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Naddik [55]
3 years ago
7

Write an expression with 4 terms, containing the coefficients 3, 6, And 9

Mathematics
1 answer:
Lynna [10]3 years ago
7 0
Let the variable used in the expression be "x".

Since the desired expression should contain 5 terms, therefore the powers of "x" will be as follows:
x^0 , x^1 , x^2 and x^3 where x^0 = 1 and x^1 = x

Now we are given only 3 coefficients to use, so we will have to assume a fourth one.
Assume the fourth coefficient to be 7

It is a convention to write the powers in an expression in descending order, so the expression will be as follows (without coefficients):
x^3 + x^2 + x + constant

Now add the coefficients in any desired order since there are no conditions, the final expression can be written as follows:
3 x^3 + 6 x^2 + 9 x + 7
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Which of the following is a polynomial?
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Read 2 more answers
A home security system is designed to have a 99% reliability rate. Suppose that nine homes equipped with this system experience
bulgar [2K]

Answer:

a) P(X \geq 1) = 1-P(X

P(X=0)=(9C0)(0.99)^0 (1-0.99)^{9-0}=1x10^{-18}

And replacing we got:

P(X \geq 1) =1 -1x10^{-18} \approx 1

b) P(X=7)=(9C7)(0.99)^7 (1-0.99)^{9-7}=0.003355

P(X=8)=(9C8)(0.99)^8 (1-0.99)^{9-8}=0.083047

P(X=9)=(9C9)(0.99)^9 (1-0.99)^{9-9}=0.913517

And adding we got:

P(X \geq 7) = 0.003355+0.083047+0.913517 =0.99992

c) P(X \leq 8) =1 -P(X>8) = 1-P(X=9)

P(X=9)=(9C9)(0.99)^9 (1-0.99)^{9-9}=0.913517

And replacing we got:

P(X \leq 8)= 1-0.913517=0.086483

Step-by-step explanation:

Let X the random variable of interest "numebr of times that an alarm is triggered", on this case we now that:

X \sim Binom(n=9, p=0.99)

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

We want to find this probability:

P(X \geq 1) = 1-P(X

P(X=0)=(9C0)(0.99)^0 (1-0.99)^{9-0}=1x10^{-18}

And replacing we got:

P(X \geq 1) =1 -1x10^{-18} \approx 1

Part b

P(X \geq 7)= P(X=7) +P(X=8)+ P(X=9)

P(X=7)=(9C7)(0.99)^7 (1-0.99)^{9-7}=0.003355

P(X=8)=(9C8)(0.99)^8 (1-0.99)^{9-8}=0.083047

P(X=9)=(9C9)(0.99)^9 (1-0.99)^{9-9}=0.913517

And adding we got:

P(X \geq 7) = 0.003355+0.083047+0.913517 =0.99992

Part c

P(X \leq 8) =1 -P(X>8) = 1-P(X=9)

P(X=9)=(9C9)(0.99)^9 (1-0.99)^{9-9}=0.913517

And replacing we got:

P(X \leq 8)= 1-0.913517=0.086483

4 0
3 years ago
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