Answer:
<h3>20</h3>
Step-by-step explanation:
Standard deviation of the mean is expressed as:
SM = s/√n
s is the standard deviation of all invoices
n is the sample size of the invoiced
Given
s = $200
n = 100
Required
Standard deviation of the mean
Substitute the given parameters into the given equation
SM = 200/√100
SM = 200/10
SM = 20
Hence the standard deviation of xbar is 20
You first have to equalize all of the denominator
2/3 will be equal to : 8/12
Total of their practice time would be :
11/12 + 8/12 = 19/12 hours
Hope this helps
c is there answer Step-by-step explanation: 4 is greater than 3 and because the aligator eats the biggest number lol
The total cost of the fencing is a function of the length of a side x (feet) is 10x + 14y.
<h3>How to find the perimeter?</h3>
The perimeter is made by adding the south, north, east, and west side.
Given that
The cost of fencing for the east and west sides is $7 per foot, and the cost of fencing for the north and south sides is only $5 per foot.
We have to find total cost of the fencing is a function of the length of a side x (feet).
The east and west fencing cost is $7/ft but the south and north fencing cost is $5/ft.
x = north and south fencing
y = the east and west sides, then you can get this equation.
Total cost= (south +north) × 5 + (east+ west) × 7
Total cost =(x + x) × 5 + (y+ y) × 7
Total cost = 5(2x) + 7(2y)
Total cost = 10x + 14y
Hence, the total cost of the fencing is a function of the length of a side x (feet) is 10x + 14y.
To know more about the Equation click the link given below.
brainly.com/question/1687505
#SPJ1
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>