Answer:
The nth term of the sequence is
<h2>5 + 2n</h2>
Step-by-step explanation:
The sequence above is an arithmetic sequence
For an nth term in an arithmetic sequence
A(n) = a + ( n - 1)d
where a is the first term
n is the number of terms
d is the common difference
From the question
a = 7
d = 9 - 7 = 2 or 11 - 9 = 2
So the nth term for the sequence is
A(n) = 7 + ( n - 1)2
= 7 + 2n - 2
<h3>A(n) = 5 + 2n</h3>
Hope this helps you
The current rate is simply equal to $175 per month, let
us call this as rate A:
A = 175
The new rate is $94 plus $4.50 per devices, let us call
this as rate B:
B = 94 + 4.50 x
where x is the number of devices connected to the network
The inequality equation for us to find x which the new
plan is less than current plan is:
94 + 4.50 x < 175
Solving for x:
4.50 x < 81
x < 18
So the number of devices must be less than 18.
- From the graph, we can actually see that the new rate
intersects the current rate at number of devices equal to 18. So it should
really be below 18 devices.
(a) Data with the eight day's measurement.
Raw data: [60,58,64,64,68,50,57,82],
Sorted data: [50,57,58,60,64,64,68,82]
Sample size = 8 (even)
mean = 62.875
median = (60+64)/2 = 62
1st quartile = (57+58)/2 = 57.5
3rd quartile = (64+68)/2 = 66
IQR = 66 - 57.5 = 8.5
(b) Data without the eight day's measurement.
Raw data: [60,58,64,64,68,50,57]
Sorted data: [50,57,58,60,64,64,68]
Sample size = 7 (odd)
mean = 60.143
median = 60
1st quartile = 57
3rd quartile = 64
IQR = 64 -57 = 7
Answers:
1. The average is the same with or without the 8th day's data. FALSE
2. The median is the same with or without the 8th day's data. FALSE
3. The IQR decreases when the 8th day is included. FALSE
4. The IQR increases when the 8th day is included. TRUE
5. The median is higher when the 8th day is included. TRUE
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
Given:
Vertices of ABCD are A(-4,4), (-2,2), C(-2,-1) and D(-4,1).
Vertices of A'B'C'D' are A'(3,-4), B'(5,-2), C'(5,1) and D'(3,-1).
To find:
The sequence of transformations that changes figure ABCD to figure A'B'C'D'.
Solution:
Part A:
The figure ABCD reflected across the x-axis, then
Using this rule, we get
Similarly, the other points are 
.
Then figure translated 7 units right to get A'B'C'D'.
Similarly, the other points are 
.
Therefore, the figure ABCD reflected across the x-axis and then translated 7 units right to get A'B'C'D'.
Part B:
Reflection and translation are rigid transformation, it means shape and size of figures remains same after reflection and translation.
Therefore, the two figures congruent.
