What is equivalent to 0.0000001

Mathematics

1 answer:

strojnjashka [21]4 years ago

4 0

10E-7

Not sure if that was what you were going for but I hope it helps!

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Prove 2^n > n for all n equal to or greater than 1. I mostly need help with how to solve the problem when it is greater than

noname [10]

If $n$ is an integer, you can use induction. First show the inequality holds for $n=1$ . You have $2^1=2>1$ , which is true.

Now assume this holds in general for $n=k$ , i.e. that $2^k>k$ . We want to prove the statement then must hold for $n=k+1$ .

Because $2^k>k$ , you have

$2^{k+1}=2\times2^k>2k$

and this must be greater than $k+1$ for the statement to be true, so we require

$2k>k+1$

for $k>1$ . Well this is obviously true, because solving the inequality gives $3k>1\implies k>\dfrac13$ . So you're done.

If you $n$ is any real number, you can use derivatives to show that $2^n$ increases monotonically and faster than $n$ .

7 0

3 years ago

How to do splitting method of x square- 2x -8

Leni [432]

Answer:

Hey mate

kindly see attached picture:)

hope it helped you ☺️

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3 years ago

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A pharmaceutical company receives large shipments of aspirin tablets. The acceptance sampling plan is to randomly select and tes

hammer [34]

Answer:

a) The probability that this whole shipment will be accepted is 30%.

b) Many of the shipments with this rate of defective aspirin tablets will be rejected.

Step-by-step explanation:

We have a shipment of 3000 aspirin tablets, with a 5% rate of defects.

We select a sample of size 48 and test for defectives.

If more than one aspirin is defective, the batch is rejected.

The amount of defective aspirin tablets X can be modeled as a binomial distribution random variable, with p=0.55 and n=48

We have to calculate the probabilities that X is equal or less than 1: P(X≤1).

$P(X\leq1)=P(X=0)+P(X=1)\\\\\\P(0)=\binom{48}{0}(0.05)^0(0.95)^{48}=1*1*0.0853=0.0853\\\\\\P(1)=\binom{48}{1}(0.05)^1(0.95)^{47}=48*0.05*0.0897=0.2154\\\\\\P(X\eq1)=0.0853+0.2154=0.3007$