Andrew has one book that is 237 inches thick and a second book that is 3.56 inches thick. If he stacks the books, about how tall
1 answer:
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If 19 out of 28 renters keep a pet, there are renters who don't keep a pet.
Whenever you have a subset of some set, and you want to know which percentage of the set the subset represents, you simply have to compute
So, in your case, you're wondering what percentage of 28 does 9 represent. So, the formula becomes
Answer:
(a) The probability of exactly three flaws in 150 m of cable is 0.21246
(b) The probability of at least two flaws in 100m of cable is 0.69155
(c) The probability of exactly one flaw in the first 50 m of cable, and exactly one flaw in the second 50 m of cable is 0.13063
Step-by-step explanation:
A random variable X has a Poisson distribution and it is referred to as Poisson random variable if and only if its probability distribution is given by
for x = 0, 1, 2, ...
where , the mean number of successes.
(a) To find the probability of exactly three flaws in 150 m of cable, we first need to find the mean number of flaws in 150 m, we know that the mean number of flaws in 50 m of cable is 1.2, so the mean number of flaws in 150 m of cable is
The probability of exactly three flaws in 150 m of cable is
(b) The probability of at least two flaws in 100m of cable is,
we know that the mean number of flaws in 50 m of cable is 1.2, so the mean number of flaws in 100 m of cable is
(c) The probability of exactly one flaw in the first 50 m of cable, and exactly one flaw in the second 50 m of cable is
The occurrence of flaws in the first and second 50 m of cable are independent events. Therefore the probability of exactly one flaw in the first 50 m and exactly one flaw in the second 50 m is
Answer:
There will be total 2048 parts of the given paper if James if able to fold the paper eleven times.
The needed function is
Step-by-step explanation:
Let us assume the piece of paper James decides to fold is a SQUARE.
Now, let us assume:
n : the number of times the paper is folded.
y : The number of parts obtained after folds.
Now, if the paper if folded ONCE ⇒ n = 1
Also, when the pap er is folded once, the parts obtained are TWO equal parts.
⇒ for n = 1 , y = 2 ..... (1)
Similarly, if the paper if folded TWICE ⇒ n = 2
Also, when the paper is folded twice, the parts obtained are FOUR equal parts.
⇒ for n = 2 , y = 4 ..... (2)
⇒
Continuing the same way, if the paper is folded SEVEN times ⇒ n = 7
So,
⇒ There are total 128 equal parts.
Lastly, if the paper is folded ELEVEN times ⇒ n = 11
So,
⇒ There are total 2048 equal parts.
Hence, there will be total 2048 parts of the given paper if James if able to fold the paper eleven times.
And the needed function is