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3 years ago

Hi, can someone please answer this questions, THANKS

Mathematics

1 answer:

aliya0001 [1]3 years ago

3 0

Answer:

supplementary angles add up to 180 degrees and complementary add up to 90 degrees

Step-by-step explanation:

You might be interested in

What is the equation of this line in slope-intercept form?

12345 [234]

Answer:

y = -2x + 5

Step-by-step explanation:

Slope-intercept form: y = mx + b

The slope of the line, m, is -2 because the line shows a negative trend and moves right 1 unit every time a point moves down 2 units.

The y-intercept, b, is 5 because the line crosses that number once on the y-axis.

7 0

2 years ago

BRAILIEST IF RIGHT Which other expression has the same value as (-14) - 8 ?

stiks02 [169]

Answer:

It's the last one (-14) +(-8)

Step-by-step explanation:

since (-14)-8 is -22 and the last one is also -22

4 0

3 years ago

Draw the image of the following triangle after a dilation centered at the origin with a scale factor of 2.

Natasha_Volkova [10]

I really don’t know but I guess 2

5 0

3 years ago

what would be a real world problem situation that you can write either a percent as a decimal or write a decimal as a percent

love history [14]

When grading papers you can convert decimals into percents

7 0

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