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3 years ago

8

The total charge for a taxi ride in NYC includes an additional fee of $3.75 plus $1.25 for every half mile traveled Jody took a

taxi and the ride cost exactly $12.50 how many miles did you travel in the taxi

1 answer:

laiz [17]3 years ago

5 0

You can write a relationship where x<span> will be the distance traveled in </span><span>12</span><span> miles:</span>
<span>3.75+1.25⋅x=12.5</span>
<span>solve for </span>x:
<span>x=<span><span>12.5−3.75</span>1.25</span>=7</span>
<span>but </span>x<span> corresponds to </span><span>12</span><span> miles so that Jodi traveled </span><span>7⋅<span>12</span>=3.5</span><span> miles</span><span />

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Read 2 more answers

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Answer:

a) There is a 18.73% probability that exactly two students use credit cards because of the rewards program.

b) There is a 71.62% probability that more than two students use credit cards because of the rewards program.

c) There is a 82% probability that between two and five students, inclusive, use credit cards because of the rewards program.

Step-by-step explanation:

There are only two possible outcomes. Either the student use credit cards because of the rewards program, or they use for other reason. So, we can solve this problem by the binomial distribution.

Binomial probability

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

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

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

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

And $\pi$ is the probability of X happening.

In this problem, we have that:

10 student are sampled, so $n = 10$

34% of college students say they use credit cards because of the rewards program, so $\pi = 0.34$

(a) exactly two

This is P(X = 2).

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

$P(X = 2) = C_{10,2}.(0.34)^{2}.(0.66)^{8} = 0.1873$

There is a 18.73% probability that exactly two students use credit cards because of the rewards program.

(b) more than two

This is $P(X > 2)$ .

Either a value is larger than two, or it is smaller of equal. The sum of the decimal probabilities must be 1. So:

$P(X \leq 2) + P(X > 2) = 1$

$P(X > 2) = 1 - P(X \leq 2)$

In which

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

So

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

$P(X = 0) = C_{10,0}.(0.34)^{0}.(0.66)^{10} = 0.0157$

$P(X = 1) = C_{10,1}.(0.34)^{1}.(0.66)^{9} = 0.0808$

$P(X = 2) = C_{10,2}.(0.34)^{2}.(0.66)^{8} = 0.1873$

$P(X \leq 2) = P(X = 0) + P(X = 1) + P(X = 2) = 0.0157 + 0.0808 + 0.1873 = 0.2838$

$P(X > 2) = 1 - P(X \leq 2) = 1 - 0.2838 = 0.7162$

There is a 71.62% probability that more than two students use credit cards because of the rewards program.

(c) between two and five inclusive

This is:

$P = P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5)$

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

$P(X = 2) = C_{10,2}.(0.34)^{2}.(0.66)^{8} = 0.1873$

$P(X = 3) = C_{10,3}.(0.34)^{3}.(0.66)^{7} = 0.2573$

$P(X = 4) = C_{10,4}.(0.34)^{4}.(0.66)^{6} = 0.2320$

$P(X = 5) = C_{10,5}.(0.34)^{5}.(0.66)^{5} = 0.1434$

$P = P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5) = 0.1873 + 0.2573 + 0.2320 + 0.1434 = 0.82$

There is a 82% probability that between two and five students, inclusive, use credit cards because of the rewards program.

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3 years ago

Find an equation of the line passing through the given points. Use function notation to write the equation. (3,8) and (4,13

Olegator [25]

Answer:

$f(x)=5x-7$

Step-by-step explanation:

The given line passes through the points;

$(x_1,y_1)=(3,8)$

and

$(x_2,y_2)=(4,13)$

Let us find the slope using;

$m=\frac{y_2-y_1}{x_2-x_1}$

$m=\frac{13-8}{4-3}$

$m=\frac{5}{1}=5$

The equation is given by;

$y-y_1=m(x-x_1)$

We substitute the values into the formula to obtain;

$y-8=5(x-3)$

$y=5x-15+8$

$y=5x-7$

There the equation of the line is

$f(x)=5x-7$

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3 years ago