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An estimated regression equation that was fit to estimate ductility in steel using its carbon content was found to be significan

AnnZ [28]

Answer:

At a level of 95%, it is expected that the interval [0.45; 11.59] contains the value of the ductility in steel when its carbon content is 0.5%.

Step-by-step explanation:

Hello!

Considering the dependent variable:

Y: Ductility in steel.

And the independent variable:

X: Carbon content of the steel.

The linear regression was estimated and a prediction interval was calculated.

The prediction interval is calculated to predict a value that the variable Y (response variable) can take for a given value of the variable X (predictor variable) in the definition range of the linear regression line. Symbolically [Y/X= $x_{0}$ ]

In this case 95% prediction interval for Y/X=0.5

At a level of 95%, it is expected that the interval [0.45; 11.59] contains the value of the ductility in steel when its carbon content is 0.5%.

I hope it helps!

6 0

3 years ago

Find the area of rectangle it its length is (3x+5) and breadth is (2y+4)

Sindrei [870]

Answer:

Solution given:

length = (3x+5)

breadth = (2y+4)

we have

area of rectangle: length* breadth

=(3x+5)(2y+4)

opening bracket

=3x(2y+4)+5(2y+4)

=6xy+12x+10y+20

=12x+10y+6xy+20unit square

4 0

2 years ago

Find the perimeter and area of rectangle ABCD.

iragen [17]

Area = Length X Width
Perimeter = l(2) x W(2)
what are the measures of this rectangle?

3 0

3 years ago

A rectangular picture frame has perimeter of 58 inches. The height of the frame is 18 inches. What is the width of the frame?

Stells [14]

Answer:

11 inches

Step-by-step explanation:

A rectangle's perimeter can be found using:

p=2l+2w

We know that the perimeter, p, is 58, and the length/height is 18. Therefore, we can substitute those values in

58=2(18)+2w

Multiply 2 and 18

58=36+2w

Since we want to find the width, we need to get w by itself. First, subtract 26 from both sides

58-36=36-36+2w

22=2w

Since w is being multiplied by 2, divide both sides by 2. This will cancel out the 2, and leave w by itself.

22/2=2w/2

11=w

So, the width is 11 inches

8 0

3 years ago

In 2008, the average household debt service ratio for homeowners was 13.2. The household debt service ratio is the ratio of debt

ankoles [38]

Answer:

$t=\frac{13.88-13.2}{\frac{3.14}{\sqrt{44}}}=1.436$

$df=n-1=44-1=43$

$p_v =P(t_{(43)}>1.436)=0.079$

We see that the p value i higher than the significance level so then we FAIL to reject the null hypothesis and we can conclude that the true mean is not significantly higher than 13.2 *the value of 2008 ).

Step-by-step explanation:

Information given

$\bar X=13.88$ represent the sample mean

$s=3.14$ represent the sample standard deviation

$n=44$ sample size

$\mu_o =13.2$ represent the value that we want to test

$\alpha=0.05$ represent the significance level

t would represent the statistic (variable of interest)

$p_v$ represent the p value for the test

Hypothesis to test

We want to conduct a hypothesis in order to check if the true mean has increased from 2008 , and the system of hypothesi are:

Null hypothesis: $\mu \leq 13.2$

Alternative hypothesis: $\mu > 13.2$

The statistic for this case is:

$t=\frac{\bar X-\mu_o}{\frac{s}{\sqrt{n}}}$ (1)

Calculating the statistic

Replacing the info given we got:

$t=\frac{13.88-13.2}{\frac{3.14}{\sqrt{44}}}=1.436$

P-value

The degrees of freedom are:

$df=n-1=44-1=43$

Since is a right tailed test the p value is:

$p_v =P(t_{(43)}>1.436)=0.079$

Decision

We see that the p value i higher than the significance level so then we FAIL to reject the null hypothesis and we can conclude that the true mean is not significantly higher than 13.2 *the value of 2008 ).

8 0

3 years ago