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

6 0

3 years ago

How many ears do twelve dogs have

Schach [20]

Answer:24

Step-by-step explanation:

5 0

3 years ago

If f(x) = x-6 and g(x)= 1/2x (x+3), find g(x) * f(x)

sertanlavr [38]

Answer:

Final answer is $g\left(x\right)\cdot f\left(x\right)=\frac{\left(x-6\right)}{2x\left(x+3\right)}$ .

Step-by-step explanation:

given functions are $f(x)=x-6$ and $g\left(x\right)=\frac{1}{2x\left(x+3\right)}$ .

Now we need to find about what is the value of $g\left(x\right)*f\left(x\right)$ .

$g\left(x\right)*f\left(x\right)$ simply means we need to multiply the value of $f(x)=x-6$ and $g\left(x\right)=\frac{1}{2x\left(x+3\right)}$ .

$g\left(x\right)\cdot f\left(x\right)=\frac{1}{2x\left(x+3\right)}\cdot\left(x-6\right)$

$g\left(x\right)\cdot f\left(x\right)=\frac{\left(x-6\right)}{2x\left(x+3\right)}$

Hence final answer is $g\left(x\right)\cdot f\left(x\right)=\frac{\left(x-6\right)}{2x\left(x+3\right)}$ .

5 0

3 years ago