3 000 000 000+200 000 000+600 000+20 000+9

Mathematics

2 answers:

Alex787 [66]2 years ago

7 0

320,062,000,9 is the answer

Leona [35]2 years ago

3 0

3200620009 is the answer to your question

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The employees of two companies were asked how many times they checked their email during weekends. The table below shows the num

tekilochka [14]

Q1 of company A = 2.5
Q3 of company A = 8
Interquatile range = (Q3 - Q1)/2 = (8 - 2.5)/2 = 5.5/2 = 2.75

Q1 of company B = 2
Q3 of company B = 5.5
Interquatile range = (5.5 - 2)/2 = 3.5/2 = 1.75

Therefore, t<span>he interquartile range for Company A employees is 2 more than the interquartile range for Company B employees.</span>

3 0

2 years ago

Read 2 more answers

A company claims that the mean weight per apple they ship is 120 grams with a standard deviation of 12 grams. Data generated fro

Veronika [31]

Answer:

There is no sufficient evidence to reject the company's claim at the significance level of 0.05

Step-by-step explanation:

Let $\mu$ be the true mean weight per apple the company ship. We want to test the next hypothesis

$H_{0}:\mu=120$ vs $H_{1}:\mu\neq 120$ (two-tailed test).

Because we have a large sample of size n = 49 apples randomly selected from a shipment, the test statistic is given by

$Z=\frac{\bar{X}-120}{\sigma/\sqrt{n}}$ which is normally distributed. The observed value is

$z_{0}=\frac{122.5-120}{12/\sqrt{49}}=1.4583$ . The rejection region for $\alpha = 0.05$ is given by RR = {z| z < -1.96 or z > 1.96} where the area below -1.96 and under the standard normal density is 0.025; and the area above 1.96 and under the standard normal density is 0.025 as well. Because the observed value 1.4583 does not fall inside the rejection region RR, we fail to reject the null hypothesis.