(21+39a)/20. Because they have to have the same denominator
1) let's convert the maximum altitude of the black kite from feet to meters. We can do it by using the following proportion:
![1 ft : 0.3048 m = 362.26 ft : x](https://tex.z-dn.net/?f=1%20ft%20%3A%200.3048%20m%20%3D%20362.26%20ft%20%3A%20x)
From which we find
![x=110.42 m](https://tex.z-dn.net/?f=x%3D110.42%20m)
This is the maximum altitude (in meters) of the black kite. The problem says that the maximum altitude of the red kite is 117.46 m, therefore the red kite is the one with higher maximum altitude.
2) Let's convert the maximum altitude of the red kite from meters to feet, again by using the proportion:
![1 ft : 0.3048 m = x : 117.46 m](https://tex.z-dn.net/?f=1%20ft%20%3A%200.3048%20m%20%3D%20x%20%3A%20117.46%20m)
From which we find
![x=385.37 ft](https://tex.z-dn.net/?f=x%3D385.37%20ft)
Therefore, the maximum altitude of the red kite in feet is 385.37 ft.
Answer: 0.0129
Step-by-step explanation:
Given : Sample size : n=32
The average amount of time spent texting over a one-month period is : ![\mu=173\text{ minutes}](https://tex.z-dn.net/?f=%5Cmu%3D173%5Ctext%7B%20minutes%7D)
Standard deviation : ![\sigma=66\text{ minutes}](https://tex.z-dn.net/?f=%5Csigma%3D66%5Ctext%7B%20minutes%7D)
We assume that the time spent texting over a one-month period is normally distributed.
z-score : ![z=\dfrac{x-\mu}{\dfrac{\sigma}{\sqrt{n}}}](https://tex.z-dn.net/?f=z%3D%5Cdfrac%7Bx-%5Cmu%7D%7B%5Cdfrac%7B%5Csigma%7D%7B%5Csqrt%7Bn%7D%7D%7D)
For x= 199
![z=\dfrac{199-173}{\dfrac{66}{\sqrt{32}}}\approx2.23](https://tex.z-dn.net/?f=z%3D%5Cdfrac%7B199-173%7D%7B%5Cdfrac%7B66%7D%7B%5Csqrt%7B32%7D%7D%7D%5Capprox2.23)
Now by using standard normal table, the probability that the average amount of time spent using text messages is more than 199 minutes will be :-
![P(x>199)=P(z>2.23)=1-P(z\leq2.23)\\\\=1- 0.9871262=0.0128738\approx0.0129](https://tex.z-dn.net/?f=P%28x%3E199%29%3DP%28z%3E2.23%29%3D1-P%28z%5Cleq2.23%29%5C%5C%5C%5C%3D1-%200.9871262%3D0.0128738%5Capprox0.0129)
Hence, the required probability = 0.0129