1,2,4belong on company sources and 3 should be on external information
Answer: point F is at <u> 0.6 </u> on the number line
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Explanation:
The ratio 2:3 scales up to 2x:3x for some positive real number x.
This means the distance from D to F is 2x units, and the distance from F to E is 3x units. Combine those two smaller distances to get 2x+3x = 5x to represent the full distance from D to E.
D is at -3 and E is at 6. This is a distance of 9 units since |-3-6| = |-9| = 9
Set this equal to the 5x from earlier and solve
5x = 9
x = 9/5
x = 1.8
This leads to 2x = 2*1.8 = 3.6
Therefore, we'll move 3.6 units from -3 to -3+3.6 = 0.6 which is the location of point F on the number line.
Notice that from 0.6 to 6 is 5.4 units and that 3x = 3*1.8 = 5.4 matches up to help confirm the answer.
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!
The solution for this problem would be:
f'(x) = 1 - 1/x^2 = (x^2 - 1)/x^2
is positive where |x| > 1
hence (-inf, -1) and (1, inf) are the regions in which f' is positive and therefore f is increasing.
Therefore, the answer is (-infinity,-1] U [1,infinity).
Firstly, you change it to miles per second and then to feet
102miles per hour(60×60s)
102 per 120s
Divide
102÷120
=0.85miles per second
Convert to feet
=4488 feet per second