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I am Lyosha [343]
3 years ago
5

The weight of an object on a particular scale is 145.2 lbs. The measured weight may vary from the actual weight by at most 0.3 l

bs. What is the range of actual weights of the object?
Mathematics
2 answers:
saveliy_v [14]3 years ago
7 0
For the answer to the question, what is the range of actual weights of the object if t<span>he weight of an object on a particular scale is 145.2 lbs. The measured weight may vary from the actual weight by at most 0.3 lbs.?
The answer to your question is </span>144.9 to 145.5
I hope my answer helped you.
Alecsey [184]3 years ago
5 0
Hello there.

<span>The weight of an object on a particular scale is 145.2 lbs. The measured weight may vary from the actual weight by at most 0.3 lbs. What is the range of actual weights of the object?

</span>144.9
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kvasek [131]
Hi!

So obviously we have to divide each side by -5. It's pretty simple! Let me show you...

-125 ≥ -135
-125/-5 ≥ -135/-5
dividing 2 negative numbers make a positive
25 ≥ 27
Now that the expression has changed, we have to flip the inequality sign.
25 ≤ 27
^the answer

Hope this helps! :)
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3 years ago
Which equation represents a circle with the center at (-3, -5) and a radius of 6 units?
bagirrra123 [75]
That would be 

( x + 3)^2 + (y + 5)^2 = 6^2

(x + 3)^2 + (y + 5)^2 = 36   answer
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Determine which value is a solution to the open sentence. 3 + 6x = 21
stealth61 [152]
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3 years ago
Families USA, a monthly magazine that discusses issues related to health and health costs, surveyed 20 of its subscribers. It fo
Harman [31]

Answer:

(a) The 90 percent confidence interval for the population mean yearly premium is ($10,974.53, $10983.47).

(b) The sample size required is 107.

Step-by-step explanation:

(a)

The (1 - <em>α</em>)% confidence interval for population mean is:

CI=\bar x\pm t_{\alpha/2, (n-1)}\times \frac{s}{\sqrt{n}}

Given:

\bar x=\$10,979\\s=\$1000\\n=20

Compute the critical value of <em>t</em> for 90% confidence level as follows:

t_{\alpha/2, (n-1)}=t_{0.10/2, (20-1)}=t_{0.05, 19}=1.729

*Use a <em>t-</em>table.

Compute the 90% confidence interval for population mean as follows:

CI=\bar x\pm t_{\alpha/2, (n-1)}\times \frac{s}{\sqrt{n}}

     =10979\pm 1.729\times \frac{1000}{\sqrt{20}}\\=10979\pm4.47\\ =(10974.53, 10983.47)

Thus, the 90 percent confidence interval for the population mean yearly premium is ($10,974.53, $10983.47).

(b)

The margin of error is provided as:

MOE = $250

The confidence level is, 99%.

The critical value of <em>z</em> for 99% confidence level is:

z_{\alpha/2}=z_{0.01/2}=z_{0.005}=2.58

Compute the sample size as follows:

MOE= z_{\alpha/2}\times \frac{s}{\sqrt{n}}

      n=[\frac{z_{\alpha/2}\times s}{MOE} ]^{2}

         =[\frac{2.58\times 1000}{250}]^{2}

         =106.5024\\\approx107

Thus, the sample size required is 107.

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3 years ago
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KatRina [158]

Answer:

23.1%

Step-by-step explanation:

11/48=0.23125

Round up for percent.

0.231x100=23.1

23.1%

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