B.)155
a1+d(n-1)
a1+d(6-1)=50
a1+5d=50
a1+d(11-1)=85
a1+10d=85
a1=50-5d
(50-5d)+10d=85
50-5d+10d=85
5d+50=85
5d=35
d=7
a1+7(6-1)=50
a1+35=50
a1=15
15+7(n-1)
15+7(21-1)
15+7(20)
15+140=155
Answer:
The herd migrated a total of 43 3/10 miles and migrated about 14 13/30 miles a day.
Step-by-step explanation:
11 9/10 + 17 1/5 + 14 1/5 =
11 9/10 + 17 2/10 + 14 2/10 =
29 1/10 + 14 2/10 =
43 3/10
43 3/10 divided by 3/1 =
433/10 x 1/3 =
433/30 =
14 13/30
Answer:
Graph the polynomial in order to determine the intervals over which it is increasing or decreasing. Increasing on: (1,∞) ( 1 , ∞ ). Decreasing on: (−∞,1) ( - ∞ , 1 ).
Step-by-step explanation:
Answer:
d
Step-by-step explanation:
d
9514 1404 393
Answer:
b) 31%
Step-by-step explanation:
95% of the normal distribution is between Z values of ±1.96.
99% of the normal distribution is between Z values of ±2.576.
The change from a 95% interval to a 99% interval widens the range by ...
(2.576/1.96 -1) × 100% ≈ 31.4%
The width of the interval increases about 31%.
__
Compare the two attachments. The tail area of 0.05 means the central area is 0.95. Similarly, the tail area of 0.01 means the central area is 0.99.
_____
<em>Additional comment</em>
As always, when you're dealing with percentages, you need to understand what the base value is. Here, when we're talking about 95% and 99% confidence intervals, we're not talking about the numbers 0.95 and 0.99 and the increase from 0.95 to 0.99.
Rather we're talking about areas under the normal probability distribution curve, and the Z-values associated with those intervals. The change in interval width refers to the change in Z-values associated with the areas of 0.95 and 0.99.