Real number, rational number
Answer:
The truck travel must to have a constant speed of 
Step-by-step explanation:
we have

where
d expresses a car's distance in feet
t is the number of seconds
<em>Find the distance d for t=8 sec</em>

<em>Find the distance d for t=8.2 sec</em>

The total distance in this interval of 0.2 sec is

<em>Find the speed of the car</em>
Divide the total distance by the time

therefore
The truck travel must to have a constant speed of 
There correct answer to that question is 22.45
The answer is 10 2/3 because 3/4 equals .75. 8 divided by .75 equals 10 2/3
Answer:
a. P(x=0)=0.2967
b. P(x=1)=0.4444
c. P(x=2)=0.2219
d. P(x=3)=0.0369
Step-by-step explanation:
The variable X: "number of meals that exceed $50" can be modeled as a binomial random variable, with n=3 (the total number of meals) and p=0.333 (the probability that the chosen restaurant charges mor thena $50).
The probabilty p can be calculated dividing the amount of restaurants that are expected to charge more than $50 (5 restaurants) by the total amount of restaurants from where we can pick (15 restaurants):

Then, we can model the probability that k meals cost more than $50 as:

a. We have to calculate P(x=0)

b. We have to calculate P(x=1)

c. We have to calcualte P(x=2)

d. We have to calculate P(x=3)
