Answer:
(12.1409, 14.0591
Step-by-step explanation:
Given that Tensile strength tests were performed on two different grades of aluminum spars used in manufacturing the wing of a commercial transport aircraft. From past experience with the spar manufacturing process and the testing procedure, the standard deviations of tensile strengths are assumed to be known.
Group Group One Group Two
Mean 87.600 74.500
SD 1.000 1.500
SEM 0.316 0.433
N 10 12
The mean of Group One minus Group Two equals 13.100
standard error of difference = 0.556
90% confidence interval of this difference:
t = 23.5520
df = 20
The equivalent equation is x² + 20x = -102
<h3>
Simplification of quadratic equation</h3>
The equivalent equation can be obtained by factorizing or simplifying the given quadratic equation as shown below;
x² + 20x + 81 = -21
<em>collect similar terms together</em>
x² + 20x + 81 + 21 = 0
x² + 20x + 102 = 0
The equation can also be written as x² + 20x = -102
Thus, the equivalent equation is x² + 20x = -102
Learn more about simplification of quadratic equation here: brainly.com/question/1214333
Answer:
C. It has been stretched horizontally.
Step-by-step explanation:
According to the diagram, the parabola has not been reflected in any way, nor has it been translated, since the parabola is pointing up in a smile, and the vertex is still at the origin.
The parabola does appear to be stretched, since the parent function had points at (1, 1), and (2, 4), but this parabola does not contain those points.
Since the parabola appears to be flatter than the parent function, we can say that C. It has been stretched horizontally.
Hope this helps!
A) The area of the rectangle ABCD is base times height. 20in x 12in = 240 in sq.
B) The area of the triangle AED is base times height divided by two. (14in x 12in)/2 = 84in sq.
C) Figure EBCD is a right trapezoid (having one pair of parallel sides, while the others are slanted and forming a right angle).
D) Area of EBCD is the area of the rectangle minus the area of the triangle. 240in sq - 84in sq = 156in sq.
For the answer to the question above, the easiest way to determine is changing every runner's speed into the same unit.
<span>First = 10 m/s </span>
<span>Second = 10 miles/min = 16090.34 / 60 m/s (As 1 mile = 1609.34 meter and 1 min = 60 sec) </span>
<span>Second = 260.82 m/s </span>
<span>Third = 10 cm/hr = 10*(0.01)/60*60 (As 1 cm = 0.01 m and 1 hr = 60*60 sec) </span>
<span>Third = 0.000028 m/s </span>
<span>Fourth = 10 km/sec = 10*1000 m/s (As 1 km = 1000 m and time is already in sec) </span>
<span>Fourth = 10000 m/s </span>
<span>So fastest would be the one who covers the largest distance in 1 sec. It would be the fourth one.</span>
