At the beginning, the chances that the teacher will choose a student who has forgotten his lunch are 2/10, or 1/5.
If the teacher now chooses another student from the remaining 9 students, the chances of his choosing a student who has forgotten his lunch are 1/9, since one of the forgetful students has already been removed from the original group of 10 students, and there are only 9 in the group remaining.
The probability that both students forgot lunch is (1/5)(1/9), or 1/45.
For the function 1,
To find the equation for this function, first you take any two points and from these points you will find the slope of line and then put it in the equation of line.
(x₁,y₁)=(0,4)
(x₂,y₂)=(1,7)
slope=m=y₂-y₁/x₂-x₁
=7-4/1-0
=3/1
=3
Equation of a line
f(x)=mx+c
where m is the slope and c is the y intercept.
f(x)=3x+4
For the function 2,
As you have given the equation of a function which is
f(x)=-2x+3
you can easily fill the table,
I do it for one value for other values you will do the same procedure.
For x=-2
f(x)=-2(-2)+3
= 7
x f(x)
-----------------
-2 7
-1 5
0 3
1 1
2 -1
For the Function 3,
You can get the values from the graph directly but you can also solve it by finding the equation first and then by putting method you get the value.
Equation of a line
f(x)=mx+c
where m is the slope and c is the y intercept.
m=(y₂-y₁)/(x₂-x₁)
(x₁,y₁)=(0,-3)
(x₂,y₂)=(6,0)
m=(0-(-3))/(6-0)
m=1/2
f(x)=1/2x-3
x f(x)
-----------------
-2 -4
-1 -7/2
0 -3
1 -5/2
2 -2
The graph of function 1 and 2 are attached below.
Perimeter (p) = 2×length (l) + 2×width (w)
p = 2l+2w
area (a) = l×w, so solve for one (I'll use l):

since p = 30, and a = 50, substitute the "a÷w" in for l in the perimeter equation:

Now plug in p and a values:


therefore width can be either 5 or 10 (but not both), so let's plug in:
l = a÷w = 50÷5 = 10
So if w = 5, then l = 10
D) 5 feet, and 10 feet
Answer:
Step-by-step explanation:
Given that EM company produces two types of laptop computer bags.
Let regular version produced be R and deluxe version be D
Total capital required
= 
Total labor hours required
=
Sales revenue = 
Solving the two constraints we have
10D
140
D can be atmost 14 and hence R can be 49
Otherwise if D is made 0, R = 65 maximum
If R is made 0, D maximum is 46
Thus corner points are (65,0) (0.,46) or (49,14)
Sales revenue for (65,0) = 2990
(0,46) is 2530
(49,14) is 3024
Maximum when R =14 and D is 49