Answer:
Determine a single event with a single outcome. ...
Identify the total number of outcomes that can occur. ...
Divide the number of events by the number of possible outcomes. ...
Determine each event you will calculate. ...
Calculate the probability of each event.
Step-by-step explanation:
CAN I PLS HAVE BRAINLIEST
v + m = 32 and v = 5 + 2m are the equations that are used to determine m, the number of stuffed animals Mariposa has
Number of stuffed animals Mariposa has is 9 and number of stuffed animals with Veronica is 23
<h3>
<u>Solution:</u></h3>
Let "v" be the number of stuffed animals with Veronica
Let "m" be the number of stuffed animals with Mariposa
Given that,
Together, they have 32 stuffed animals
Therefore,
v + m = 32 --------- eqn 1
Veronica has 5 more than double the number of stutted animals as her friend Mariposa
Therefore,
Number of stuffed animals with Veronica = 5 + 2(number of stuffed animals with Mariposa)
v = 5 + 2m ---------- eqn 2
Thus eqn 1 and eqn 2 can be used to determine m, the number of stuffed animals Mariposa has
Let us solve eqn 1 and eqn 2
Substitute eqn 2 in eqn 1
5 + 2m + m = 32
5 + 3m = 32
3m = 32 - 5
3m = 27
<h3>m = 9</h3>
Substitute m = 9 in eqn 2
v = 5 + 2(9)
v = 5 + 18
<h3>v = 23</h3>
Thus number of stuffed animals Mariposa has is 9 and number of stuffed animals with Veronica is 23
Answer:
Step-by-step explanation:
d = 
d = 
d =
Answer:
33.49 cubic units
Step-by-step explanation:
Answer:
c
Step-by-step explanation:
come on ! you can literally see that in the chart.
how many parts of the gray 3/8 are covered by the gray 1/4 ?
2 parts = 2/8 are clearly covered by 1/4.
2/8 is what part of 3/8 ?
it is the same question as "2 is what part of 3" ?
is 2 a quarter (1/4) of 3 ? no, 1/4×3 = 3/4 and not 2.
is 2 one third (1/3) of 3 ? no, 1/3 of 3 = 1/3×3 = 1 and not 2.
is 2 two thirds (2/3) of 3 ? ah, 2/3 × 3 = 2. that is correct !
is 2 three quarters (3/4) of 3 ? no, 3/4×3 = 9/4 and not 2.
once you have the same denominator, you can easily compare the numerators and ignore the denominators for such problems.
