Answer:
6 and half for 27
Step-by-step explanation:
128 = 8d, because the product calls for multiplication to be performed.
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.
For the answer to this questions,<span>a. P (z ≤ z0) = 0.0401=P(z ≥ z0) = 1-0.0401 = 0.9599 = P(z ≤ -z0) = 0.9599
From tables z0 = -1.75
b. P (-z0 ≤ z ≤ z0) = .95 = P (z ≤ z0)- P (z ≤ -z0) = P (z ≤ z0)- P (z ≥ z0) =
P (z ≤ z0)-(1- P (z ≤ z0))
P (z ≤ z0) = (0.95+1)/2=0.975
From tables z0 = 1.96
c. P (-z0 ≤ z ≤ z0) = 0.90
the procedure is the same that exercise b P (z ≤ z0) = (0.9+1)/2=0.95
From tables the nearest value is z0 = 1.64
</span>d. P (-z0 ≤ z ≤ 0) = 0.2967= P (z ≤ 0) - P (z ≤ -z0) = P (z ≤ 0) - P (z ≥ z0) =
<span>P (z ≤ 0) - (1- P (z ≤ z0)) </span>
<span>P (z ≤ z0) = 0.2967 + 1 - P (z ≤ 0)= 0.2967 + 1 - 0.5 = 0.7967 </span>
<span>From tables z0 = 0.83
</span><span>
I hope my answer helped you
</span>
Answer:
x = 4
Step-by-step explanation:
6 : 4.5 = 4 : 3
x : 3 = 4: 3
x= 4