(A)
P(<em>X</em> < 61.25) = P((<em>X</em> - 55.4)/4.1 < (61.25 - 55.4)/4.1)
… ≈ P(<em>Z</em> ≤ 0.1427)
… ≈ 0.5567
(B)
P(<em>X</em> > 46.5) = P((<em>X</em> - 55.4)/4.1 > (46.5 - 55.4)/4.1)
… ≈ P(<em>Z</em> > -2.1707)
… ≈ 1 - P(<em>Z</em> ≤ -2.1707)
… ≈ 0.9850
Answer:
8 and 5
Step-by-step explanation:
If a quadrilateral is a rhombus then:
i) all its sides have equal length,
and
ii) the 2 diagonals are perpendicular and bisect each other.
The 2 diagonals of rhombus JKLM intersect at N.
Then m(MNJ)=90°, |JN|=4 and |NM|=3 units
thus, by the Pythagorean theorem,

The perimeter of the rhombus is 4*5 units=20 units.
Answer: 20 units
Answer:
They'll reach the same population in approximately 113.24 years.
Step-by-step explanation:
Since both population grows at an exponential rate, then their population over the years can be found as:

For the city of Anvil:

For the city of Brinker:

We need to find the value of "t" that satisfies:
![\text{population brinker}(t) = \text{population anvil}(t)\\21000*(1.04)^t = 7000*(1.05)^t\\ln[21000*(1.04)^t] = ln[7000*(1.05)^t]\\ln(21000) + t*ln(1.04) = ln(7000) + t*ln(1.05)\\9.952 + t*0.039 = 8.8536 + t*0.0487\\t*0.0487 - t*0.039 = 9.952 - 8.8536\\t*0.0097 = 1.0984\\t = \frac{1.0984}{0.0097}\\t = 113.24](https://tex.z-dn.net/?f=%5Ctext%7Bpopulation%20brinker%7D%28t%29%20%3D%20%5Ctext%7Bpopulation%20anvil%7D%28t%29%5C%5C21000%2A%281.04%29%5Et%20%3D%207000%2A%281.05%29%5Et%5C%5Cln%5B21000%2A%281.04%29%5Et%5D%20%3D%20ln%5B7000%2A%281.05%29%5Et%5D%5C%5Cln%2821000%29%20%2B%20t%2Aln%281.04%29%20%3D%20ln%287000%29%20%2B%20t%2Aln%281.05%29%5C%5C9.952%20%2B%20t%2A0.039%20%3D%208.8536%20%2B%20t%2A0.0487%5C%5Ct%2A0.0487%20-%20t%2A0.039%20%3D%209.952%20-%208.8536%5C%5Ct%2A0.0097%20%3D%201.0984%5C%5Ct%20%3D%20%5Cfrac%7B1.0984%7D%7B0.0097%7D%5C%5Ct%20%3D%20113.24)
They'll reach the same population in approximately 113.24 years.