If we assume the given segments are those from the vertices to the point of intersection of the diagonals, it seems one diagonal (SW) is 20 yards long and the other (TR) is 44 yards long. The area (A) of the kite is half the product of the diagonals:
... A = (1/2)·SW·TR = (1/2)·(20 yd)·(44 yd)
... A = 440 yd²
Answer: Undefined slope
Step-by-step explanation: Undefined slope is the a vertical line The x-coordinate never changes no matter what the y-coordinate which means there no run .
Answer:
When X = 24
Step-by-step explanation:
Use derivative to locate the turning point.
48 - 2X = 0, => X = 24
98 days = (98 ⁄ 7) weeks = 14 weeks
<span>Po = initial population = 5 </span>
<span>Ƭ = doubling time in weeks </span>
<span>t = elapsed time in weeks </span>
<span>P{t} = population after "t" weeks </span>
<span> P{t} = (Po)•2^(t ⁄ Ƭ) </span>
<span> P{t} = (Po)•2^(t ⁄ 4) </span>
<span> P{t} = 5•2^(t ⁄ 4) </span>
<span> P{14} = (5)•2^(14 ⁄ 4) … t = 14 weeks = 98 days </span>
<span> P{14} = 56 … population after 14 weeks</span>
Answer:
Correct option: B
Step-by-step explanation:
The professor can perform a One-mean <em>t</em>-test to determine whether the average score of the students in his class is more than the average score of all the students attending university.
A <em>t</em>-test will be used instead of the <em>z</em>-test because the population standard deviation is not provided instead it is estimated by the sample standard deviation.
The hypothesis for this test can be defined as follows:
<em>H₀</em>: The average score of the students in his class is not more then the entire university, i.e. <em>μ ≤ 35</em>.
<em>Hₐ</em>: The average score of the students in his class is more then the entire university, i.e. <em>μ > 35</em>.
Given:
The test statistic is:
Thus, the correct option is (B).
