9514 1404 393
Answer:
3.16 ≈ 3.2
Step-by-step explanation:
Put the numbers in the distance formula and do the arithmetic.
d = √((x2 -x1)² +(y2 -y1)²)
d = √((2-(-1))² +(3-2)²) = √(3² +1²) = √10
d ≈ 3.16 ≈ 3.2
The distance rounded to hundredths is 3.16. Rounded to tenths, it is 3.2.
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: 10
Step-by-step explanation:
f(x)= 3 (2) - x + 6
f(x)= 6 - 2 + 6
f(x)= 4 + 6
f(x) = 10
Since she is rolling 1 six sided dice the chances of her rolling a 3 are 1:6 because, she is trying to land on 3 not 3 numbers.