Let t be the number of tolls they crossed.
Amount they spent at each toll = $1.75.
Amount they spent at gas station = $28.
Let C be the total amount they spent on gas and tolls.
If they crossed 1 toll, then
C = 28 + 1.75(1).
If they crossed 3 tolls, then,
C = 28 + 1.75(3)
If they crossed t tolls, then,
C = 28 + 1.75t
Here, the terms are 28 and 1.75t and the factors are 1.75 and t.
Answer:
Step-by-step explanation:
Given that minimum is 8 and maximum equals 82
Range = 
No of classes =6
Class width = 76/6 ~13
But not given whether variable is discrete or continuous.
If discrete, we have classes as
8-20, 21-33, 34-46, 47-59, 60-72, 73-85
If continuous, we have classes as
8 to <21
21 to <34
and ... ending 73-<86
Answer:
I'm pretty sure it is d, 5/22
Answer:
A.
Step-by-step explanation:
you have to find the discriminant
b²-4ac for each equation
if discriminant < 0 no real solutions because will be negative under the squareroot whhan you try to find the roots
if discriminant = 0 there is only one solution
if discriminant > 0 two real solutions
for your given problems
A. discriminant =(-2)²-4*2*15 will be negative
Answer:
A die is thrown which means that it can land on either 1-6 on the die.So if you want to find the probability of the multiple of 2 or 3....we have multiplies of 2 in the die from 1-6 which is 2,4 and 6 if you count them they are just 3 numbers and since there is a probability of it to land on any of them it will be 3/6 which is 1/2 if yiu divide by 3 to it's lowest term....Then multiple of 3 is 3,and 6.which will be 2/6 which is 1/3 if u cut to its lowest term. Or in the question means addition sign in probability then it is 2 or 3 which is (1/2)+(1/3)=5/6.Thank you for the question
