Val does because he covers 10.4 miles. Hope that helps.
No, if you multiply the factors in 3(5b-2) the equivalent expression would be 15b-6
f(2) is three just look at where it says 2 on the x side of the table.
Let the total number of cars = 100
Total of sedan, van and sports cars = 39+33+7 = 79
Number of other cars = 100 - 79 = 21.
Let A be the event of owning a sedan,
B be the event of owning a van and
C be the event of owning a sports car.
D be the event of owning other cars.
Then,
and
Now,
p(non selection of van) = p(A ∪ C ∪ D)
= p(A) + p(C) + p(D)
= 0.67
The probability of none of the three selections would be a van is:
0.67 × 0.67 × 0.67
= 0.300763
=0.301
Hence, the correct option is (D).
#33 is 0.3 times 42, because 0.3 is basically 30%, and you always multiply the percent by the number to get the answer. It could also be 3/10 times 42.