PLEASE HELP (look at pictures)
1 answer:
<h2>Answer:</h2>
This is a theorem called Converse of Alternate Exterior Angles that states that <em>if two lines are cut by a transversal and the alternate exterior angles are congruent, then the lines are parallel</em>. Moreover, this theorem is based upon the corresponding Angles Converse Postulate that states that<em> if two lines are cut by a transversal and corresponding angles are congruent, then the lines are parallel. </em>We don't need to prove this postulate, it's assume to be true. So our goal is to get corresponding angles congruent in order to use the corresponding Angles Converse Postulate,
1.
Reason: Given
Statement:
2.
Reason: Def of vertical
Statement:
3.
Reason: Def of vertical
Statement:
4.
Reason: Transitive Property
Statement:
5.
Reason: corresponding Angles Converse Postulate
Statement:
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Answer:
<em>Explicación más abajo</em>
Step-by-step explanation:
Integración Indefinida
La integral
Se resuelve con el cambio de variables:
t=arcsen(x)
Una vez hechos los cambios, la integral se resuelve en función de t:
Hay que devolver los cambios para mostrarla en función de x.
El cambio de variables también se puede escribir:
x=sen(t)
Recordando que
Entonces:
Devolviendo los cambios:
Es la respuesta correcta
Answer:
C) 515 hours.
D) 500 hours
c) sample 3
Step-by-step explanation:
1. Sample 2 mean = x2`= ∑x2/n2= 2060/4= 515 hours
Sample Service Life (hours)
1 2 3
495 525 470
500 515 480
505 505 460
<u>500 515 470 </u>
<u>∑2000 2060 1880</u>
x1`= ∑x1/n1= 2000/4= 500 hours
x2`= ∑x2/n2= 2060/4= 515 hours
x3`= ∑x3/n3= 1880/4= 470 hours
2. The mean of the sampling distribution of sample means for whenever service life is in control is 500 hours . It is the given mean in the question and the limits are determined by using μ ± σ , μ±2 σ or μ ± 3 σ.
In this question the limits are determined by using μ ± σ .
3. Upper control limit = UCL = 520 hours
Lower Control Limit= LCL = 480 Hours
Sample 1 mean = x1`= ∑x1/n1= 2000/4= 500 hours
Sample 2 mean = x2`= ∑x2/n2= 2060/4= 515 hours
Sample 3 mean = x3`= ∑x3/n3= 1880/4= 470 hours
This means that the sample mean must lie within the range 480-520 hours but sample 3 has a mean of 470 which is out of the given limit.
We see that the sample 3 mean is lower than the LCL. The other two means are within the given UCL and LCL.
This can be shown by the diagram.
