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2 years ago

A. Going 20 miles per hour over the speed limit will result in a fine of $ 2

1 answer:

5 0

Assuming that the cost of the fine is proportionate per each 20mph increase, then a 40mph speeding ticket should be $4, and if they received a fine of $150 would be a crazy amount of speed, is there something you did not attach?

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13y-{-14y+[9y-(-8y+17)]}

Lesechka [4]

Answer:

Step-by-step explanation:

6 0

2 years ago

Marks test scores for this semester were 83,80,91,74,66,95,87,72,95,97 what is the mode of his test scores

Vlada [557]

Answer:

Step-by-step explanation:

Mode is what is the most frequent number that occurred; 95 occurred the most from the data.

3 0

2 years ago

What is the greatest common factor of 32a and 48b

SpyIntel [72]

32=16x2=2x2x2x2x2=2^5

48=24*2=6x4x2=2x3x2x2x2

Pick the highest number that occurs. In this case it is 2. Now we have to see how many times it appears in both. It appears 5 times in 32 and 4 times in 48. 4 is the highest number of times it appears in the numbers so:

2^4=2x2x2x2=16

The Greatest Common Factor (GCF) of 32 and 48 is 16.

5 0

2 years ago

for every 5 pounds of pasta served , 15 people can be fed. how many people can be fed with 4 pounds of pasta?

Sidana [21]

Answer:

Step-by-step explanation:

its direct proportion. 15 divided by 5 is 3 so you must multiply 3 by 4 to get the amount of people, which is 12

7 0

3 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.

6 0

3 years ago