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FrozenT [24]
3 years ago
6

Please help me thank you very much

Mathematics
2 answers:
olasank [31]3 years ago
5 0
Question 1:

Let's use the Pythagorean theorem.
a=12
b=15
c=20

If c^2\ \textless \ a^2+b^2, then the angle is acute.
If c^2\ \textgreater \ a^2+b^2, then the angle is obtuse.
If c^2=a^2+b^2, then the angle is right.

c^2=400
a^2+b^2=144+225=369

\boxed{c^2\ \textgreater \ a^2+b^2}

Thus, the angle is obtuse.

Question 2:

Use the Pythagorean theorem.
a=b=8
c=?
a^2+b^2=c^2

8^2+8^2=c^2
c^2=64+64
c^2=128

c= \sqrt{128}
c= \sqrt{64*2}
\boxed{c=8 \sqrt{2} }

That's your answer.
Hope this helps! :)
8_murik_8 [283]3 years ago
4 0
Here u go
1.B obtuse
2. C. 8/2
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3 positive integers have a mean of 5 and range of 10, what are the integers
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The three positive integer that has mean of 5 and range of 10 will be 0, 5, 10.

<h3>What is Algebra?</h3>

The analysis of mathematical representations is algebra, and the handling of those symbols is logic.

Let the three positive number be (a – d), (a), and (a + d).

Then the mean of the number will be 5. Then we have

5 = (a – d + a + a + d) / 3

5 = (3a) / 3

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Then the range will be 10. Then we have

(a + d) – (a – d) = 10

   a + d – a + d = 10

                   2d = 10

                     d = 5

Then the three positive integer that has mean of 5 and range of 10 will be 0, 5, 10.

More about the Algebra link is given below.

brainly.com/question/953809

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A machine is used to fill plastic bottles with bleach. A sample of 18 bottles had a mean fill volume of 2.007 L and a standard d
abruzzese [7]

Answer:

The 99% confidence interval for the difference between the mean fill volumes at the two locations is;

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Step-by-step explanation:

The number of bottles in the sample at the first location, n₁ = 18 bottles

The mean fill volume, \bar{x}_{1} = 2.007 L

The standard deviation, σ₁ = 0.010 L

The number of bottles in the sample at the second location, n₂ = 10 bottles

The mean fill volume, \bar{x}_{2} = 2.001 L

The standard deviation, σ₂ = 0.012 L

The nature of the variance of the two samples = Equal variance

The confidence interval of the statistics, C = 99%

The difference between the mean

\mu_1 - \mu_2 = \left (\bar{x}_{1}- \bar{x}_{2}  \right )\pm t_{\alpha /2} \times \sqrt{\dfrac{\sigma _{1}^{2}}{n_{1}}+\dfrac{\sigma _{2}^{2}}{n_{2}}}

(1 - C)/2 = (1 - 0.99)/2 = 0.005, the degrees of freedom, f = n₁ - 1 = 10 - 1 = 9

∴ t_{\alpha /2} = 3.25

Therefore, we have;

\mu_1 - \mu_2 = \left (2.007- 2.001  \right )\pm 3.25 \times \sqrt{\dfrac{0.01^{2}}{18}+\dfrac{0.12^{2}}{10}}

Therefore, we have the difference of the two means given as follows;

-0.1175665 L < μ₁ - μ₂ < 0.1295665 L

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