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3 years ago

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The mean yearly rainfall in Sydney, Australia, is about 134 mm and the standard deviation is about 66 mm ("Annual maximums of,"

2013). Assume rainfall is normally distributed. How many yearly mm of rainfall would there be in the top 15%? Round answer to 2 decimal places.

1 answer:

svetlana [45]3 years ago

5 0

Answer:

At least 202.44 mm in the top 15%.

Step-by-step explanation:

When the distribution is normal, we use the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$ , the zscore of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

In this question, we have that:

$\mu = 134, \sigma = 66$

How many yearly mm of rainfall would there be in the top 15%?

At least X mm.

X is the 100-15 = 85th percentile, which is X when Z has a pvalue of 0.85. So X when Z = 1.037.

$Z = \frac{X - \mu}{\sigma}$

$1.037 = \frac{X - 134}{66}$

$X - 134 = 66*1.037$

$X = 202.44$

At least 202.44 mm in the top 15%.

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Find the vertex of the function h(x) = x2 – 4x – 21

lakkis [162]

Answer: (2, -25)

Step-by-step explanation:

The x coordinate of the vertex is $x=-\frac{-4}{2(1)}=2$ .

When x=2, $h(2)=2^2 - 4(2)-21=-25$ .

So, the vertex is (2, -25).

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2 years ago

2) Fill in an angle measure greater than 100 degrees for the missing angle in yellow below. You will fill

timama [110]

Answer:

m∠a = 110°

m∠b = 110°

m∠c = 70°

m∠d = 70°

m∠i = 110°

m∠h = 110°

m∠j = 70°

m∠g = 110°

m∠f = 70°

m∠e = 70°

m∠k = 57°

m∠m = 70°

m∠l = 167°

Step-by-step explanation:

Let the measure of the yellow angle = 110°

Therefore;

m∠a = 110° by vertically opposite angles

m∠b = m∠a = 110° by opposite interior angles of a parallelogram

m∠c = 180° - 110° = 70° by same side interior angles

m∠d = m∠c = 70° by opposite interior angles of a parallelogram

m∠i and m∠c are supplementary angles, therefore, m∠i = 110°

m∠h = m∠i = 110° by vertically opposite angles

m∠j = m∠c = 70° by vertically opposite angles

m∠g and m∠d are supplementary angles, therefore, m∠g = 110°

m∠f = m∠d = 70° by alternate interior angles

m∠f = m∠e = 70° by vertically opposite angles

m∠k = 180° - (m∠h + 13°) = 180° - (110° + 13°) = 57°

m∠b and m∠m are supplementary angles, therefore, m∠m = 70°

m∠l and 13° are supplementary angles, therefore, m∠l = 167°

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3 years ago

1. A report from the Secretary of Health and Human Services stated that 70% of single-vehicle traffic fatalities that occur at n

Nuetrik [128]

Using the binomial distribution, it is found that there is a 0.7215 = 72.15% probability that between 10 and 15, inclusive, accidents involved drivers who were intoxicated.

For each fatality, there are only two possible outcomes, either it involved an intoxicated driver, or it did not. The probability of a fatality involving an intoxicated driver is independent of any other fatality, which means that the binomial distribution is used to solve this question.

Binomial probability distribution

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$C_{n,x} = \frac{n!}{x!(n-x)!}$

The parameters are:

x is the number of successes.
n is the number of trials.
p is the probability of a success on a single trial.

In this problem:

70% of fatalities involve an intoxicated driver, hence $p = 0.7$ .

A sample of 15 fatalities is taken, hence $n = 15$ .

The probability is:

$P(10 \leq X \leq 15) = P(X = 10) + P(X = 11) + P(X = 12) + P(X = 13) + P(X = 14) + P(X = 15)$

Hence

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 10) = C_{15,10}.(0.7)^{10}.(0.3)^{5} = 0.2061$

$P(X = 11) = C_{15,11}.(0.7)^{11}.(0.3)^{4} = 0.2186$

$P(X = 12) = C_{15,12}.(0.7)^{12}.(0.3)^{3} = 0.1700$

$P(X = 13) = C_{15,13}.(0.7)^{13}.(0.3)^{2} = 0.0916$

$P(X = 14) = C_{15,14}.(0.7)^{14}.(0.3)^{1} = 0.0305$

$P(X = 15) = C_{15,15}.(0.7)^{15}.(0.3)^{0} = 0.0047$

Then:

$P(10 \leq X \leq 15) = P(X = 10) + P(X = 11) + P(X = 12) + P(X = 13) + P(X = 14) + P(X = 15) = 0.2061 + 0.2186 + 0.1700 + 0.0916 + 0.0305 + 0.0047 = 0.7215$

0.7215 = 72.15% probability that between 10 and 15, inclusive, accidents involved drivers who were intoxicated.

A similar problem is given at brainly.com/question/24863377

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3 years ago

You take a quiz with 6 multiple choice questions. After you studied your estimated that you would have about an 80 % chance of g

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Answer:

1.15%

Step-by-step explanation:

To get the probability of m independent events you multiply the individual probability of each event. In this case we have m independent events, each one with the same probability, therefore:

$p^{m}$

$0.8^{20} = 1.15\%$

This is a particlar scenario of binomial distribution problem. So the binomial distribution questions are about the number of success of m independent events, where every individual event has the same p probability. In the question we have 20 events and each event has a probability of 80%. The binomial distribution formula is:

$\binom{n}{k} * p^{k} * (1-p)^{n-k}$

n is the number of events

k is the number of success

p is the probability of each individual event

$\binom{n}{k}$ is the binomial coefficient

the binomial coefficient allows to find the subsets of k elements in a set of n elements. In this case there is only one subset possible since the only way to get 20 of 20 correct questions is to getting right all questions (for getting 19 of 20 questions there are many ways, for example getting the first question wrong and all the other questions right, or getting second questions wrong and all the other questions right, etc).

$\binom{n}{k} = \frac{n!}{k!(n-k)!}$

therefore, for this questions we have:

$\frac{20!}{20!(20-20)!} * 0.8^{20} * (1-0.8)^{0} = 1.15\%$

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