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aleksandr82 [10.1K]

3 years ago

6

Determine whether each equation has one solution, no solution, or infinite solutions.

1 answer:

Sati [7]3 years ago

6 0

Answer:

a. has one solution

b. infinite solution

Step-by-step explanation:

a.

2(x - 1) + 6 = 4x - 22

2x - 2 + 6 = 4x - 22

2x - 4x = 2 - 6 - 22

-2x = -26

x = 26/2

x = 13

b.

6(2x + 1) – 2 = 12x + 4

12x + 6 - 2 = 12x + 4

12x - 12x = 4 + 2 - 6

0 = 0

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8=1/3(a+5)<br> What does a equal?

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Answer: a=19

Step-by-step explanation: In order to solve this question, we must first take out the parenthesis, we can do this by taking 1/3 and multiply it by a and 5. This will give us 8=1/3a+5/3, with this we can subtract 5/3 on both sides of the equation, which will then give us, 19/3=1/3a. Last, we can divide 19/3 by 1/3 giving us 19 as our final anwser. a=19

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2 years ago

Multiply. 5 x 1 1/3<br> Answer with a mixed number in simplest form.

Crank

Answer:

6 2/3

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2 years ago

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2 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:

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

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3 years ago