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mash [69]
3 years ago
9

Unit 2 Lesson 11 Cool Down Graph

Mathematics
2 answers:
pantera1 [17]3 years ago
7 0

Answer:

2.4

Step-by-step explanation:

12/5=2.4

tatuchka [14]3 years ago
3 0

Answer:2.4

Step-by-step explanation:

because 12by 5 would equal 2.4

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Please help with all of question 4.​
olga55 [171]

Answer:

Step-by-step explanation:

Combine like terms: Like terms have same variable with same power and to combine the like terms, add/subtract the co efficient of the variables.

<h3>Perimeter:</h3>

      Perimeter = sum of all sides

a) Perimeter of ΔABC = AB + BC + CA

                                    = x + 14 + x + 14 + x + 14

                                    = x + x + x + 14 + 14 + 14    {Combine like terms}

                                   = 3x + 42

b) EF = DI - GH

        = 2x + 3 - x

       = 2x - x + 3

       = x + 3

c) FG = HI - ED

         = 12 + 2x - (x + 5)

         = 12 + 2x - x - 5 {To open the brackets, (-1) is distributed to x and 5}

         = 12 - 5 + 2x - x

         = 7 + x

d) Perimeter of DEFGHI = DE +  EF + FG + GH + HI + ID

                                       =  x + 5 + x + 3 + 7 +x  + x + 12 +2x + 2x + 3

                                       = x +x + x + x + 2x + 2x + 5 + 3 + 7 + 3 + 12  

                                       = 8x + 30

6 0
2 years ago
According to the article "Characterizing the Severity and Risk of Drought in the Poudre River, Colorado" (J. of Water Res. Plann
mihalych1998 [28]

Answer:

(a) P (Y = 3) = 0.0844, P (Y ≤ 3) = 0.8780

(b) The probability that the length of a drought exceeds its mean value by at least one standard deviation is 0.2064.

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:

P(Y=y)=(1-p)^{y}p;\ y=0,12...

(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:

\sigma=\sqrt{\frac{1-p}{p^{2}}}=\sqrt{\frac{1-0.409}{(0.409)^{2}}}=1.88

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

P (Y ≥ μ + σ) = P (Y ≥ 1.445 + 1.88)

                    = P (Y ≥ 3.325)

                    = P (Y ≥ 3)

                    = 1 - P (Y < 3)

                    = 1 - P (X = 0) - P (X = 1) - P (X = 2)

                    =1-[(1-0.409)^{0}\times 0.409+(1-0.409)^{1}\times 0.409\\+(1-0.409)^{2}\times 0.409]\\=1-[0.409+0.2417+0.1429]\\=0.2064

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

6 0
3 years ago
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