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3 years ago

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After plotting the data where x represents the length of the side of a polygon, and f(x) represents the area of the polygon, Jac

k used technology and determined the appropriate model to approximate the area of the polygon, f(x) = x2 + 2x + 5. Use the model Jack created to predict the area of a polygon that has a side length of 3. 10 18 19 20

2 answers:

konstantin123 [22]3 years ago

7 0

Answer:

Option d. 20 $unit^{2}$

Step-by-step explanation:

After plotting data determined the appropriate model to approximate the area of polygon is

$f(x)=x^{2}+2x+5$ , here x = length of the side of a polygon

We have to find the area of a polygon that has a side length = 3

$f(3)=3^{2}+2*3+5$

= 9 + 6 + 5

= 15 + 5 = 20

Option d. 20 $unit^{2}$ is the answer.

Zigmanuir [339]3 years ago

3 0

Answer:

See the attachment

Step-by-step explanation:

It is convenient to use a spreadsheet or graphing calculator for this.

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-2(-4x + 2y - 6)

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Answer:

a) 2.43% probability that exactly 8 of the 20 Internet browser users use Chrome as their Internet browser.

b) 80.5% probability that at least 3 of the 20 Internet browser users use Chrome as their Internet browser.

c) 4.074

d) Variance 3.24, standard deviation 1.8

Step-by-step explanation:

For each internet browser user, there are only two possible outcomes. Either they use chrome, or they do not. They are chosen at random, which means that the probability of an user using chrome is independent from other users. So we use the binomial probability distribution to solve this question.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

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

In which $C_{n,x}$ is the number of different combinations of x objects from a set of n elements, given by the following formula.

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

And p is the probability of X happening.

The expected value of the binomial distribution is:

$E(X) = np$

The variance of the binomial distribution is:

$V(X) = np(1-p)$

20.37% share of the browser market

This means that $p = 0.2037$

Group of 20 Internet browser users

This means that $n = 20$

(a) Compute the probability that exactly 8 of the 20 Internet browser users use Chrome as their Internet browser.

This is $P(X = 8)$

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

$P(X = 8) = C_{20,8}.(0.2037)^{8}.(0.7963)^{12} = 0.0243$

2.43% probability that exactly 8 of the 20 Internet browser users use Chrome as their Internet browser.

(b) Compute the probability that at least 3 of the 20 Internet browser users use Chrome as their Internet browser.

Either less than 3 users use Chrome, or at least 3 do. The sum of the probabilities of these events is decimal 1. So

$P(X < 3) + P(X \geq 3) = 1$

So

$P(X \geq 3) = 1 - P(X < 3)$

In which

$P(X < 3) = P(X = 0) + P(X = 1) + P(X = 2)$

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

$P(X = 0) = C_{20,0}.(0.2037)^{0}.(0.7963)^{20} = 0.0105$

$P(X = 1) = C_{20,1}.(0.2037)^{1}.(0.7963)^{19} = 0.0538$

$P(X = 2) = C_{20,2}.(0.2037)^{2}.(0.7963)^{18} = 0.1307$

$P(X < 3) = P(X = 0) + P(X = 1) + P(X = 2) = 0.0105 + 0.0538 + 0.1307 = 0.195$

Finally

$P(X \geq 3) = 1 - P(X < 3) = 1 - 0.195 = 0.805$

80.5% probability that at least 3 of the 20 Internet browser users use Chrome as their Internet browser.

(c) For the sample of 20 Internet browser users, compute the expected number of Chrome users.

$E(X) = np = 20*0.2037 = 4.074$

(d) For the sample of 20 Internet browser users, compute the variance and standard deviation for the number of Chrome users.

$V(X) = np(1-p)$

$V(X) = 20*0.2037*0.7963 = 3.24$

The standard deviation is the square root of the variance. SO

$\sqrt{V(X)} = \sqrt{3.24} = 1.8$

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