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alexira [117]
3 years ago
14

Write the equation of a line that is perpendicular to the line; y – 2 = -1/6 (x + 1), and that passes through the given point (–

1, 2).
Mathematics
1 answer:
creativ13 [48]3 years ago
3 0

the equation would be y=6x+8

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Daniel had $57 to spend on gifts he spent $10.51 on a gift for Isaac $8.73 on a gift for madison and $8.47 on a gift for Abigail
mina [271]
57-10.51=46.49
46.49-8.73=37.76
37.76-8.47=29.29
Therefore Daniel has $29.29 left.
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PLEASE HELP! Is my answer correct?
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We know angle P is 90 degrees because of Thales Theorem, and angle N is 60 degrees. Because the angles in a triangle have to add up to 180 degrees, angle PMN has to be 30 degrees.

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Find the value of x. Round to the nearest tenth
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36.869 or 36.87

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7 0
3 years ago
A _____ is accepted to be true without proof, while a _____ is an assertion that can be proven using the rules of logic.A.theore
Luba_88 [7]

Answer:

  C.  postulate; theorem

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While a "given" is taken as true without proof, and a "problem" sometimes involves an assertion that can be proven, the answer choices involving these terms are not correct here.

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3 years ago
The amount of lateral expansion (mils) was determined for a sample of n = 8 pulsed-power gas metal arc welds used in LNG ship co
Alona [7]

Answer:

95% Confidence interval for the variance:

3.6511\leq \sigma^2\leq 34.5972

95% Confidence interval for the standard deviation:

1.9108\leq \sigma \leq 5.8819

Step-by-step explanation:

We have to calculate a 95% confidence interval for the standard deviation σ and the variance σ².

The sample, of size n=8, has a standard deviation of s=2.89 miles.

Then, the variance of the sample is

s^2=2.89^2=8.3521

The confidence interval for the variance is:

\dfrac{ (n - 1) s^2}{ \chi_{\alpha/2}^2} \leq \sigma^2 \leq \dfrac{ (n - 1) s^2}{\chi_{1-\alpha/2}^2}

The critical values for the Chi-square distribution for a 95% confidence (α=0.05) interval are:

\chi_{0.025}=1.6899\\\\\chi_{0.975}=16.0128

Then, the confidence interval can be calculated as:

\dfrac{ (8 - 1) 8.3521}{ 16.0128} \leq \sigma^2 \leq \dfrac{ (8 - 1) 8.3521}{1.6899}\\\\\\3.6511\leq \sigma^2\leq 34.5972

If we calculate the square root for each bound we will have the confidence interval for the standard deviation:

\sqrt{3.6511}\leq \sigma\leq \sqrt{34.5972}\\\\\\1.9108\leq \sigma \leq 5.8819

6 0
3 years ago
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