7.998/0.7008=11.41267123287671
That's a big answer haha...
Well I hope this helps! :D
Answer:
a)120
b)6.67%
Step-by-step explanation:
Given:
No. of digits given= 6
Digits given= 1,2,3,5,8,9
Number to be formed should be 3-digits, as we have to choose 3 digits from given 6-digits so the no. of combinations will be
6P3= 6!/3!
= 6*5*4*3*2*1/3*2*1
=6*5*4
=120
Now finding the probability that both the first digit and the last digit of the three-digit number are even numbers:
As the first and last digits can only be even
then the form of number can be
a)2n8 or
b)8n2
where n can be 1,3,5 or 9
4*2=8
so there can be 8 three-digit numbers with both the first digit and the last digit even numbers
And probability = 8/120
= 0.0667
=6.67%
The probability that both the first digit and the last digit of the three-digit number are even numbers is 6.67% !
Answer: 180.5
9.5
X19.0
=180.5
Answer:
Step-by-step explanation:
We can use the quadratic formula or factor, This looks hard to factor so we should use the quadratic formula.
ax^2 + bx + c
so
a=4
b=-9
c=2
you can try it by hand or use this:
https://www.calculatorsoup.com/calculators/algebra/quadratic-formula-calculator.php
you get x=2 and x=0.25 these are the x intercepts (where Y is zero and where the graph crosses the x axis)
so mark x=2 and x=0.25 with a dot on a number line and you can draw a straight line between them since that is the part of the graph that is below the x axis (because the equation has <0) it is a positive parabola because the "a" value is positive and the presence of a "b" value means part of the graph will be below the x axis
6x-4y=-14
2(-3x-3y=12)
6x-4y=-14
-6x-6y=24
-10y=10
Divide both sides by 10
-1y=1 divide by -1
y=-1
6x-4(-1)=-14
6x+4=-14
-4 -4
6x=-18
Divide both sides by 6
x=-3
