Answer:
Step-by-step explanation:
StartFraction 2 Over 5 EndFraction plus p equals StartFraction 4 Over 5 EndFraction plus StartFraction 3 Over 5 EndFraction
- 2/5 + p= 4/5 + 3/5p
- p - 3/5p = 4/5 - 2/5
- 2/5p = 2/5
- 2p = 2
- p = 1
Answer is 1
Answer:
72.3 - 39.1 = 4tens - 7ones + 2tenth
Step-by-step explanation:
Give the expression 72.3 + (-39.1)
opening the parenthesis:
= 72.3 + (-39.1)
= 72.3 - 39.1
Breaking the decimal values into place values
72.3 = 7tens + 2units + 3tenth
72.3 = 7(10)+2(1)+3(1/10)
72.3 =70+2+0.3
Similarly for 39.1
39.1 = 3tens + 9units + 1tenth
39.1 = 3(10)+9(1)+1(1/10)
39.1 =30+9+0.1
72.3 - 39.1 = 70+2+0.3 - (30+9+0.1)
72.3 - 39.1 = 70+2+0.3 - 30-9-0.1
72.3 - 39.1 = 70-30+2-9+0.3-0.1
72.3 - 39.1 = 40 - 7 +0.2
72.3 - 39.1 = 4tens - 7ones + 2tenth
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.