70x^4y^2+14xy
14(5x^4y^2+xy)
14xy(5x^3y+1)
Just keep taking out things that both terms have in common, to find the answer.
In the first survery
![\frac{336}{500}](https://tex.z-dn.net/?f=%20%5Cfrac%7B336%7D%7B500%7D%20)
considered cola.
Assuming we can extrapolate this ratio, all we have to do is get the same ratio, but out of 2000 people.
![\frac{336}{500} * \frac{4}{4} \\ \frac{1344}{2000}](https://tex.z-dn.net/?f=%20%5Cfrac%7B336%7D%7B500%7D%20%2A%20%5Cfrac%7B4%7D%7B4%7D%20%20%5C%5C%20%20%5Cfrac%7B1344%7D%7B2000%7D%20)
The answer is 1344 people.
Answer:
Step-by-step explanation:
The position function is
and if we are looking for the time t it takes for the ball to hit the ground, we are looking for the height of the ball when it is on the ground. Of course the height of anything on the ground is 0, so if we set s(t) = 0 and solve for t, we will find our answer.
and factor that however you are currently factoring in class to get that
t = -.71428 seconds or
t = 1.42857 seconds (neither one of those is rational so they can't be expressed as fractions).
We all know that time will never be a negative value, so the time it takes this ball to hit the ground is
1.42857 seconds (round how you need to).
Answer:
Time with 4.2MB/S speed = 7314.3 seconds
Time with 900 KB/S speed = 34952.5 seconds
Step-by-step explanation:
Given that:
Internet download speed = 4.2 MB/S
File to download = 30 GB
1 GB = 1024 MB
30 GB = 30*1024 = 30720 MBS
Time required = ![\frac{30720}{4.2}](https://tex.z-dn.net/?f=%5Cfrac%7B30720%7D%7B4.2%7D)
Time required = 7314.3 seconds
With 900 KB/S
1 MB = 1024 KB
30720 MB = 30720 * 1024 = 31457280 KBS
Time = ![\frac{31457280}{900}=34952.5\ seconds](https://tex.z-dn.net/?f=%5Cfrac%7B31457280%7D%7B900%7D%3D34952.5%5C%20seconds)
Hence,
Time with 4.2MB/S speed = 7314.3 seconds
Time with 900 KB/S speed = 34952.5 seconds
Step-by-step explanation:
In his last 30 throws, Tom hit the target 20 times and missed 10 times.
p = probability he misses the target = ⅓
q = probability he hits the target = ⅔
Using binomial probability:
P = nCr (p)^r (q)^(n-r)
Given n = 3, r = 1, p = ⅓, and q = ⅔:
P = ₃C₁ (⅓)¹ (⅔)³⁻¹
P = (3) (⅓) (⅔)²
P = 4/9
There is a 4/9 probability that he misses exactly 1 of his next 3 throws.