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Ainat [17]
3 years ago
9

On one highway, Gabriela noticed that they passed mile marker 123 at 1:00. She then saw that they reached mile marker 277at 3:00

. Since Mr. Morales was driving at a constant speed, their mile-marker location over time can be represented by a line where the time in hours is the independent variable and the mile marker is the dependent variable. The points (1,123) and (3,277) are two points on this line.
What is the value of the slope of this line?
Mathematics
1 answer:
Elena L [17]3 years ago
6 0

<u>Given</u>:

Let time be the independent variable.

Let mile marker be the dependent variable.

On one highway, Gabriela noticed that they passed mile marker 123 at 1:00. She then saw that they reached mile marker 277 at 3:00 and Mr. Morales was driving at a constant speed.

The coordinates of the points on this line are (1,123) and (3,277)

We need to determine the slope of the line.

<u>Slope of the line:</u>

The slope of the line can be determined using the formula,

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

Substituting the points (1,123) and (3,277) in the above formula, we get;

m=\frac{277-123}{3-2}

m=\frac{154}{1}

m=154

Therefore, the value of the slope of this line is 154.

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Answer:

(a) The probability that only one goblet is a second among six randomly selected goblets is 0.3888.

(b) The probability that at least two goblet is a second among six randomly selected goblets is 0.1776.

(c) The probability that at most five must be selected to find four that are not seconds is 0.9453.

Step-by-step explanation:

Let <em>X</em> = number of seconds in the batch.

The probability of the random variable <em>X</em> is, <em>p</em> = 0.31.

The random variable <em>X</em> follows a Binomial distribution with parameters <em>n</em> and <em>p</em>.

The probability mass function of <em>X</em> is:

P(X=x)={n\choose x}p^{x}(1-p)^{n-x};\ x=0,1,2,3...

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P(X=1)={6\choose 1}0.13^{1}(1-0.13)^{6-1}\\=6\times 0.13\times 0.4984\\=0.3888

Thus, the probability that only one goblet is a second among six randomly selected goblets is 0.3888.

(b)

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P (X ≥ 2) = 1 - P (X < 2)

              =1-{6\choose 0}0.13^{0}(1-0.13)^{6-0}-{6\choose 1}0.13^{1}(1-0.13)^{6-1}\\=1-0.4336+0.3888\\=0.1776

Thus, the probability that at least two goblet is a second among six randomly selected goblets is 0.1776.

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If goblets are examined one by one then to find four that are not seconds we need to select either 4 goblets that are not seconds or 5 goblets including only 1 second.

P (4 not seconds) = P (X = 0; n = 4) + P (X = 1; n = 5)

                            ={4\choose 0}0.13^{0}(1-0.13)^{4-0}+{5\choose 1}0.13^{1}(1-0.13)^{5-1}\\=0.5729+0.3724\\=0.9453

Thus, the probability that at most five must be selected to find four that are not seconds is 0.9453.

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