Star = 5
square =3
or the other way around
Answer:
ANSWER: ADC=82
bda = 144
(7x+34)+(9x+46)=144
16x+80=144
16x=64
x=4
9(4)+46=82
adc=82
Step-by-step explanation:
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
Answer:
Second option: (-2,-3) and (1,0)
Step-by-step explanation:
Given the system of equations 
, you can rewrite them in this form:
Simplify:
Factor the quadratic equation. Choose two number whose sum be 1 and whose product be -2. These are: 2 and -1, then:
Substitute each value of "x" into any of the original equation to find the values of "y":
Then, the solutions are:
(-2,-3) and (1,0)
Answer:
a) x = 1225.68
b) x = 1081.76
c) 1109.28 < x < 1198.72
Step-by-step explanation:
Given:
- Th random variable X for steer weight follows a normal distribution:
X~ N( 1154 , 86 )
Find:
a) the highest 10% of the weights?
b) the lowest 20% of the weights?
c) the middle 40% of the weights?
Solution:
a)
We will compute the corresponding Z-value for highest cut off 10%:
Z @ 0.10 = 1.28
Z = (x-u) / sd
Where,
u: Mean of the distribution.
s.d: Standard deviation of the distribution.
1.28 = (x - 1154) / 86
x = 1.28*86 + 1154
x = 1225.68
b)
We will compute the corresponding Z-value for lowest cut off 20%:
-Z @ 0.20 = -0.84
Z = (x-u) / sd
-0.84 = (x - 1154) / 86
x = -0.84*86 + 1154
x = 1081.76
c)
We will compute the corresponding Z-value for middle cut off 40%:
Z @ 0.3 = -0.52
Z @ 0.7 = 0.52
[email protected] < x < [email protected]
-.52*86 + 1154 < x < 0.52*86 + 1154
1109.28 < x < 1198.72
