The type of sampling method in which a sample frame is <em>divided into groups,</em> which are highly similar to the other
- <u>Simple random sampling.</u>
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According to the given question, we are asked to show the name which is given to the type of sampling method in which a sample frame is divided into groups, which are highly similar to the other.
As a result of this, when a sample frame is divided into groups which are very similar to each other, then this is a simple random sampling.
In this method, we can see that there are different types of <em>random sampling techniques</em> which includes:
- <u>Simple random sampling</u>
- <u>Stratified random sampling</u>
- <u>Cluster random sampling</u>
- <u>Systematic random sampling</u>
Therefore, the correct answer is Simple random sampling.
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Answer:
60 students
Step-by-step explanation:
96/6=16=96-16=80/4=20=80-20=60
Answer:
Funciones trigonométricas de μ
sen (μ) = b/ √(a^2+ b^2)
cos (μ) = a/ √(a^2+ b^2)
tan (μ) = b/ a
cot (μ) = a/ b
sec (μ) = √(a^2+ b^2) / a
csc (μ) = √(a^2+ b^2) /b
Step-by-step explanation:
Ya que no proportionaste el valor de cot μ podemos suponer un valor = a/b para que tengas una respuesta general y reemplaces el valor de a y b de acuerdo con tu caso.
cot (μ) = a/b
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Funciones trigonométricas
sen (μ) = cateto opuesto/ hipotenusa
cos (μ) = cateto adyacente/ hipotenusa
tan (μ) = cateto opuesto/ cateto adyacente
cot (μ) = cateto adyacente/ cateto opuesto
sec (μ) = hipotenusa / cateto adyacente
csc (μ) = hipotenusa /cateto opuesto
cot (μ) = cateto adyacente/ cateto opuesto = cot (μ) = a/ b
Por lo tanto:
Cateto adyacente = a
Cateto opuesto = b
Hipotenusa
H^2 = Cateto adyacente^2 + Cateto opuesto^2
H= √(a^2+ b^2)
Reemplaznado los valores de los catetos y la hipotenusa se obteienen los valores de las funciones trigonométricas de μ.
Elimination, Multiply the second equation by -1, then add the equations together.