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Sati [7]
3 years ago
11

Based on FAA estimates the average age of the fleets of the 10 largest U.S. commercial passenger carriers is 13.4 years with a s

tandard deviation of 1.7 years. Suppose that 40 airplanes were randomly selected from the fleets of these 10 carriers and were inspected for cracks in these airplanes that are considered too large for flying. What is the probability that the average age of these 40 airplanes is at least 14 years old
Mathematics
1 answer:
Aliun [14]3 years ago
6 0

Answer : 0.0129

Step-by-step explanation:

Given : Based on FAA estimates the average age of the fleets of the 10 largest U.S. commercial passenger carriers is \mu=13.4 years and standard deviation is \sigma=1.7 years.

Sample size : n=40

Let X be the random variable that represents the age of fleets.

We assume that the ages of the fleets of the 10 largest U.S. commercial passenger carriers are normally distributed.

For z-score,

z=\dfrac{x-\mu}{\dfrac{\sigma}{\sqrt{n}}}

For x=14

z=\dfrac{14-13.4}{\dfrac{1.7}{\sqrt{40}}}\approx2.23

By using the standard normal distribution table , the probability that the average age of these 40 airplanes is at least 14 years old will be :-

P(x\geq 14)=P(z\geq2.23)\\\\=1-P(z

Hence, the required probability = 0.0129

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dybincka [34]

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Below.

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The amount of profit, p, you earn by selling knives,k, can be determined by:p=200k-500. A) determine the constraints on profit a
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3 years ago
Please Help. 25 points. <br>​
vodomira [7]

Answer:

\boxed{x = 7, y = 9, z = 68}

Step-by-step explanation:

We must develop three equations in three unknowns.

I will use these three:

\begin{array}{lrcll}(1) & 8x + 13y +7 & = & 180 & \\(2)& 9x - 7 + 13y +7 & = & 180 & \\(3)& 8x + 5y - 11 + z & = & 180 &\text{We can rearrange these to get:}\\(4)& 8x + 13y & = & 173 &\\(5) & 9x + 13y & = & 180 & \\(6)& 8x + 5y + z & = & 169 & \\(7)& x & = & \mathbf{7} & \text{Subtracted (4) from (5)} \\\end{array}

\begin{array}{lrcll}& 8(7) + 13y & = & 173 & \text{Substituted (7) into (4)} \\& 56 + 13y & = & 173 & \text{Simplified} \\& 13y & = & 117 & \text{Subtracted 56 from each side} \\(8)& y & =& \mathbf{9}&\text{Divided each side by 13}\\& 8(7) + 5(9) + z & = & 169 & \text{Substituted (8) and (7) into (6)} \\& 56 + 45 + z& = & 169 & \text{Simplified} \\& 101 + z& = & 169 & \text{Simplified} \\&z& = & \mathbf{68} & \text{Subtracted 101 from each side}\\\end{array}

\boxed{\mathbf{ x = 7, y = 9, z = 68}}

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3 years ago
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