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Ann [662]
3 years ago
13

24 divided by 111 in a decimal number

Mathematics
1 answer:
MissTica3 years ago
6 0

Answer:

.216

Step-by-step explanation:

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Two cars leave the same location traveling in opposite directions. One car leaves at 3:00 p.m. traveling at an average rate of 5
sergey [27]
Recall your d = rt,   distance = rate * time

thus \bf \begin{array}{lccclll}
&distance&rate&time(hrs)\\
&-----&-----&-----\\
\textit{first car}&d&55&x\\
\textit{second car}&380-d&75&x+1
\end{array}\\\\
-----------------------------\\\\

\begin{cases}
\boxed{d}=(55)(x)\\\\
380-d=(75)(x+1)\\
----------\\
380-\left( \boxed{(55)(x)} \right)=(75)(x+1)
\end{cases}

notice, the first car leaves at "x" time, the other leaves on hour later, or x + 1

the first car travels some distance "d", whatever that is, thus
the second car, picks up the slack, or the difference, they're 380 miles
apart, thus the difference is 380-d
8 0
3 years ago
PLEASE HELP ME!! :)​
masha68 [24]

Answer:

In order: There are potentially many solutions, but the one I know that will work with this is 2 blue spaces, 1 Green Space, 2 Orange Spaces, and 5 Red Spaces.

Step-by-step explanation:

How I came to my answer was simple. I divided the greatest common factor into every number of spins, and that number would be the amount of spaces for each.

With this, we get the following equations, as listed below.

16 divided by 8, which accounts for both Orange and Blue.

40 divided by 8, which will account for Red.

8 divided by 8, which accounts for Green.

This gives us our solutions as listed in the Answer Section.

Hope this helped.

7 0
3 years ago
What is the probability that all three cards are face cards when you do not replace each card before selecting the next card rou
Ipatiy [6.2K]
Define three events A,B,C such that
A = first card is a face card
B = second card is a face card (assuming event A happens)
C = third card is a face card (assuming events A & B happen)
No replacements are made

There are 3 face cards (J, K, Q) per suit. There are 4 suits. In total, there are 3*4 = 12 face cards out of 52 total
P(A) = probability that event A happens
P(A) = 12/52
P(A) = 3/13

After event A happens, we have 12-1 = 11 face cards out of 52-1 = 51 total
P(B) = 11/51

After B happens, we have 11-1 = 10 face cards out of 51-1 = 50 left over
P(C) = 10/50
P(C) = 1/5

Based on how the events A,B,C are set up, we can form this equation
P(A and B and C) = P(A)*P(B)*P(C)
P(A and B and C) = (3/13)*(11/51)*(1/5)
P(A and B and C) = 11/1105
P(A and B and C) = 0.00995475113122

Rounded to the nearest tenth, the answer is simply 0.0 which is a lousy answer in my opinion because it implies that the answer is truly 0 instead of something really small. Though to be fair, the result is very close to 0. A much better approach is to round to the nearest hundredth to get 0.01; though I would ask your teacher for clarification and guidance.

note: a probability of 0 means that the event is impossible to happen, yet it is not impossible to pull out three face cards in a row. Is it unlikely? Yes. Because the result is roughly 0.00995 which is a really small decimal close to 0. But it's not 0 itself. 

6 0
3 years ago
4x^0y^-3 for x= -3 and y=2​
exis [7]

Answer:

1/2

Step-by-step explanation:

Plugging in the values

4(-3)^0*2^-3

4(1)*1/8

4/8

1/2

6 0
3 years ago
Jake sold 42 tickets to the school fair and Jeanne sold 9 tickets. What is the ratio, in simplest form, of the number of tickets
MrRissso [65]

Answer:

Step-by-step explanation:

10

i think

7 0
2 years ago
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