15\x and 80\100
15 times 100=1500
1500=8x
then you divide 1500 by 80 so x is by itself
x=18.75
If the speed of the ball after kicking is 48 feet per second then the ball will return on the ground after 3 seconds.
Given that the speed of the ball after kicking is 48 feet per second and the function that represents the height of the ball is
.
We are required to find the time that the ball took to travel before returning the ground.
We know that speed is the distance a thing covers in a particular time period.
The height of the ball after t seconds is as follows:
h(t)=![-16t^{2} +48t](https://tex.z-dn.net/?f=-16t%5E%7B2%7D%20%2B48t)
It is at ground at the instants of t.
Hence,
=0
-16t(t-3)=0
We want value of t different of 0, hence :
t-3=0
t=3.
Hence if the speed of the ball after kicking is 48 feet per second then the ball will return on the ground after 3 seconds.
Learn more about speed at brainly.com/question/4931057
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Answer:
Provide the statements please
Step-by-step explanation:
Answer:
Event is not unusual(p>0.05).
Step-by-step explanation:
Given that :
![p=0.10\\\\n=1100\\\\x=121](https://tex.z-dn.net/?f=p%3D0.10%5C%5C%5C%5Cn%3D1100%5C%5C%5C%5Cx%3D121)
#The sample proportion is calculated as:
![\hat p=\frac{x}{n}\\\\=\frac{121}{1100}\\\\=0.1100](https://tex.z-dn.net/?f=%5Chat%20p%3D%5Cfrac%7Bx%7D%7Bn%7D%5C%5C%5C%5C%3D%5Cfrac%7B121%7D%7B1100%7D%5C%5C%5C%5C%3D0.1100)
#Mathematically, the z-value is the value decreased by the mean then divided the standard deviation :
![z=\frac{\hat p- p}{\sqrt{\frac{p(1-p)}{n}}}\\\\\\\\=\frac{0.11-0.10}{\sqrt{\frac{0.10(1-0.10)}{1100}}}\\\\\\=1.1055](https://tex.z-dn.net/?f=z%3D%5Cfrac%7B%5Chat%20p-%20p%7D%7B%5Csqrt%7B%5Cfrac%7Bp%281-p%29%7D%7Bn%7D%7D%7D%5C%5C%5C%5C%5C%5C%5C%5C%3D%5Cfrac%7B0.11-0.10%7D%7B%5Csqrt%7B%5Cfrac%7B0.10%281-0.10%29%7D%7B1100%7D%7D%7D%5C%5C%5C%5C%5C%5C%3D1.1055)
#We use the normal probability table to determine the corresponding probability;
![P(X\geq 121)=P(Z>1.1055)\\\\=0.1314](https://tex.z-dn.net/?f=P%28X%5Cgeq%20121%29%3DP%28Z%3E1.1055%29%5C%5C%5C%5C%3D0.1314)
Hence, the probaility is more than 0.05, thus the event not unusual and thus this is not necessarily evidence that the proportion of Americans who are afraid to fly has increased.