To identify the corresponding equation, you can follow these steps:
ax^2 + bx + c = 0
where a = -2
b = 1
c = 3
-2x^2 + x + 3 = 0
The right result will show as a: 0 = <span>-2x^2 + x + 3.</span>
The translation rule can be expressed as T -3,1(x,y). The translation can also be indicated as (x,y)➡️(x-3,y+1).
1
2
3
Step-by-step explanation: Generally, during the roll of two fair 6-sided dice, the doubles result in (1, 1), (2, 2), (3, 3), (4, 4), (5, 5), (6, 6). Therefore, the total for doubles is N = 6. The outcome of rolling two fair 6-sided dice yields n = 36. Thus, the probability of rolling doubles (matching numbers on both dice) is calculated mathematically. When rolling two fair dice, outcomes that sum to 4 or less are (1, 1), (1, 2), (1, 3), (2, 1), (2, 2), (3, 1). Observing this, we see two doubles present. Consequently, the conditional probability of rolling doubles is represented mathematically. Lastly, when rolling the two fair dice, outcomes that show different numbers result in L = 30, while outcomes where at least one die shows a 1 give W = 10. Hence, the conditional probability of having at least one die show a 1 is presented mathematically.
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
