Answer:
Therefore the ball strikes the ground 5.98 s after kick.
Step-by-step explanation:
Given that , a ball is kicked in the air. The height of the ball can be calculated using the equation 
where h is height in feet and t is time in second.
When the ball touches the ground then the height will be zero.
i.e h(t)=0
Therefore
Since -2≠0 then 
[The solution of a quadratic equation ax²+bx+c=0 is
here a=8, b=-32 and c=-95]
![\Rightarrow t =\frac{-(-32)\pm\sqrt{(-32)^2-[4\times8\times(-95)]} }{2\times8}](https://tex.z-dn.net/?f=%5CRightarrow%20t%20%3D%5Cfrac%7B-%28-32%29%5Cpm%5Csqrt%7B%28-32%29%5E2-%5B4%5Ctimes8%5Ctimes%28-95%29%5D%7D%20%7D%7B2%5Ctimes8%7D)
Therefore t= 5.98 or -1.98
since time can not be negative So , t=5.98 s
Therefore the ball strikes the ground 5.98 s after kick.
A)
150 were surveyed, possible outcomes, and 132 said they prefer dogs, favorable outcomes, 132/150 or 22/25 or 0.88, now 0.88 * 100 is just 88, namely 88% is the probability.
b)
since we know is 88%, what is 88% of 100? (88/100) * 100, 0.88 * 100, or just 88 students.
c)
18/150 or 3/25 or 0.12 namely 12%,
what's 12% of 300? (12/100) * 300, or 0.12 * 300, so just 36 students.
<span>A+B)^2 is the largest. It is A^2+2AB+B^2, which is clearly greater than the last two options. To compare (A+B)^2 and 2(A+B), we remove one A+B so that we're just comparing A+B and 2. As A+B must be at least 3 (as both must be positive integers, and one must be greater than the other, leading to a minimum value of A=2, B=1), A+B is greater than 2, and as a result, (A+B)^2 is always the largest.</span>
The <em>correct answer</em> is:
The median is less than the mean.
Explanation:
This histogram is skewed right; the data "peaks" further to the left than the center.
If a histogram is skewed right, the mean is greater than the median.
This is because skewed-right data have a few large values that increase the mean but do not affect where the exact middle of the data is.
Answer:
b) 5
Step-by-step explanation:
6+10=16
40 times 2 = 80
80 div by 15 = 5
