You roll a standard number cube. Find P (number greater than 4). A. 2/3. B. 1/2. C. 3/5. D. 1/3

2 answers:

8 0

Probablity=(number of desired outcomes)/(total possible outcomes)

desired outcomes is greater than 4
5 and 6
2 options

total possible outcomes
6 faces, 6 outcomes

2/6=1/3
D is answer

damaskus [11]3 years ago

4 0

Total Outcome = 6

Number greater than 4 = 2 (5&6)

So, probability would be 2/6 = 1/3

(D)OPTION D IS YOUR ANSWER

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Please help me solve inequalities and show work. Will get brainiest!!

Dahasolnce [82]

The answer depends on whether you're looking for the x or the y. If y, the answers are the ones on the left, and the x on the right.

1) -4x + 3y ≤ 15 -4x + 3y ≤ 15
 +4x +4x -3y -3y

3y ≤ 4x + 15 -4x ≤ -3y + 15
 --- --------- ---- ------------
3 3 -4 -4

y ≤ 4/3x + 5 x ≥ 3/4y + 11

2) -6x - 4y < 12 -6x - 4y < 12
+6x +6x +4y +4y

-4y < 6x +12 -6x < 4y + 12
---- ---------- ---- -----------
-4 -4 -6 -6

y > -3/2 - 3 x > -2/3y - 2

8 0

2 years ago

I don't know how to do this or what i'm doing plz help

mote1985 [20]

recalling that d = rt, distance = rate * time.

we know Hector is going at 12 mph, and he has already covered 18 miles, how long has he been biking already?

$\bf \begin{array}{ccll} miles&hours\\ \cline{1-2} 12&1\\ 18&x \end{array}\implies \cfrac{12}{18}=\cfrac{1}{x}\implies 12x=18\implies x=\cfrac{18}{12}\implies x=\cfrac{3}{2}$

so Hector has been biking for those 18 miles for 3/2 of an hour, namely and hour and a half already.

then Wanda kicks in, rolling like a lightning at 16mph.

let's say the "meet" at the same distance "d" at "t" hours after Wanda entered, so that means that Wanda has been traveling for "t" hours, but Hector has been traveling for "t + (3/2)" because he had been biking before Wanda.

the distance both have travelled is the same "d" miles, reason why they "meet", same distance.

$\bf \begin{array}{lcccl} &\stackrel{miles}{distance}&\stackrel{mph}{rate}&\stackrel{hours}{time}\\ \cline{2-4}&\\ Hector&d&12&t+\frac{3}{2}\\[1em] Wanda&d&16&t \end{array}\qquad \implies \begin{cases} \boxed{d}=(12)\left( t+\frac{3}{2} \right)\\[1em] d=(16)(t) \end{cases}$

$\bf \stackrel{\textit{substituting \underline{d} in the 2nd equation}}{\boxed{(12)\left( t+\frac{3}{2} \right)}=16t}\implies 12t+18=16t \\\\\\ 18=4t\implies \cfrac{18}{4}=t\implies \cfrac{9}{2}=t\implies \stackrel{\textit{four and a half hours}}{4\frac{1}{2}=t}$