Find the missing factor B that makes the equality true.-35^6= (-5x^2)(b)
1 answer:
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Answer:
x=1/3
Step-by-step explanation:
A function f is given as
in the interval [0,1]
This function f being an algebraic polynomial is continuous in the interval [0,1] and also f is differntiable in the open interval (0,1)
Hence mean value theorem applies for f in the given interval
The value
Find derivative for f
Equate this to -5 to check mean value theorem
We find that 1/3 lies inside the interval (0,1)
Answer:
a) There is 17.64% probability that exactly two of the selected bulbs are rated 23-watt.
b) There is a 8.65% probability that all three of the bulbs have the same rating.
c) There is a 12.45% probability that one bulb of each type is selected.
Step-by-step explanation:
There are 24 compact fluorescent lightbulbs in the box, of which:
8 are rated 13-watt
9 are rated 18-watt
7 are rated 23-watt
(a) What is the probability that exactly two of the selected bulbs are rated 23-watt?
There are 7 rated 23-watt among 23. There are no replacements(so the denominators in the multiplication decrease). Then can be chosen in different orders, so we have to permutate.
It is a permutation of 3(bulbs selected) with 2(23-watt) and 1(13 or 18 watt) repetitions. So
There is 17.64% probability that exactly two of the selected bulbs are rated 23-watt.
(b) What is the probability that all three of the bulbs have the same rating?
is the probability that all three of them are 13-watt. So:
is the probability that all three of them are 18-watt. So:
is the probability that all three of them are 23-watt. So:
There is a 8.65% probability that all three of the bulbs have the same rating.
(c) What is the probability that one bulb of each type is selected?
We have to permutate, permutation of 3(bulbs), with (1,1,1) repetitions(one for each type). So
There is a 12.45% probability that one bulb of each type is selected.