From the Figure :
Point A is (-3 , -3)
Point B is (6 , 6)
We know that, The Mid Point of Two Points (x₁ , y₁) and (x₂ , y₂) is given by :
![\mathsf{\implies [(\frac{x_1 + x_2}{2})\;,\;(\frac{y_1 + y_2}{2})]}](https://tex.z-dn.net/?f=%5Cmathsf%7B%5Cimplies%20%5B%28%5Cfrac%7Bx_1%20%2B%20x_2%7D%7B2%7D%29%5C%3B%2C%5C%3B%28%5Cfrac%7By_1%20%2B%20y_2%7D%7B2%7D%29%5D%7D)
![\mathsf{\implies Midpoint\;of\;Line\;AB\;is\;[(\frac{-3 + 6}{2})\;,\;(\frac{-3 + 6}{2})]}](https://tex.z-dn.net/?f=%5Cmathsf%7B%5Cimplies%20Midpoint%5C%3Bof%5C%3BLine%5C%3BAB%5C%3Bis%5C%3B%5B%28%5Cfrac%7B-3%20%2B%206%7D%7B2%7D%29%5C%3B%2C%5C%3B%28%5Cfrac%7B-3%20%2B%206%7D%7B2%7D%29%5D%7D)
![\mathsf{\implies Midpoint\;of\;Line\;AB\;is\;[(\frac{3}{2})\;,\;(\frac{3}{2})]}](https://tex.z-dn.net/?f=%5Cmathsf%7B%5Cimplies%20Midpoint%5C%3Bof%5C%3BLine%5C%3BAB%5C%3Bis%5C%3B%5B%28%5Cfrac%7B3%7D%7B2%7D%29%5C%3B%2C%5C%3B%28%5Cfrac%7B3%7D%7B2%7D%29%5D%7D)
Step-by-step explanation:
Let
x
be the kg of coffee of brand A in the mix and
y
be the kg of coffee of brand B in the mix.
The total kg must be
50
.
x
+
y
=
50
The cost per kg of the mix must br
$
7.20
. For this, the total cost of the mix will be
6
x
+
8
y
, so the total cost per kg of the mix will be
6
x
+
8
y
50
.
6
x
+
8
y
50
=
7.20
Now that we have our two equations, we can solve.
6
x
+
8
y
=
7.20
⋅
50
6
x
+
8
y
=
360
From the first equation, we can multiply both sides by
6
to get:
6
x
+
6
y
=
300
Subtracting, we get:
2
y
=
60
y
=
30
Thus, we need
30
kg of brand B in our mix. This means that
50
−
30
=
20
kg will be of brand A.
Answer:
-9/14
Step-by-step explanation:
This is my simplest-
Do the keep-change-flip method for this problem (and also any problem involving this)
KEEP -3/8
CHANGE the division sign to multiplication
FLIP the last fraction. 7/12 to 12/7
Now solve the problem by multiplying the fractions
-3/8 × 12/7 =
-3 × 12 = -36
8 × 7 = 56
-36 / 56 =
Simplify it by dividing the numerator (up of the line) and denominator (below the line) to get -9/14
Answer: 1/5, 1/2, 0.
Step-by-step explanation:
given data:
no of cameras = 6
no of cameras defective = 3
no of cameras selected = 2
Let p(t):=P(X=t)
p(2)=m/n,
m=binomial(3,2)=3!/2!= 3
n=binomial(6,2)=6!/2!/4! = 15
p(3)= 3/15
= 1/5.
p(1)=m/n,
m=binomial(6,1)*binomial(2,2)=6!/1!/4!*2!/2!/0!= 7.5
n=binomial(6,2)= 15
p(2)= 7.5/15
= 1/2
p(0)=m/n,
m=0
p(0)=0
