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kolbaska11 [484]
3 years ago
11

A bucket was 7/10 full with water. After Vick washed the car, the bucket wad only 3/10 full. What fraction of the water in the b

ucket did vick use to wash the car?

Mathematics
1 answer:
Phoenix [80]3 years ago
3 0
Hey there!


Vick had 7/10 bucket of water. . . .

She used some of the water and now has 3/10 left . . .

Subtracts the numbers 7 and 3 to solve this:

7 - 3 = 4 <----

The correct answer to your question should be 4/10 of a bucket.

Vick used 4/10 of the bucket to wash the car.

Hope this helps you.
Have a great day!

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Answer:

0; 10; 20

Step-by-step explanation:

x is the independent variable

y is the dependent variable

y is dependent on x

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y=0.3x−6

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0=0.3x

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Therefore, if the independent variable is 0, the value of the function will be -6.

b) For what value of the independent variable will the value of the function be equal to −3

y=0.3x−6

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Therefore, if the independent variable is 10, the value of the function will be -3.

c) For what value of the independent variable will the value of the function be equal to 0.

y=0.3x−6

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Therefore, if the independent variable is 20, the value of the function will be 0.

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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
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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)

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