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Neko [114]
2 years ago
5

A banana peel weighs 1/12 of the total weight of a banana. If an unpeeled banana balances a peeled banana of the same​ weight, p

lus 11/12 of an​ ounce, how much does the banana weigh with the​ peel?
(NEED HELP ASAP 100PTS + Brainliest)
Mathematics
2 answers:
Advocard [28]2 years ago
7 0

Answer:

11 ounces

Step-by-step explanation:

Suppose, weight of the banana with the​ peel is x ounce.

x-11x/12

As the banana peel weighs  1/12 of the total weight of a banana, so the weight of the peel  1/12 ounce and the weight of a peeled banana11/12 ounce

12/12 ounce= weight 11 ounces

Ilia_Sergeevich [38]2 years ago
4 0

Answer:

11 ounces

Step-by-step explanation:

Unpeeled = X

Peeled = (1 - 1/12)X = (11/12)X

X = (11/12)X + (11/12)

X - (11/12)X = 11/12

(1/12)X = 11/12

X = 11 ounces

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Answer:

a) P(X=0)=(4C0)(0.75)^0 (1-0.75)^{4-0}=0.0039  

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d) P(X \geq 3) \geq 0.98

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Step-by-step explanation:

Previous concepts

The binomial distribution is a "DISCRETE probability distribution that summarizes the probability that a value will take one of two independent values under a given set of parameters. The assumptions for the binomial distribution are that there is only one outcome for each trial, each trial has the same probability of success, and each trial is mutually exclusive, or independent of each other".

Let X the random variable of interest, on this case we now that:

X \sim Binom(n=4, p=1-0.25=0.75)

The probability mass function for the Binomial distribution is given as:

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nCx=\frac{n!}{(n-x)! x!}

Part a

P(X=0)=(4C0)(0.75)^0 (1-0.75)^{4-0}=0.0039  

P(X=1)=(4C1)(0.75)^1 (1-0.75)^{4-1}=0.0469  

P(X=2)=(4C2)(0.75)^2 (1-0.75)^{4-2}=0.211  

P(X=3)=(4C3)(0.75)^2 (1-0.75)^{4-3}=0.422  

P(X=4)=(4C4)(0.75)^2 (1-0.75)^{4-4}=0.316

Part b

The expected value is givn by:

E(X) = np = 4*0.75=3

Part c

For the standard deviation we have this:

Sd(X) =\sqrt{np(1-p)}=\sqrt{4*0.75*(1-0.75)}=0.866

Part d

For this case the sample size needs to be higher or equal to 9. Since we need a value such that:

P(X \geq 3) \geq 0.98

And the dsitribution that satisfy this is X\sim Binom(n=9,p=0.75

We can verify this using the following code:

"=1-BINOM.DIST(3,9,0.75,TRUE)" and we got 0.99 and the condition is satisfied.

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