Answer:they would meet at 1 pm
Step-by-step explanation:
The distance between Quebec City and New York City is 520 miles.
At the point where Kate and Will meet, they would have travelled 520 miles.
Let t represent the time it takes Kate to travel a certain distance before they met.
Distance = speed × time
If Kate drives at 55mph, the distance covered would be
55 × t = 55t.will drives at 75 mph
Since Will left same that that Kate left, distance travelled by Will in t hours would be
75 × t = 75t
Since the total distance covered is 520 miles, it means that
55t +75t = 520
130t = 520
t = 520/130 = 4
9 am + 4 hours = 1 pm
Answer:
Idk give me the options.....
Step-by-step explanation:
Answer:
Step-by-step explanation:
a) Frequency table:
Category Frequency
Dog 4
Cat 3
Fish 2
Hamster 1
b) Relative frequencies of each animal type
Dog: 4/10 = 0.4
Cat: 3/10 = 0.3
Fish: 2/10 = 0.2
Hamster: 1/10 = 0.1
c) Popularity
Dog is the most popular because it has the highest relative frequency.
Hamster is the least popular because it has the lowest relative frequency.
Answer:
180$
Step-by-step explanation:
If two-thirds of Sam's weekly income is $480
so total is $720
720$/4= 180$