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sveticcg [70]
2 years ago
14

Solve: 2a + 6 = 12 a =

Mathematics
1 answer:
Alex Ar [27]2 years ago
5 0

Answer:

8

Step-by-step explanation:

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13.12

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What is x divided by 6.2=-2.1
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Find the volume of a cylinder if it has a diameter of 8 and a height of 12. Leave your answer in terms of 70.
rewona [7]
If we need to choose the closest answer from the above given ones it is 6471.

SOLUTION:
Diameter(d)= 8
Radius(r)= d/2=8/2 =4,,
Height of cylinder (h)= 12


So,
Volume of cylinder = Pi r^2*h
=22/7 * 4^2 * 12
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6 0
3 years ago
An airline finds that 5% of the persons making reservations on a certain flight will not show up for the flight. If the airline
Greeley [361]

Answer:

0.5438 = 54.38% probability that a seat will be available for every person holding a reservation and planning to fly

Step-by-step explanation:

We 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 = 1 - 0.05 = 0.95

I use p = 0.95 because i consider a success a person showing up to the flight. 5% probability that a person misses the flight, so 100-5 = 95% probability that a person shows up to the flight.

For the approximation:

\mu = E(X) = np = 105*0.95 = 99.75

\sigma = \sqrt{V(X)} = \sqrt{np(1-p)} = \sqrt{105*0.95*0.05} = 2.23

What is the probability that a seat will be available for every person holding a reservation and planning to fly?

Probability of 100 or less people showing up, which is the pvalue of Z when X = 100. So

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

Z = \frac{100 - 99.75}{2.23}

Z = 0.11

Z = 0.11 has a pvalue of 0.5438

0.5438 = 54.38% probability that a seat will be available for every person holding a reservation and planning to fly

4 0
3 years ago
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