Response: 8n
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Clarification:
The two sides, 4n+5 and 5n+6, comprise terms 4n and 5n which total to 9n. We require 8n to add to this to achieve 17n. In simpler terms, 4n + 5n + 8n = 9n + 8n = 17n
That is the reason why the outcome is 8n. There are no additional terms because the "+5" and "+6" in the two given expressions (4n+5 and 5n+6) sum to 5+6 = 11, matching what we aim for in the perimeter expression 17n + 11
Side 1 = 4n + 5
Side 2 = 5n + 6
Side 3 = 8n
Perimeter = (side1) + (side2) + (side3)
Perimeter = (4n + 5) + (5n + 6) + (8n)
Perimeter = 4n + 5 + 5n + 6 + 8n
Perimeter = (4n + 5n + 8n) + (5 + 6)
Perimeter = 17n + 11
So this confirms we possess the correct expression for the third side that is missing.
Answer:
14.2 hectares
Step-by-step explanation:
Fund Mexico reports that monarch butterflies occupied 1.13 hectares in one winter and 4.01 hectares the next winter (one year later)
The Geometric growth formula is
(Pt/ Po)^1/t
Where Pt = Size after t years = 4.01 hectares
Po = Initial size = 1.13
t = time = 1
=( 4.01/1.13 )^1/1
= 3.5486725664
Thus, the geometric growth rate is 3.5486725664.
The area they will occupy in hectares after one more year = Current area × Geometric growth rate
= 4.01 ×3.5486725664
= 14.230176991 hectares
Approximately equal to 14.2
c) Step-by-step breakdown: The collision rate is 1.2 incidents per 4 months, which can be expressed as 0.3 incidents monthly. Therefore, the Poisson distribution for the variable X representing monthly collisions is defined as P(X = x) =... for x ∈ N ∪ {0} = 0 otherwise. (1) Where X = 0 denotes no collisions during a 4-month timeframe, substituting gives P(X = 0) =... (2). For a 4-month period, P(No collision in 4 month period) =... (3). Two collisions in a 2-month span translate to 1 per month, thus P(X =1) =... (4). Over 2 months, P(2 collisions in a 2 month period) =... (5). One collision over a 6-month period equates to P(1 collision in 6 months period) =... (6). Consequently, P(1 collision in 6 month period) results in... (7). For no collisions in a 6-month period, P(No collision in 6 months period) =... (8). Finally, the probability of 1 or fewer collisions over six months is P(1 or fewer collision in 6 months period) = (8) + (7) = 0.0785 + 0.1653.
We consider all workers as either full-time or part-time.
36 = 24 + 12
If there are 24 or fewer full-time workers, there must be at least 12 part-time workers. (This conclusion is based on the understanding of sums.)
You can formulate the inequality in two steps. First, present and resolve an equation for full-time workers in relation to part-time workers. Then, apply the specified limit on full-time workers. This results in an inequality that can be solved for part-time workers.
Let f and p represent full-time and part-time positions, respectively.
f + p = 36... given
f = 36 - p... subtract p to express f in terms of p.
f ≤ 24......... given
(36 - p) ≤ 24.... substitute for f. This gives your inequality in terms of p.
36 - 24 ≤ p.... rearranging gives p ≥ 12........ this is the solution to the inequality
<span><span>1. </span>We have two boxes with weights:
=> 9.4 lb and 62.6 lb.
To find a rough estimate of their total weight, we will round and use compatible numbers.
For 9.4 lbs, rounding gives us 9 lbs
And for 62.6 lbs, it rounds to 63 lbs
=> Adding these two rounded numbers yields:
=> 9 + 63
=> 72, the estimated total is 72.
Let’s verify if this is close to the actual weight
=> 9.4 + 62.6
=> 72</span>
