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Travka [436]
3 years ago
8

A minister claims that more than 60% of the adult population attends a religious service at least once a month. The null and alt

ernative hypotheses you'd use to test this claim would be:
a. H0 : μ = .6, Ha: μ > .6
b. H0 : p = .6, Ha: p > .6
c. H0 : p = .6, Ha: p ≠ .6
d. H0 : pˆ = .6, Ha: pˆ > .6
e. H0 : x = .6, Ha: x > .6
Mathematics
1 answer:
LuckyWell [14K]3 years ago
6 0

Answer:

d. H0 : pˆ = .6, Ha: pˆ > .6

Step-by-step explanation:

Given that a minister claims that more than 60% of the adult population attends a religious service at least once a month.

Here the subject of interest about a proportion of adult population who attends a religious service at least once a month.

The claim is  this proportion is greater than 60% and we want to check  whether this claim is true.

Hence hypotheses would be for proportion only and not mean.

Since population proportion is to be checked we use p hat

Correct answer is

d. H0 : pˆ = .6, Ha: pˆ > .6

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Even if they both spent all of their money, they would still be $0.43 short.

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3 years ago
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4 years ago
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3 years ago
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3 years ago
Se arrastra un bloque de 18 kg a través de una superficie con una fuerza de 300 N en donde el coeficiente de fricción de dicha s
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Answer:

10.9\ \text{m/s}^2

Step-by-step explanation:

m = Masa del bloque = 18 kg

F = Fuerza en el bloque = 300 N

\mu = Coeficiente de fricción = 0.6

El equilibrio de fuerzas del sistema viene dado por

F-f=ma\\\Rightarrow 300-0.6\times 18\times 9.81=18a\\\Rightarrow a=\dfrac{300-0.6\times 18\times 9.81}{18}\\\Rightarrow a=10.9\ \text{m/s}^2

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