Answer:
20 grape lollipops
Step-by-step explanation:
From the 30 lollipops, 6 were grape
P(grape) = grape/total
=6/30
=1/5
If we get 100 lollipop, to determine the number that would be grape, we multiply the number of lollipops by the probability that they are grape
grape = number * probability
= 100 * 1/5
=20
Answer:
It depends in the go kart. But my guess is about 8 miles?
Step-by-step explanation:
I dont know.
Answer:
the rate of each train is 139 km/h and 159 km/h.
Step-by-step explanation:
let
v = the rate of the slower train
v + 20 = the rate of the faster train
d = 1490 km the distance that the two train will travel
t = 5 hr the time of the travel
since speed = distance/time => distance = speed*time
v*t + (v+20)t = d
5v + 5(v + 20) = 1490
5v+5(v+20)=1490
Step 1: Simplify both sides of the equation.
5v+5(v+20)=1490
5v+(5)(v)+(5)(20)=1490(Distribute)
5v+5v+100=1490
(5v+5v)+(100)=1490(Combine Like Terms)
10v+100=1490
10v+100=1490
Step 2: Subtract 100 from both sides.
10v+100−100=1490−100
10v=1390
Step 3: Divide both sides by 10.
10v/10 = 1390/10
v = 139km/h
Next
we add 139 to 20
139 + 20 = 159 km/h
the rate of each train is 139 km/h and 159 km/h.
1.
Make the fractions improper.
72/12 x 19/6 = 19.
2.
Area = L x W
L = 3 1/6 x
W = 6 7/12
= 19.
#1 = 19 feet
#2 = <span>9 3/4 feet</span>
In three dimensions, the cross product of two vectors is defined as shown below

Then, solving the determinant

In our case,

Where we used the formula for AxB to calculate ixj.
Finally,

Thus, (i+j)x(ixj)=i-j