Option D is indeed correct, as it ensures that the post's point is equidistant from the ground, maintaining a perpendicular angle at two points on the surface.
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
1.29(30) +2 Step-by-step explanation: Begin by calculating 1.29 multiplied by 30, which results in 38.7. Next, add 2 to this sum to arrive at 40.7. Consequently, the total expense amounts to $40.70.
1.) <span>The ratio of note A compared to middle C is 440.0 / 261.6 = 1.6820 ≈ 1.6818
2.) </span><span>The D note's ratio against middle C is 293.7 / 261.6 = 1.1227
3.) </span><span>D# = 293.6 multiplied by 1.0595 equals 311.1
4.) </span><span>The frequency ratio of G compared to C is 1: 262/392 = 1: 0.6683 ≈ 3: 2
5.) </span><span>The frequency ratio of E to C is 1: 262/330 = 1: 0.7939 ≈ 5: 4
6.) The </span><span>element in a musical note referred to as pitch is the frequency.</span>
