SOMEONE PLEASE HELP ME OUT THIS IS DUE In 20 MINUTES (PICTURE)

Identify the pairs of corresponding angles

topjm [15]

Answer: A

Step-by-step explanation:

If you flip the second triangle straight upwards so that x is on top and z,y are on the bottom, you could visualize the points and angles in the same locations.

6 0

3 years ago

An equation _____ has one solution. 1. never 2. always 3.sometimes

Tomtit [17]

Answer:

Option 3

Step-by-step explanation:

An equation is sometimes one solution because it can also have infinitely many solution or no solutions.

3 0

3 years ago

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Write 3 equivalent expressions for 12x-36

morpeh [17]

Answer:

1) 24x-72; 2) 6x-18; 3) 2x-9

Step-by-step explanation:

You can divide it or multiply it by whatever number it is to make it a whole number, not a decimal. Or else it will be even more difficult.

4 0

3 years ago

Need help with this question!

rosijanka [135]

Not sure:((((((( sorrryyyy

3 0

3 years ago

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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 Y is defined as the number of consecutive time intervals in which the water supply remains below a critical value y₀.

The random variable Y follows a Geometric distribution with parameter p = 0.409.

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 Y as follows:

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

Compute the standard deviation of the random variable Y 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

4 years ago