Please help we’ve been stuck in this for a while

Mathematics

1 answer:

In-s [12.5K]3 years ago

6 0

If you plug in 5 for x, the radical ends up being (5-1) which is 4. The square root of 4 is 2, and the plus/minus symbol means the possible answers are 2 OR -2. When you plug in 1 for x the radical becomes (1-1) or zero. The square root of 0 is zero, and plus/minus zero are the same. This IS NOT a function because for x=10 and x=5, there are TWO DIFFERENT y values (each “x” can only have one y in a function)

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Calculate the area, with the appropriate number of significant figures, of a rectangular room that is 14.10 m wide by 15.0 m lon

Paladinen [302]

<h3>Answer: Choice A) 212 square meters</h3>

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Explanation:

14.10 has four sig figs. The 0 at the end is significant. This measurement is accurate to the hundredths place. If the author wrote 14.1, then it would be accurate to the tenths place with three sig figs.

15.0 has three sig figs. The 0 is significant. If they wrote 15 or 15. then we'd be dealing with 2 sig figs instead; however they placed the 0 there to tell the reader "this measurement is accurate to the tenths place"

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Multiply the values given

14.10*15.0 = 211.5

Round that to three sig figs. We pick the smaller sig fig count to round to since we cannot be certain it is accurate to four sig figs. In a sense, we pick the weakest link and use that to determine rounding when it comes to multiplication and division.

This means 211.5 rounds to 212

6 0

4 years ago

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.