March, April, and May correspond to 3,4, and 5th months. Each month the amount of money he saves increases by $40, so this is a constant rate of change, or constant velocity, which means that this is a linear equation. Let y=the amount of money he has and x=the month number and model it with the line of form y=mx+b, where m=slope, or rate, and b=y-intercept...
y=40x+b, we know that he has $40 on the third month (February) so:
40=40(3)+b
40=120+b
-80=b so the equation is:
y=40x-80 (note that the domain is x=[3, +oo) only for the function to have meaining)
Now you want to know when he has $320...
320=40x-80
400=40x
10=x
So he will have $320 on the 10th month, which is October.
Answer:
980.71875
Step-by-step explanation:
Answer:
h = 5
Step-by-step explanation:
For there to be 1 car, we consider two possible outcomes:
The first door opened has a car or the second door opened has a car.
P(1 car) = 2/6 x 4/5 + 4/6 x 2/5
P(1 car) = 8/15
For there to be no car in either door
P(no car) = 4/6 x 3/5
P(no car) = 2/5
Probability of at least one car is the sum of the probability of one car and probability of two cars:
P(2 cars) = 2/6 x 1/5
= 1/15
P(1 car) + P(2 cars) = 8/15 + 1/15
= 3/5
9:45 plus 3 hours goes to 12:45 pm. add 45 minutes you get 1:30PM as your answet
