39 ft/min is how many yd/s
2 answers:
0.0181 yd/s. To solve this problem you use dimensional analysis. You convert the feet and minutes to yards and seconds by using this process. The units of feet and minutes cancel out, so you are left with yards per second. You should get 0.0180555555 when you multiply and divide everything, but you can just round it to 0.0181 or whatever place your teacher wants.
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Answer:
Probability that the sample average is at most 3.00 = 0.98030
Probability that the sample average is between 2.65 and 3.00 = 0.4803
Step-by-step explanation:
We are given that the sediment density (g/cm) of a randomly selected specimen from a certain region is normally distributed with mean 2.65 and standard deviation 0.85.
Also, a random sample of 25 specimens is selected.
Let X bar = Sample average sediment density
The z score probability distribution for sample average is given by;
Z = ~ N(0,1)
where, = population mean = 2.65
= standard deviation = 0.85
n = sample size = 25
(a) Probability that the sample average sediment density is at most 3.00 is given by = P( X bar <= 3.00)
P(X bar <= 3) = P( <=
) = P(Z <= 2.06) = 0.98030
(b) Probability that sample average sediment density is between 2.65 and 3.00 is given by = P(2.65 < X bar < 3.00) = P(X bar < 3) - P(X bar <= 2.65)
P(X bar < 3) = P( <
) = P(Z < 2.06) = 0.98030
P(X bar <= 2.65) = P( <=
) = P(Z <= 0) = 0.5
Therefore, P(2.65 < X bar < 3) = 0.98030 - 0.5 = 0.4803 .
So first thing we have to realise is that we have to make them have equal denominators. So,
.
Now order the fractions....so,
Now we can see why is greater than
but less than