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2 years ago

The khan's car averages 22 miles per gallon of gas. how far they can travel on 5 gallons of gas. (Don't make a prediction)

Mathematics

1 answer:

ANTONII [103]2 years ago

6 0

They can travel 110 miles for 5 gallons of gas.

22 • 5 = 110 miles on 5 gallons of gas

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The z-score associated with 95% is 1.96. If the sample mean is 200 and the standard deviation is 30, find the upper limit of the

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Answer:

The upper limit of the 95% confidence interval is:

C.I_u = 200 + (58.8/ $\sqrt{n}$ )

Step-by-step explanation:

The formula is given as:

C.I = μ ± Z*σ/ $\sqrt{n}$

The upper limit => C.I_u = μ + Z*σ/ $\sqrt{n}$

The lower limit => C.I_l = μ - Z*σ/ $\sqrt{n}$

The sample size (n) is not stated in the question. Hence, we calculate the upper limit with respect to n.

The upper limit => C.I_u = 200 + 1.96*(30/ $\sqrt{n}$ )

= 200 + (1.96*30)/ $\sqrt{n}$

= 200 + 58.8/ $\sqrt{n}$

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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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2 years ago