Answer:
a)
b)d(t)=500t
c)
Step-by-step explanation:
d = Horizontal distance
s = the distance between the plane and the radar station
The horizontal distance (d), the one mile altitude, and s form a right triangle.
So, use Pythagoras theorem
a)
(b) Express d as a function of the time t (in hours) that the plane has flown.
d(t)=500t
(c) Use composition to express s as a function of t.
using b
Hey
So my brother posted this on Yahoo
Draw a line from the center of the circle to one of the ends of the chord (water surface) and another to the point at greatest depth. A right-angled triangle is formed. Length of side to the water-surface is 5 cm, the hypot is 7 cm.
<span>What you do now is the following: </span>
<span>Calculate the angle θ in the corner of the right-angled triangle by: cos θ = 5/7 ⇒ θ = cos ˉ¹ (5/7) </span>
<span>So θ is approx 44.4°, so the angle subtended at the center of the circle by the water surface is roughly 88.8° </span>
<span>The area shaded will then be the area of the sector minus the area of the triangle above the water in your diagram. </span>
<span>Shaded area ≃ 88.8/360*area of circle - ½*7*7*sin88.8° </span>
<span>= 88.8/360*π*7² - 24.5*sin 88.8° </span>
<span>≃ 13.5 cm² </span>
<span>(using area of ∆ = ½.a.b.sin C for the triangle) </span>
<span>b) </span>
<span>volume of water = cross-sectional area * length </span>
<span>≃ 13.5 * 30 cm³ </span>
<span>≃ 404 cm³</span>
Hoped it Helped
Answer:
Volume A= one third
Step-by-step explanation:
Use volume B
Answer:
Step-by-step explanation:
Given a set of data sorted from smallest to largest, i.e. arranged in ascending order we are to find out the median, I and III quartiles
We know that the median is the middle entry of data arranged in ascending order
Q1 is the entry below which 25% lie and Q3 is one above which 25% lie
Hence proper definition would be
d. The first quartile is the median of the lower half of the data below the overall median.
The second quartile is the overall median
The third quartile is the median of the upper half of the data above the overall median.
Option b is wrong becuase mean is not necessary here. Option a is wrong because this has nothing to do with std deviation
Option c is wrong since minimum value cannot be q1
Option e is wrong because we have exactly 25% lie below Q1
It is
12000 x (1.06)^12 + 50000 x (1.06)^6
= 95,072.31
start of 4th year to end of 6th year = 6 semi-annual periods where interest is compounded for the second deposit
