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3 years ago

How to solve for 93,94, and 95

Mathematics

1 answer:

kvasek [131]3 years ago

3 0

Answer:

93- not a triangle

94- right triangle

95- I dont know sorry!

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Evaluate −b2−2bx2−x when x=−2.

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The answer...... is 6x+2

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3 years ago

I need help with #2 please I don’t understand

kipiarov [429]

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4. Pentagon
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3 years ago

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If 90 percent of automobiles in Orange County have both headlights working, what is the probability that in a sample of eight au

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Answer:

$P(X \geq 7) = P(X=7) +P(X=8)$

And we can find the individual probabilities using the probability mass function

$P(X=7)=(8C7)(0.9)^7 (1-0.9)^{8-7}=0.3826$

$P(X=8)=(8C8)(0.9)^8 (1-0.9)^{8-8}=0.4305$

And replacing we got:

$P(X \geq 7) = P(X=7) +P(X=8)=0.3826 +0.4305=0.8131$

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 automobiles with both headligths working", on this case we now that:

$X \sim Binom(n=8, p=0.9)$

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:

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

And for this case we want to find this probability:

$P(X \geq 7) = P(X=7) +P(X=8)$

And we can find the individual probabilities using the probability mass function

$P(X=7)=(8C7)(0.9)^7 (1-0.9)^{8-7}=0.3826$

$P(X=8)=(8C8)(0.9)^8 (1-0.9)^{8-8}=0.4305$

And replacing we got:

$P(X \geq 7) = P(X=7) +P(X=8)=0.3826 +0.4305=0.8131$

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2 years ago

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Answer:

Step-by-step explanation:

4.56 $

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3 years ago

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Express the product in simplest form.

Romashka [77]

$\frac{8}{2x+8} * \frac{ x^{2}-16 }{4}$

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2. Multiply numerators
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3. Multiply denominators
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