Answer:
a) X ~ 
b) μ = 100/3
c) 
d) A battery is expected to last 100/3 months (33 months and 10 days approximately).
e) For seven batteries, i would expect them to last 700/3 months (approximately 19 years, 5 months and 10 days).
Step-by-step explanation:
a) The life of a battery is usually modeled with an exponential distribution X ~
b) The mean of X is μ = 1/0.03 = 100/3
c) The standard deviation is 
d) The expected value of the bateery life is equal to its mean, hence it is 100/3 months.
e) The expected value of 7 (independent) batteries is the sum of the expected values of each one, hence it is 7*100/3 = 700/3 months.
Answer:
Starting time: 1:37
Step-by-step explanation:
Instead of proof to solve the problem, let us prove that the starting time above is correct;
1. There are 60 minutes in an hour, so 60 - 37 = the minutes 'till 2:00, or 23 minutes
2. Knowing that the time the truck stopped driving through the neighborhood was 5 minutes past 2:00, the minute difference being 23 + 5, or <em>28 minutes</em>
<em>Proved: 28 minutes</em>
Answer:
m<N = 76°
Step-by-step explanation:
Given:
∆JKL and ∆MNL are isosceles ∆ (isosceles ∆ has 2 equal sides).
m<J = 64° (given)
Required:
m<N
SOLUTION:
m<K = m<J (base angles of an isosceles ∆ are equal)
m<K = 64° (Substitution)
m<K + m<J + m<JLK = 180° (sum of ∆)
64° + 64° + m<JLK = 180° (substitution)
128° + m<JLK = 180°
subtract 128 from each side
m<JLK = 180° - 128°
m<JLK = 52°
In isosceles ∆MNL, m<MLN and <M are base angles of the ∆. Therefore, they are of equal measure.
Thus:
m<MLN = m<JKL (vertical angles are congruent)
m<MLN = 52°
m<M = m<MLN (base angles of isosceles ∆MNL)
m<M = 52° (substitution)
m<N + m<M° + m<MLN = 180° (Sum of ∆)
m<N + 52° + 52° = 180° (Substitution)
m<N + 104° = 180°
subtract 104 from each side
m<N = 180° - 104°
m<N = 76°
Remark
Let the pigs = p
Let the chickens = c
Givens
30 heads is another way of saying that there were 30 creatures in the barnyard. In addition each chicken has 2 legs and each pig has 4
c + p = 30
2C + 4P = 100 (2) Divide (2) by 2
C + 2P = 50 (3) Subtract (1) from (3)
<u>C + P = 30</u>
P = 20
So there were 20 pigs.
The probability that his number will be 4 is one in eleven, or 1/11.
