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Alex
2 years ago
13

Jessica and Mike are making presentations for a class project. Jessica's slideshow starts with a verbal introduction that is 16

seconds long, and then each slide is left up for 2 seconds. Mike leaves each slide on screen for 3 seconds, and his introduction lasts 10 seconds. Jessica and Mike notice that their presentations have both the same number of slides and the same duration. How many slides are in each presentation? How long is each presentation? Explain why.
Mathematics
1 answer:
uysha [10]2 years ago
6 0

Answer:

they both has 7 slides and the presentation had a duration of 28 seconds

Step-by-step explanation:

we have to setup equation that makes them equal to each other if you know what i mean. 2x+16=3x+10

then we just solve for x

2x+16=3x+10

subtract 10 from each side

2x+6=3x

step 2 subtract 2x from each side

x=6

so both of their presentations are 7 slides long

then to find the length of their presentations we just plug in 6 to one of the equations

2(6)+16

2x6=12 12+16=28 so both of their presentations were 28 seconds

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2 years ago
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Step-by-step explanation:

The random variable <em>Y</em> is defined as the number of consecutive time intervals in which the water supply remains below a critical value <em>y₀</em>.

The random variable <em>Y</em> follows a Geometric distribution with parameter <em>p</em> = 0.409<em>.</em>

The probability mass function of a Geometric distribution is:

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(a)

Compute the probability that a drought lasts exactly 3 intervals as follows:

P(Y=3)=(1-0.409)^{3}\times 0.409=0.0844279\approx0.0844

Thus, the probability that a drought lasts exactly 3 intervals is 0.0844.

Compute the probability that a drought lasts at most 3 intervals as follows:

P (Y ≤ 3) =  P (Y = 0) + P (Y = 1) + P (Y = 2) + P (Y = 3)

              =(1-0.409)^{0}\times 0.409+(1-0.409)^{1}\times 0.409+(1-0.409)^{2}\times 0.409\\+(1-0.409)^{3}\times 0.409\\=0.409+0.2417+0.1429+0.0844\\=0.8780

Thus, the probability that a drought lasts at most 3 intervals is 0.8780.

(b)

Compute the mean of the random variable <em>Y</em> as follows:

\mu=\frac{1-p}{p}=\frac{1-0.409}{0.409}=1.445

Compute the standard deviation of the random variable <em>Y</em> as follows:

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The probability that the length of a drought exceeds its mean value by at least one standard deviation is:

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Thus, the probability that the length of a drought exceeds its mean value by at least one standard deviation is 0.2064.

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