Will Mark Brainliest to the smartest and best answer!!!!! What is the measure of each interior angle in a regular 24-gon? If nec
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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=]
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!
Answer:
1)
Minimum is a 4th Degree
2)
Positive; Even
3)
Step-by-step explanation:
Part 1)
The minimum degree of our function will be 4.
Looking at the graph, we know that the graph crosses the x-axis at -4 and -1. Since it <em>crosses through</em> the x-axis at these two points, these two factors must have an odd multiplicity.
So, it can be anything 1, 3, 5, 7, etc.
However, we will choose the lowest one, 1.
Next, we know that the graph <em>bounces off</em> at 3.
So, it must have an even multiplicity. In other words, 2, 4, 6, 8, etc.
We choose the lowest one, 2.
Therefore, the minimum degree of our function will be 1+1+2 or 4.
Part 2)
The degree of our polynomial is (and will always be) even. Therefore, both ends of the graph will go in the same direction.
Recall the simplest even polynomial, the parent quadratic function. When the leading coefficient is positive, both of the ends go straight up.
This applies to all polynomials with even degrees.
Therefore, since the arms of the graph is going straight up towards positive infinity, the leading coefficient of our graph must be positive.
Part 3)
This is similar to Part 1.
We can see that the graph touches the x-axis at -4, -3, and 1. So, the zeros of the function is:
We know that it<em> passes through</em> . So, these two factors must have an odd multiplicity.
However, since the graph <em>bounces off</em> , this factor must have an even multiplicity.
And we're done!