Question 1: car a began a journey from a point at 9 am, traveling at 30 mph. at 10 am car b started traveling from the same point at 40 mph in the same direction as car a. at what time will car b pass car a?
solution:
let after t hours the distances d1 traveled by car a
=> d1 = 30 t
car b starts at 10 am and will therefore have spent one hour less than car a when it passes it.
after (t - 1) hours, distance d2 traveled by car b
=> d2 = 40 (t - 1)
when car b passes car a, they are at the same distance from the starting point and therefore d1 = d2
=> 30 t = 40 (t - 1)
solve the equation for t,

if you are saying purchasing 3 social media ads and one newspaper ad counts as one purchase then there are:

6 purchases

if you are asking how many of each then:

18 social media ads

6 newspaper ads

the solution is the point

step-by-step explanation:

we have

> equation a

> equation b

adds equation a and equation b

substitute the value of x in equation b to find the value of y

the solution is the point

answer

a) the midrange of the data set shown is 85.5

step by step explanation

let's find the midrange.

the midrange is nothing but add the greatest value and the least value, then divide it by 2.

here the greatest value = 99

the least value = 72

midrange = (99 + 72)/2

= 171/2

midrange = 85.5

you.

so mark-up is 0.4 multiplied by 180 = 72