Answer:
(a) The probability that a single randomly selected value lies between 158.6 and 159.2 is 0.004.
(b) The probability that a sample mean is between 158.6 and 159.2 is 0.0411.
Step-by-step explanation:
Let the random variable <em>X</em> follow a Normal distribution with parameters <em>μ</em> = 155.4 and <em>σ</em> = 49.5.
(a)
Compute the probability that a single randomly selected value lies between 158.6 and 159.2 as follows:

*Use a standard normal table.
Thus, the probability that a single randomly selected value lies between 158.6 and 159.2 is 0.004.
(b)
A sample of <em>n</em> = 246 is selected.
Compute the probability that a sample mean is between 158.6 and 159.2 as follows:

*Use a standard normal table.
Thus, the probability that a sample mean is between 158.6 and 159.2 is 0.0411.
Answer:
Is there more to the question?
Step-by-step explanation:
1a) A = 4πpw
/4πw = /4πw
A / 4πw = p
1b) A = 4πpw
22 = 4πp(2)
p = 11/4π (≈0.87)
2a) P = 2πr + 2x
P - 2x = 2πr
/2π /2π
P-2x / 2π = r
2b) P = 2πr + 2x
440 = 2πr + 2(110)
r = 110/π (≈35.014)
Answer: quotient is 2x^2 + 10x - 5
Solution:
The first polynomial is miswritten.
The right one is: 2x^3 + 4x^2 - 35x + 15.
So, the division is [2x^3 + 4x^2 - 35x + 15] / (x - 3)
The synthetic division uses the coeffcients and obviate the letters, but you have to be sure to respect the place of the coefficient.
So, in this case it is:
3 | 2 4 -35 15
---------------------------------
2 10 - 5 0
So, the quotient is 2x^2 + 10x - 5, and the remainder is 0.
I like to show it in this other way:
| 2 4 -35 15
|
|
3 | +6 +30 -15
--------------------------------
2 10 - 5 0
Of course they are the same coefficients and the answer continue being quotien 2x^2 + 10x - 5, remainder 0.