B and C cant be it because its not a whole number and it is .4 so it cant go up one making D the only possible choice.
Answer:
0.3 to 2.3 min
Step-by-step explanation:
n1=n2=10
x1=48
x2=49
s1=4
s2=1
Determine the deegres of freedom.

t=2.037 (student's appendix)



We are 95% confident that average commuting time for rute A is between 0.3 and 2.3 min shorter than for rute B.
Answer:
A) 68.33%
B) (234, 298)
Step-by-step explanation:
We have that the mean is 266 days (m) and the standard deviation is 16 days (sd), so we are asked:
A. P (250 x < 282)
P ((x1 - m) / sd < x < (x2 - m) / sd)
P ((250 - 266) / 16 < x < (282 - 266) / 16)
P (- 1 < z < 1)
P (z < 1) - P (-1 < z)
If we look in the normal distribution table we have to:
P (-1 < z) = 0.1587
P (z < 1) = 0.8413
replacing
0.8413 - 0.1587 = 0.6833
The percentage of pregnancies last between 250 and 282 days is 68.33%
B. We apply the experimental formula of 68-95-99.7
For middle 95% it is:
(m - 2 * sd, m + 2 * sd)
Thus,
m - 2 * sd <x <m + 2 * sd
we replace
266 - 2 * 16 <x <266 + 2 * 16
234 <x <298
That is, the interval would be (234, 298)
Answer:
4x109
Step-by-step explanation: 4x109=436 and that is the closest to that number.
Answer:
2/10 and 0.2
Step-by-step explanation: