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3 years ago

14

(x2-13x-48)÷(x+3) dividing polybomials

1 answer:

-BARSIC- [3]3 years ago

5 0

The final awser is x-16
hope this is what u needed

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consider the exponential function g(x)=100 x (2). which of the following situations could be modeled by the function?

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The initial value of 100 that doubles over each interval.
without the answer choices, I can only describe it and give you an example of the graph.
I'm assuming the function is 100*(2)^x because if it is as listed it would be a quadratic function with a vertical stretch of 100.

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Use the explicit formula = 3 − 7( − 1) and

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3 years ago

On a map the distance between Juneau and Fairbanks is 5 inches. The approximate distance is 625 miles. The

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2437.5 Miles

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3 years ago

A public health organization reports that 40%of baby boys 6-8 months old in the United

Sveta_85 [38]

Using the binomial distribution, the probabilities are given as follows:

0.3675 = 36.75% probability that more than 4 weigh more than 20 pounds.

0.1673 = 16.73% probability that fewer than 3 weigh more than 20 pounds.

Since P(X > 7) < 0.05, it would be unusual if more than 7 of them weigh more than 20 pounds.

<h3>What is the binomial distribution formula?</h3>

The formula is:

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

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

The parameters are:

x is the number of successes.

n is the number of trials.

p is the probability of a success on a single trial.

The values of the parameters for this problem are:

n = 10, p = 0.4.

The probability that more than 4 weigh more than 20 pounds is:

$P(X > 4) = 1 - P(X \leq 4)$

In which:

$P(X \leq 4) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4)$

Then:

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 0) = C_{10,0}.(0.4)^{0}.(0.6)^{10} = 0.0061$

$P(X = 1) = C_{10,1}.(0.4)^{1}.(0.6)^{9} = 0.0403$

$P(X = 2) = C_{10,2}.(0.4)^{2}.(0.6)^{8} = 0.1209$

$P(X = 3) = C_{10,3}.(0.4)^{3}.(0.6)^{7} = 0.2150$

$P(X = 4) = C_{10,4}.(0.4)^{4}.(0.6)^{6} = 0.2502$

Hence:

$P(X \leq 4) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) = 0.0061 + 0.0403 + 0.1209 + 0.2150 + 0.2502 = 0.6325$

$P(X > 4) = 1 - P(X \leq 4) = 1 - 0.6325 = 0.3675$

0.3675 = 36.75% probability that more than 4 weigh more than 20 pounds.

The probability that fewer than 3 weigh more than 20 pounds is:

P(X < 3) = P(X = 0) + P(X = 1) + P(X = 2) = 0.0061 + 0.0403 + 0.1209 = 0.1673

0.1673 = 16.73% probability that fewer than 3 weigh more than 20 pounds.

For more than 7, the probability is:

$P(X > 7) = P(X = 8) + P(X = 9) + P(X = 10)$

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 8) = C_{10,8}.(0.4)^{8}.(0.6)^{2} = 0.0106$

$P(X = 9) = C_{10,9}.(0.4)^{9}.(0.6)^{1} = 0.0016$

$P(X = 10) = C_{10,10}.(0.4)^{10}.(0.6)^{0} = 0.0001$

Since P(X > 7) < 0.05, it would be unusual if more than 7 of them weigh more than 20 pounds.

More can be learned about the binomial distribution at brainly.com/question/24863377

#SPJ1

4 0

2 years ago