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3 years ago

Range includes all but what

Mathematics

1 answer:

Romashka [77]3 years ago

6 0

The range would consist of all real numbers EXCEPT for zero

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Select the correct answer from each drop-down menu.

umka21 [38]

Answer:

A)-4,8

B)-2,11

Step-by-step explanation:

A) ( -1-3)(2+5)

B) (1-3)(5+6)

7 0

3 years ago

What is the value of b?

son4ous [18]

The answer is 128. Therefore it’s C.

6 0

2 years ago

Read 2 more answers

I need help solving this I just can’t get the right answer

hodyreva [135]

Answer:

-1.99, 1.99

Step-by-step explanation:

Use a table to find the z-scores.

For the first z-score:

P(x < z) = 0.0233

z = -1.99

For the second z-score:

P(x > z) = 0.0233

P(x < z) = 1 − 0.0233 = 0.9767

z = 1.99

7 0

3 years ago

Which of the following statements is true

Eva8 [605]

The salary range at company B is greater than Company A is the only right answer

6 0

3 years ago

(b) how many measurements must be averaged to get a margin of error of ±0.0001 with 98% confidence? (round your answer up to the

goblinko [34]

Given that the scale readings of the laboratory scale is normally distributed with an unknown mean and a standard deviation of 0.0002 grams.

Part A:

If the weight is weighted 5 times and the mean weight is 10.0023 grams, the 98% confidence interval for μ, the true mean of the scale readings is given by:

$98\% \ C. \ I.=\bar{x}\pm z_{\alpha/2}\left(\frac{\sigma}{\sqrt{n}}\right) \\ \\ =10.0023\pm2.33\left(\frac{0.0002}{\sqrt{5}}\right) \\ \\ =10.0023\pm2.33(0.000089) \\ \\ =10.0023\pm0.00021 \\ \\ =(10.0023-0.00021, \ 10.0023+0.00021) \\ \\ =(10.0021, \ 10.0025)$

Thus, we are 98% confidence that the true mean of the scale readings is between 10.0021 and 10.0025.

Part B:

To get a margin of error of +/- 0.0001 with a 98% confidence, then

$\pm z_{\alpha/2}\left(\frac{\sigma}{\sqrt{n}}\right)=\pm0.0001 \\ \\ \Rightarrow2.33\left(\frac{0.0002}{\sqrt{n}}\right)=0.0001 \\ \\ \Rightarrow\left(\frac{0.0002}{\sqrt{n}}\right)= \frac{0.0001}{2.33} =0.0000429 \\ \\ \Rightarrow\sqrt{n}= \frac{0.0002}{0.0000429} =4.66 \\ \\ \Rightarrow n=4.66^2=21.7156\approx22$

Therefore, the number of <span>measurements that must be averaged to get a margin of error of ±0.0001 with 98% confidence is 22.</span>

8 0

3 years ago