Answer:
13
Step-by-step explanation:
If he reads 119 pgs in 7 days that mean 119/7 = 17 he reads 17 pgs per day
next 340-119 = 221 then 221/17= 13
Answer:
32.8
Step-by-step explanation:
C^2 - B^2 = A^2
34.6 ^2 - 11^2 = A^2
1197.16 - 121 = A^2
1076.16 = A^2
Answer:
Step-by-step explanation:
In this particular case we have the following system of equations:
y
=
−
3
x
+
4
[
E
q
.
1
]
x
+
4
y
=
−
6
[
E
q
.
2
]
Substituting
[
E
q
.
1
]
in
[
E
q
.
2
]
:
x
+
4
(
−
3
x
+
4
)
=
−
6
Applying the distributive property on the left side:
x
−
12
x
+
16
=
−
6
Simplifying
:
−
11
x
=
−
22
Solving for
y
:
x
=
−
22
−
11
=
2
Substituting
x
=
2
in
[
E
q
.
1
]
:
y
=
−
3
(
2
)
+
4
=
−
2
Therefore
, the solutions are
x
=
2
and
y
=
−
2
Answer:
uhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhh
Step-by-step explanation:
Answer:
For first lamp ; The resultant probability is 0.703
For both lamps; The resultant probability is 0.3614
Step-by-step explanation:
Let X be the lifetime hours of two bulbs
X∼exp(1/1400)
f(x)=1/1400e−1/1400x
P(X<x)=1−e−1/1400x
X∼exp(1/1400)
f(x)=1/1400 e−1/1400x
P(X<x)=1−e−1/1400x
The probability that both of the lamp bulbs fail within 1700 hours is calculated below,
P(X≤1700)=1−e−1/1400×1700
=1−e−1.21=0.703
The resultant probability is 0.703
Let Y be a lifetime of another lamp two bulbs
Then the Z = X + Y will follow gamma distribution that is,
X+Y=Z∼gamma(2,1/1400)
2λZ∼
X+Y=Z∼gamma(2,1/1400)
2λZ∼χ2α2
The probability that both of the lamp bulbs fail within a total of 1700 hours is calculated below,
P(Z≤1700)=P(1/700Z≤1.67)=
P(χ24≤1.67)=0.3614
The resultant probability is 0.3614
