0.3d−1 3 1 3 =0.84: 7 15

In a class of 10, there are 2 students who forgot their lunch. If the teacher chooses 2 students, what is the probability that b

Anuta_ua [19.1K]

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.

7 0

3 years ago

Fill in the graph function 2 and 3

nordsb [41]

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.

7 0

3 years ago

A rectangle has a perimeter of 30 feet and an area of 50 square feet. What are the dimensions of the rectangle?

REY [17]

Perimeter (p) = 2×length (l) + 2×width (w)
p = 2l+2w
area (a) = l×w, so solve for one (I'll use l):
$a = l \times w \\ l = a \div w \\ p = 2l + 2w = 2(l + w)$
since p = 30, and a = 50, substitute the "a÷w" in for l in the perimeter equation:
$p = 2(l + w) = 2((a \div w) + w) \\ = 2((a \div w) + ( {w}^{2} \div w)) \\ p= 2((a + {w}^{2}) \div w$
Now plug in p and a values:
$p= 2((a + {w}^{2}) \div w) \\ 30 = 2((50 + {w}^{2}) \div w) \\ 30 \div 2 = (50 + {w}^{2}) \div w \\ 15w = 50 + {w}^{2}$
$15w = 50 + {w}^{2} \\ {w}^{2} - 15w + 50 = 0 \\ (w - 5)(w - 10) = 0$
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

3 0

3 years ago

EM Company produces two types of laptop computer bags. The regular version requires $32 in capital and 4 hours of labor and sell

expeople1 [14]

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

= $32R+38D\leq 2100$