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3 years ago

How do you find the answer to 8/10 divided by 1/5

2 answers:

8 0

When fractions are divided by each other, you're really multiplying by the "multiplicative inverse". This is the fraction, but flipped.

$\frac{8}{10} \div \frac{1}{5} = \frac{8}{10} \times \frac{5}{1} \\\\So\\\frac{8}{10} \times \frac{5}{1} = \frac{40}{10} = 4$

irakobra [83]3 years ago

6 0

1/5 x 2 to get 2/10 so now they have common denominators and 8 divides by 2 is 4 so your answer is 4

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3 years ago

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3 years ago

Find an equation for the line with the given properties. Perpendicular to the line x - 6y = 8; containing the point (4,4) O 1) y

Pepsi [2]

Answer:

Option 3 - $y=-6x+28$

Step-by-step explanation:

Given : Perpendicular to the line $x - 6y = 8$ ; containing the point (4,4).

To Find : An equation for the line with the given properties ?

Solution :

We know that,

When two lines are perpendicular then slope of one equation is negative reciprocal of another equation.

Slope of the equation $x - 6y = 8$

Converting into slope form $y=mx+c$ ,

Where m is the slope.

$y=\frac{x-8}{6}$

$y=\frac{x}{6}-\frac{8}{6}$

The slope of the equation is $m=\frac{1}{6}$

The slope of the perpendicular equation is $m_1=-\frac{1}{m}$

The required slope is $m_1=-\frac{1}{\frac{1}{6}}$

$m_1=-6$

The required equation is $y=-6x+c$

Substitute point (x,y)=(4,4)

$4=-6(4)+c$

$4=-24+c$

$c=28$

Substitute back in equation,

$y=-6x+28$

Therefore, The required equation for the line is $y=-6x+28$

So, Option 3 is correct.

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3 years ago

72÷p=8 world problems

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<span>72÷p=8

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3 years ago

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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

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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