Answer:
The range of the 95% data (X) = 238.3 days < X < 289.9 days
Step-by-step explanation:
Given;
mean of the normal distribution, m = 264.1 days
standard deviation, d = 12.9 days
between two standard deviation below and above the mean is 96% of all the data.
two standard deviation below the mean = m - 2d
= 264.1 - 2(12.9)
= 238.3 days
two standard deviation above the mean = m + 2d
= 264.1 + 2(12.9)
= 289.9 days
The middle of the 95% of most pregnancies would be found in the following range;
238.3 days < X < 289.9 days
Answer:
x
4
−
5
x
3
−
3
x
2
+
22
x
+
20
Step-by-step explanation:
<span>With 1026 being the mean score on the SAT and StDev of 209, Jessica has a score of 799/209 z-scores above the mean. Her z-score would be 3.823. The mean of the ACT and StDev are 20.8 and 4.8, respectively. A 28 ACT score would be 7.2/4.8 z-scores above the mean, for a z = 1.500. This means that Jessica has the higher z-score.</span>
Answer:
C = 125S + $3750
Step-by-step explanation:
$ 3750 = truck rental; $125 per ton of sugar transported. C is the cost; S is the number of tons transported. The equation relating C to S would be a linear equation like y = mx + b. This equation would be graphed in the first quadrant only. You would start with your y-intercept at (0, 3750). As x increases by 1, your y increases by 125 yielding these points:
(1, 3875) (2, 4000) (3, 4125) etc.
This shows that for each increase by one ton of sugar, the cost goes up $125