Answer:
Around 0.22
<u>FIRST THOUGH, I FEEL LIKE 90 MIGHT HAVE MEANT TO BEEN 0.9, SO IF SO SUBSTITUTE THE 90 BELOW FOR 0.9!</u>
<u></u>
Step-by-step explanation:
Okay, so first.
Company A will be 50 a day plus 0.4 a mile
Company B will be 30 a day plus 90 per mile
We can write an equation like this:
50 + 0.4m = 30 + 90m
m = miles they are the same
Then we solve for m.
50 + 0.4m = 30 + 90m
- 0.4m - 0.4m
50 = 30 + 89.6m
- 30 - 30
20 = 89.6m
Divide both sides by 89.6
Around 0.22!
Whats the question exactly??
Answer:
The Ballwin Bears are taller on average, and the Aviston Aces have players whose heights are more consistent.
Step-by-step explanation:
for derby dragon;
mean height = 72 inches
standard deviation = 1.2 inches
for aviston aces:
mean height = 70.8
standard deviation = 0.7 inches
for balwin bears;
mean height = 73 inches
standard deviation = 1.0 inches
the mean height of aviston aces < mean height of derby < mean height of balwin bears
So on average the balwin bears are taller.
The standard deviation of derby dragon is > that of balwin bears > aviston aces.
So the more consistent height is that of aviston aces players.
Answer:
Step-by-step explanation:
Step 1: Convert into improper fractions
Step 2: Find least common denominator and find equivalent fractions
Step 3 : Do the required operation

Least common denominator of 2 , 4 = 4

The Sample space for three cards is


There are two cards red(R) and black(B) and a six-sided die.
The event is one card is picked randomly and then rolls the die.
So, the Sample space is as follows.

<h3>Find Sample space if there are three cards.</h3>
Consider there were three cards red(R), black(B) and green(G) a six-sided die.
The event is one card is picked randomly and then rolls the die.
So, the sample space will be as follows,

.
Learn more about sample space here :-
brainly.com/question/24273864
#SPJ1