{\frac{0.15*(1-0.15)}{80} }$ ,

CI(95%) = [0.0718 < p < 0.2282]

c) The hypothesis:

$H_{0}$ : The proportion of returns at the Houston store is not significantly different from the returns for the nation as a whole.

$H_{a}$ : The proportion of returns at the Houston store is significantly different from the returns for the nation as a whole.

The test statistics:

Z = $\frac{\hat{p} - p_{0}}{\sqrt{\frac{p*(1-p)}{n} }}$ , where $p_{0}$ is the proportion of nation returns.

Z calculated = 2.2545, and this gives us the p-value = 0.0121. We assumed 95% confident interval. Hence, the level of significance (α) = 5%. Conclusively, since the p-value ==> 0.0121 is less than (α) = 5%, the test is significant. Hence, the proportion of returns at the Houston store is significantly different from the returns for the nation as a whole.

6 0

3 years ago