Answer:
1458/3456 reduces to 27/64 (after simplification)
The cube root of 27/64 equals 3/4
The square of 3/4 calculates to 9/16
Multiplying 9/16 by 1024 results in 576.
THIS IS THE ACTUAL WORK AND RESULT
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Answer:
Answer and Explanation:
We have:
Population mean,
μ
=
3
,
000
hours
Population standard deviation,
σ
=
696
hours
Sample size,
n
=
36
1) The standard deviation for the sampling distribution:
σ
¯
x
=
σ
√
n
=
696
√
36
=
116
2) By the central limit theorem, the sampling distribution's expected value matches the population mean.
Thus:
The expected value of the sampling distribution equals the population mean,
μ
¯
x
=
μ
=
3
,
000
The standard deviation of the sampling distribution,
σ
¯
x
=
116
The sampling distribution of
¯
x
is roughly normal due to a sample size greater than
30
.
3) The likelihood that the average lifespan of the sample falls between
2670.56
and
2809.76
hours:
P
(
2670.56
<
x
<
2809.76
)
=
P
(
2670.56
−
3000
116
<
z
<
2809.76
−
3000
116
)
=
P
(
−
2.84
<
z
<
−
1.64
)
=
P
(
z
<
−
1.64
)
−
P
(
z
<
−
2.84
)
=
0.0482
In Excel: =NORMSDIST(-1.64)-NORMSDIST(-2.84)
4) The probability of the average life in the sample exceeding
3219.24
hours:
P
(
x
>
3219.24
)
=
P
(
z
>
3219.24
−
3000
116
)
=
P
(
z
>
1.89
)
=
0.0294
In Excel: =NORMSDIST(-1.89)
5) The likelihood that the sample's average life is lower than
3180.96
hours:
P
(
x
<
3180.96
)
=
P
(
z
<
3180.96
−
3000
116
)
=
P
(
z
<
1.56
)
=
0.9406
Class B exhibits the most consistent sleep patterns since there's a smaller variance between 6.87 and 3.65 compared to the other classes.
Respuesta: Los contratos de opciones pueden ser valuados empleando modelos matemáticos tales como el modelo de precios Black-Scholes o el modelo Binomial. El costo de una opción se divide principalmente en dos componentes: su valor intrínseco y su valor temporal.... El valor temporal depende de la volatilidad anticipada del activo subyacente y del tiempo restante hasta que la opción expire.
Explicación paso a paso: ¡espero que esto ayude!
Por cierto, ¡también hablo inglés!