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3 years ago

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The table shows the cost of a game from 2000 to 2004, which has been increasing in a quadratic fashion. Let x = 0 in 2000, and f

ind the best-fit quadratic equation. What will game cost in 2010? A) $417 Eliminate B) $746 C) $960 D) $1,586

1 answer:

Burka [1]3 years ago

5 0

Answer: Here we are going to use the equation y=15x2+3x+56.

x=0 in 2,000 so x=10 in 2010.

Substitute 10 for x.

After this we have the answer: D. $1,586

<u>Hope this helps</u>

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The proportion of high school seniors who are married is 0.02. Suppose we take a random sample of 300 high school seniors; a.) F

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Answer:

a) Mean 6, standard deviation 2.42

b) 10.40% probability that, in our sample of 300, we find that 8 of the seniors are married.

c) 14.85% probability that we find less than 4 of the seniors are married.

d) 99.77% probability that we find at least 1 of the seniors are married

Step-by-step explanation:

For each high school senior, there are only two possible outcomes. Either they are married, or they are not. The probability of a high school senior being married is independent from other high school seniors. 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 standard deviation of the binomial distribution is:

$\sqrt{V(X)} = \sqrt{np(1-p)}$

In this problem, we have that:

$n = 300, p = 0.02$

a.) Find the mean and standard deviation of the sample count X who are married.

Mean

$E(X) = np = 300*0.02 = 6$

Standard deviation

$\sqrt{V(X)} = \sqrt{np(1-p)} = \sqrt{300*0.02*0.98} = 2.42$

b.) What is the probability that, in our sample of 300, we find that 8 of the seniors are married?

This is P(X = 8).

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

$P(X = 8) = C_{300,8}.(0.02)^{8}.(0.98)^{292} = 0.1040$

10.40% probability that, in our sample of 300, we find that 8 of the seniors are married.

c.) What is the probability that we find less than 4 of the seniors are married?

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

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

$P(X = 0) = C_{300,0}.(0.02)^{0}.(0.98)^{300} = 0.0023$

$P(X = 1) = C_{300,1}.(0.02)^{1}.(0.98)^{299} = 0.0143$

$P(X = 2) = C_{300,2}.(0.02)^{2}.(0.98)^{298} = 0.0436$

$P(X = 3) = C_{300,3}.(0.02)^{3}.(0.98)^{297} = 0.0883$

$P(X < 4) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) = 0.0023 + 0.0143 + 0.0436 + 0.0883 = 0.1485$

14.85% probability that we find less than 4 of the seniors are married.

d.) What is the probability that we find at least 1 of the seniors are married?

Either no seniors are married, or at least 1 one is. The sum of the probabilities of these events is decimal 1. So

$P(X = 0) + P(X \geq 1) = 1$

From c), we have that $P(X = 0) = 0.0023$ . So

$0.0023 + P(X \geq 1) = 1$

$P(X \geq 1) = 0.9977$

99.77% probability that we find at least 1 of the seniors are married

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Factor this expression completely and then place the factors in the proper location. Note: Place factors alphabetically!

Usimov [2.4K]

To solve this problem you must apply the proccedure shown below:

1. You have the following expression given in the problem above:

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2. The first thing you must do to factor the expression is to group the terms, as following:

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What is the slope of the line that passes through (3, −7) and (−1, 1)? (1 point)

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Answer:

The slope of the line is -2

Step-by-step explanation:

<em>The rule of the slope of a line is</em>:

$m=\frac{y_{2}-y_{1}}{x_{2}-x_{1}}$ , where

$(x_{1},y_{1})$ and $(x_{2},y_{2})$ are two points on the line

<em>Let us find the slope using the rule above</em>

∵ The line passes through points (3, -7) and (-1 , 1)

∴ $x_{1}$ = 3 and $x_{2}$ = -1

∴ $y_{1}$ = -7 and $y_{2}$ = 1

<em>Substitute these values in the rule above to find m</em>

∴ $m=\frac{1--7}{-1-3}=\frac{1+7}{-4}=\frac{8}{-4}=-2$

∴ The slope of the line that passes through the given points is -2

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4 years ago