a) <u>n² + 6 </u>
~Here n is the number of term. So, put n = 1,2,3,4.... to get the term.
- If n = 1, first term = 1² + 6 = 7
- n = 2, second term = 10
- n = 3, third temm = 15
- n = 4, fourth term = 22
- n = 10, tenth term = 106
b) <u>3n²</u>
- If n = 1, first term = 3
- n = 2, second term = 12
- n = 3, third term = 27
- n = 4, fourth term = 48
- n = 10, tenth term = 300
see the attached figure with the letters
1) find m(x) in the interval A,BA (0,100) B(50,40) -------------- > p=(y2-y1(/(x2-x1)=(40-100)/(50-0)=-6/5
m=px+b---------- > 100=(-6/5)*0 +b------------- > b=100
mAB=(-6/5)x+100
2) find m(x) in the interval B,CB(50,40) C(100,100) -------------- > p=(y2-y1(/(x2-x1)=(100-40)/(100-50)=6/5
m=px+b---------- > 40=(6/5)*50 +b------------- > b=-20
mBC=(6/5)x-20
3)
find n(x) in the interval A,BA (0,0) B(50,60) -------------- > p=(y2-y1(/(x2-x1)=(60)/(50)=6/5
n=px+b---------- > 0=(6/5)*0 +b------------- > b=0
nAB=(6/5)x
4) find n(x) in the interval B,CB(50,60) C(100,90) -------------- > p=(y2-y1(/(x2-x1)=(90-60)/(100-50)=3/5
n=px+b---------- > 60=(3/5)*50 +b------------- > b=30
nBC=(3/5)x+30
5) find h(x) = n(m(x)) in the interval A,B
mAB=(-6/5)x+100
nAB=(6/5)x
then
n(m(x))=(6/5)*[(-6/5)x+100]=(-36/25)x+120
h(x)=(-36/25)x+120
find <span>h'(x)
</span>h'(x)=-36/25=-1.44
6) find h(x) = n(m(x)) in the interval B,C
mBC=(6/5)x-20
nBC=(3/5)x+30
then
n(m(x))=(3/5)*[(6/5)x-20]+30 =(18/25)x-12+30=(18/25)x+18
h(x)=(18/25)x+18
find h'(x)
h'(x)=18/25=0.72
for the interval (A,B) h'(x)=-1.44
for the interval (B,C) h'(x)= 0.72
<span> h'(x) = 1.44 ------------ > not exist</span>
Answer:
X=24
Step-by-step explanation:
1. (4x + 11) + (3x + 1) = 180 (Combine like terms)
2. 7x + 12 = 180 (Subtract 12 from both sides)
3. 7x = 168 (Divide both sides by 7)
4. x = 24
Answer:
The values of the box-plot are: B = {8.2, 8.3, 8.4, 8.5 and 8.9}.
Step-by-step explanation:
The data provided is as follows:
X Frequency
8 0
8.1 0
8.2 2
8.3 2
8.4 2
8.5 3
8.6 0
8.7 1
8.8 0
8.9 1
9 0
So, the actual data is:
S = {8.2
, 8.2
, 8.3
, 8.3
, 8.4
, 8.4
, 8.5
, 8.5
, 8.5
, 8.7
, 8.9
}
A boxplot, also known as a box and whisker plot is a method to demonstrate the distribution of a data-set based on the following 5 number summary,
- Minimum (shown at the bottom of the chart)
- First Quartile (shown by the bottom line of the box)
- Median (or the second quartile) (shown as a line in the center of the box)
- Third Quartile (shown by the top line of the box)
- Maximum (shown at the top of the chart).
The data set is arranged in ascending order.
The minimum value is, Min. = 8.2
The lower quartile is,
![[\frac{n+1}{4}]^{th}\ obs.=[\frac{11+1}{4}]^{th}\ obs.=3^{rd }\ obs. =8.3](https://tex.z-dn.net/?f=%5B%5Cfrac%7Bn%2B1%7D%7B4%7D%5D%5E%7Bth%7D%5C%20obs.%3D%5B%5Cfrac%7B11%2B1%7D%7B4%7D%5D%5E%7Bth%7D%5C%20obs.%3D3%5E%7Brd%20%7D%5C%20obs.%20%3D8.3)
,
Q₁ = 8.3.
The median value is,
![[\frac{n+1}{2}]^{th}\ obs.=[\frac{11+1}{2}]^{th}\ obs.=6^{th}\ obs.=8.4](https://tex.z-dn.net/?f=%5B%5Cfrac%7Bn%2B1%7D%7B2%7D%5D%5E%7Bth%7D%5C%20obs.%3D%5B%5Cfrac%7B11%2B1%7D%7B2%7D%5D%5E%7Bth%7D%5C%20obs.%3D6%5E%7Bth%7D%5C%20obs.%3D8.4)
Median = 8.4
The upper quartile is,
![[\frac{3(n+1)}{4}]^{th}\ obs.=[\frac{3(11+1)}{4}]^{th}\ obs.=9^{th }\ obs. =8.5](https://tex.z-dn.net/?f=%5B%5Cfrac%7B3%28n%2B1%29%7D%7B4%7D%5D%5E%7Bth%7D%5C%20obs.%3D%5B%5Cfrac%7B3%2811%2B1%29%7D%7B4%7D%5D%5E%7Bth%7D%5C%20obs.%3D9%5E%7Bth%20%7D%5C%20obs.%20%3D8.5)
,
Q₃ = 8.5.
The maximum value is, Max. = 8.9.
So, the values of the box-plot are: B = {8.2, 8.3, 8.4, 8.5 and 8.9}.
So circumference is the distance around the wheel
that is the part that is actually touching the ground
so therefor the wheele travels 3 inches per rotations
5 rotations per minute
we have to find the distance
3 times (number of roations) times 1 minute=3 times 5 times 1=15
answer is 15 in per miute
I=15
