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mihalych1998 [28]

4 years ago

13

A store ladders.The retail price was a 40 percent makeup over the manfacturer price. A mounth later the store ruduced thr retail

price of the ladder by 25 percent what percent makeup is thr new retail price over yhe manufacturer price?

1 answer:

likoan [24]4 years ago

3 0

Could Be About 30Ish Maybe.

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1Step-by-step explanation:

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F(x) = |x|; Find f(8)

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<h2><u>Answer:</u></h2>

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<h2><u>Step-by-step explanation:</u></h2>

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3 years ago

(CO 3) Fifty-four percent of US teens have heard of a fax machine. You randomly select 12 US teens. Find the probability that th

11111nata11111 [884]

Answer:

a) $P(X=6)=(12C6)(0.54)^6 (1-0.54)^{12-6}=0.217$

b) $P(X> 8) = P(X\geq 9)= P(X=9)+P(X=10)+P(X=11)+P(X=12)$

$P(X=9)=(12C9)(0.54)^9 (1-0.54)^{12-9}=0.0836$

$P(X=10)=(12C10)(0.54)^{10} (1-0.54)^{12-10}=0.0294$

$P(X=11)=(12C11)(0.54)^{11} (1-0.54)^{12-11}=0.00628$

$P(X=12)=(12C12)(0.54)^{12} (1-0.54)^{12-12}=0.000615$

And adding the values we got:

$P(X> 8) = P(X\geq 9)= P(X=9)+P(X=10)+P(X=11)+P(X=12)= 0.0836+0.0294+0.00628+0.000615 = 0.120$

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".

Solution to the problem

Let X the random variable of interest "number of teens that have heard of a fax machine", on this case we now that:

$X \sim Binom(n=12, p=0.54)$

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

$P(X)=(nCx)(p)^x (1-p)^{n-x}$

Where (nCx) means combinatory and it's given by this formula:

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

Part a

For this case we want this probability:

$P(X=6)=(12C6)(0.54)^6 (1-0.54)^{12-6}=0.217$

Part b

For this case we want this probability:

$P(X> 8) = P(X\geq 9)= P(X=9)+P(X=10)+P(X=11)+P(X=12)$

$P(X=9)=(12C9)(0.54)^9 (1-0.54)^{12-9}=0.0836$

$P(X=10)=(12C10)(0.54)^{10} (1-0.54)^{12-10}=0.0294$

$P(X=11)=(12C11)(0.54)^{11} (1-0.54)^{12-11}=0.00628$

$P(X=12)=(12C12)(0.54)^{12} (1-0.54)^{12-12}=0.000615$

And adding the values we got:

$P(X> 8) = P(X\geq 9)= P(X=9)+P(X=10)+P(X=11)+P(X=12)= 0.0836+0.0294+0.00628+0.000615 = 0.120$

5 0

3 years ago

Read 2 more answers