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sveticcg [70]
3 years ago
8

whats another way to write (s-6)(s+1) when s represents the side of a square and the equation represents the area of a rectangle

?
Mathematics
1 answer:
viktelen [127]3 years ago
8 0

The another way to write (s - 6)(s + 1) is s² - 5s - 6

Step-by-step explanation:

Let us revise how multiply two binomial (a + b)(x + y)

  • Multiply the 1st terms
  • Multiply the 2nd terms
  • Multiply the nears and extremes, where nears are the 2nd term of first bracket and 1st term in the second bracket, the extremes are the 1st term in the first bracket and the 2nd term in the second bracket
  • Add the like terms if necessary

∵ The area of the square = (s - 6)(s + 1)

- To find the another way multiply the two brackets

∵ s × s = s²

∵ -6 × 1 = -6

∵ -6 × s = - 6s ⇒ nears

∵ s × 1 = s ⇒ extremes

- The terms - 6s and s are like terms, then add them

∵ - 6s + s = - 5s

∴ (s - 6)(s + 1) = s² - 5s - 6

∴ The area of the square = s² - 5s - 6

The another way to write (s - 6)(s + 1) is s² - 5s - 6

Learn more:

You can learn more about the binomials in brainly.com/question/2334388

#LearnwithBrainly

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There is a 70 percent chance that an airline passenger will check bags. In the next 16 passengers that check in for their flight
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Answer:

a) P(X=16)=(16C16)(0.7)^{16} (1-0.7)^{16-16}=0.00332

b) P(X

And we can find the individual probabilities like this:

P(X=10)=(16C10)(0.7)^{10} (1-0.7)^{16-10}=0.1649

P(X=11)=(16C11)(0.7)^{11} (1-0.7)^{16-11}=0.2099

P(X=12)=(16C12)(0.7)^{12} (1-0.7)^{16-12}=0.2040

P(X=13)=(16C13)(0.7)^{13} (1-0.7)^{16-13}=0.1465

P(X=14)=(16C14)(0.7)^{14} (1-0.7)^{16-14}=0.0732

P(X=15)=(16C15)(0.7)^{15} (1-0.7)^{16-15}=0.0228

P(X=16)=(16C16)(0.7)^{16} (1-0.7)^{16-16}=0.0033

And replacing we got:

P(X

c) P(x \geq 10) = P(X=10)+P(X=11)+P(X=12)+P(X=13)+P(X=14)+P(X=15)+P(X=16)

And replacing we got 0.825

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, on this case we now that:

X \sim Binom(n=17, p=0.7)

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

And we want to find this probability:

P(X=16)=(16C16)(0.7)^{16} (1-0.7)^{16-16}=0.00332

Part b fewer than 10 will check bags

We want this probability:

P(X

We can use the complement rule and we have:

P(X

And we can find the individual probabilities like this:

P(X=10)=(16C10)(0.7)^{10} (1-0.7)^{16-10}=0.1649

P(X=11)=(16C11)(0.7)^{11} (1-0.7)^{16-11}=0.2099

P(X=12)=(16C12)(0.7)^{12} (1-0.7)^{16-12}=0.2040

P(X=13)=(16C13)(0.7)^{13} (1-0.7)^{16-13}=0.1465

P(X=14)=(16C14)(0.7)^{14} (1-0.7)^{16-14}=0.0732

P(X=15)=(16C15)(0.7)^{15} (1-0.7)^{16-15}=0.0228

P(X=16)=(16C16)(0.7)^{16} (1-0.7)^{16-16}=0.0033

And replacing we got:

P(X

Part c at least 10 bags

We can find this probability like this:

P(x \geq 10) = P(X=10)+P(X=11)+P(X=12)+P(X=13)+P(X=14)+P(X=15)+P(X=16)

And replacing we got 0.825

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