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3 years ago

Help me out please will mark you BRAINLIST

Mathematics

1 answer:

melisa1 [442]3 years ago

6 0

Answer:

D. 18.68

Step-by-step explanation:

$27\div \frac{5}{3}-6+53\times (\frac{2}{5})^2$

Applying PEMDAS as order of operations.

Solving the exponents first $[(\frac{2}{5})^2=\frac{4}{25}]$

$=27\div \frac{5}{3}-6+53\times \frac{4}{25}$

Multiplying $[53\times \frac{4}{25}=8.48]$

$=27\div \frac{5}{3}-6+8.48$

Dividing $[27\div \frac{5}{3}=16.2]$

$=16.2-6+8.48$

Adding and subtracting.

$=18.68$

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An airplane has 100 seats for passengers. Assume that the probability that a person holding a ticket appears for the flight is 0

zmey [24]

Answer:

96.33% probability that everyone who appears for the flight will get a seat

Step-by-step explanation:

I am going to use the binomial approximation to the normal to solve this question.

Binomial probability distribution

Probability of exactly x sucesses on n repeated trials, with p probability.

Can be approximated to a normal distribution, using the expected value and the standard deviation.

The expected value of the binomial distribution is:

$E(X) = np$

The standard deviation of the binomial distribution is:

$\sqrt{V(X)} = \sqrt{np(1-p)}$

Normal probability distribution

Problems of normally distributed samples can be solved using 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.

When we are approximating a binomial distribution to a normal one, we have that $\mu = E(X)$ , $\sigma = \sqrt{V(X)}$ .

In this problem, we have that:

$n = 105, p = 0.9$

$\mu = E(X) = np = 105*0.9 = 94.5$

$\sigma = \sqrt{V(X)} = \sqrt{np(1-p)} = \sqrt{105*0.9*0.1} = 3.07$

What is the probability that everyone who appears for the flight will get a seat

100 or less people appearing to the flight, which is the pvalue of Z when X = 100. So

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

$Z = \frac{100 - 94.5}{3.07}$

$Z = 1.79$

$Z = 1.79$ has a pvalue of 0.9633

96.33% probability that everyone who appears for the flight will get a seat

7 0

2 years ago

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In my opinion, Darrin's inference is wrong because according to given question, "Darrin surveyed a random sample of 10 students from his science class about their favorite types of TV shows."

This line provides the information that the survey is taken randomly. Also, if Darrin had taken some other students, then the ineference of other new students compared with previously surveyed students will be different.

This frankly tells that the probability is different always.

Therefore, Darrin's inference is wrong or invalid.

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2 years ago

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Answer:

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