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2 years ago

T/15=3 what operation is that?

Mathematics

1 answer:

stich3 [128]2 years ago

5 0

Answer:

t = 45

Step-by-step explanation:

$\frac{t}{15} = \frac{3}{1}$

t × 1 = 15 × 3

t = 45

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3 years ago

In ΔABC, BC = 13, CA = 20 and AB = 19. Which statement about the angles of ΔABC must be true?

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Answer:

m∠B > m∠C > m∠A

Step-by-step explanation:

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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 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.

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3 years ago

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3 years ago

There are two major tests of readiness for college: the act and the sat. act scores are reported on a scale from 1 to 36. the di

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Solution: We are given:

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Now, let's find the z score corresponding to Joe's SAT score 1351.

$z=\frac{x-\mu}{\sigma}$

$=\frac{1351-1026}{209}$

$=\frac{325}{209}$

$=1.56$

Therefore, Joe's SAT score is 1.56 standard deviations above the mean.

Now, we have find the Joe's ACT score, which will be 1.56 standard deviations above the mean.

Therefore, we have:

$z=\frac{x-\mu}{\sigma}$

$1.56=\frac{x-20.9}{4.8}$

$1.56 \times 4.8 = x - 20.9$

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3 years ago