Given that <span>For
a certain model of car the distance

required to stop the vehicle if
it is traveling at

mi/h is given by the formula
![d=v+\frac{v^2}{20}, where [tex]d](https://tex.z-dn.net/?f=d%3Dv%2B%5Cfrac%7Bv%5E2%7D%7B20%7D%2C%20where%20%5Btex%5Dd%20)
is measured in feet.
If Kerry wants her stopping distance not to exceed 75
ft, then the range of speeds (in mi/h) can she travel is obtained as follows:

Therefore, the range of speed she can travel is

</span>
I uploaded the answer to a file hosting. Here's link:
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Answer:
Hence the required probability is, 3/4
Step-by-step explanation:
At the shelter, he likes :
a male coolie, a female coolie, a male boxer, a female boxer, a male beagle, a female beagle, a male Labrador, and a female Labrador.
Let, A denote the event of selecting a male coolie and B denote the event of selecting a male Labrador.
P(A) = 1/8 = P(B)
Here the probability of selecting a puppy except A & B is,
P(AUB)c = 1 - P(AUB) = 1 - { P(A) + P(B) } = 1 - 1/8 - 1/8 = 3/4
1 gallon = 8 pints, so 2 gallons = 16 pints.
If each container can hold one pint, then she will need 16 containers.
<span>its C. Alan has 13 CDs, Tom has 16 CDs, Barbara has 23 CDs !</span>