Assume that you only include whole numbers (1,2,3,4,5,6,7,8,9) and not 3.5 and such
so if 1 is odd and less than 5 then it is
1 or 3, since 5 isn't included
then the other number, to be less than 5 when added,
must be
1+x<5
3+x<5
solve each
1+x<5
subtract 1
x<4
set of answers are 1,2,3
3+x<5
subtract 3
x<2
set of answer is 1
so the possible numbers are
1,2,3
that is 3 numbesr out of 9 so
probability=(total desired outcomes)/(total possible outcomes) so
disred outcomes=3
total possible=9
3/9=1/3
the probabiltiy is 1/3
Answer:
Option C
Step-by-step explanation:
When you put the write the scenario in an equation, it'll look like
(5q+3)(5q-3) for the new width times the new length
Factor these, and you shall get 25q^2+15q-15q-9. Simplify that into
25q^2-9
To model the scenario, the following will be the numerical coefficients for the variables: 0.05 for nickels (5 cents), 0.10 for dime (10 cents) and 0.25 for quaters (25 cents). Thus the model/equation is, $1.87 = 0.25q + 0.10d + 0.05n + 60.
Answer:
t = -1
Step-by-step explanation:
15(t+2)+9t=6
15t + 30 + 9t = 6
combine like terms
24t +30 = 6
subtract 30 both sides
24t = -24
t = -1
22-5(6v-1) = -63
22- 30v + 5 = -63
27 -30v = -63
subtract both sides by 27
-30v = -90
divide it by -30
v = 3
