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2 years ago

A single, standard cube is tossed. What is the probability of getting a number greater than 3?

Mathematics

1 answer:

sertanlavr [38]2 years ago

7 0

Answer:

1/2 or 50%

Step-by-step explanation:

numbers on a dice:
1, 2, 3, 4, 5, 6

4 5 and 6 are above 3
therefore, there are 3 numbers that are greater than 3 on a standard dice

3/6 can be simplified to 1/2 which is also the same as 50%.

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4(x + 3) = 6-x what is the answer I need help please

Amanda [17]

4(x + 3) = 6 - x

First, expand to remove parentheses.
$4x + 12 = 6 - x$
Second, subtract '6' from both sides.
$4x + 12 - 6 = -x$
Third, subtract '12 - 6' to get 6.
$4x + 6 = -x$
Fourth, subtract '4x' from both sides.
$6 = -x - 4x$
Fifth, since 'x' can be referred to as '1', add it to '4x' to get '-5x'.
$6 = -5x$
Sixth, divide both sides by '-5'.
$\frac{6}{-5} =x$
Seventh, change the whole fraction into a negative.
$-\frac{6}{5} =x$
Eighth, switch your sides.
$x = -\frac{6}{5}$

Answer as fraction: $-\frac{6}{5}$
Answer as decimal: -1.2

4 0

3 years ago

What is the inequality of the problem and the symbol

Illusion [34]

48 is greater than or equal too 40in I think so < with a little line under it

3 0

3 years ago

Read 2 more answers

710 pts;10% decrease

allochka39001 [22]

10%of 710 is 71

now we have 710-71=639

3 0

3 years ago

A car park is full. 1/3 of the cars leave.60% of the cars are red. There are 174 red cars. How many cars are left in the car par

juin [17]

Answer:

290

Step-by-step explanation:

60% of ⅔X = 174

60/100 × 2/3 × X = 174

⅖X = 174

X = 435

⅔X = ⅔×435

290

290 cars are left in the parking,

145 cars left

5 0

3 years ago

Read 2 more answers

Gina looked at the architectural plan of a room with four walls in which the walls meet each other at right angles. The length o

Lerok [7]

For a better understanding of the solution provided here, please find the diagram attached.

In the diagram, ABCD is the room.

AC is the diagonal whose length is 18.79 inches.

The length of wall AB is 17 inches.

From the given information, we have to determine the length of the BC, which is depicted a $x$ , because for the room to be a square, the length of the wall AB must be equal to the length of the wall BC.

In order to determine the length of the wall BC, or $x$ , we will have to employ the Pythagoras' Theorem here. Thus:

$x=\sqrt{(AC)^2-(AB)^2}$

$x=\sqrt{(18.79)^2-(17)^2}\approx\sqrt{64.06} \approx8$

Thus, $BC\approx 8$ inches

$\therefore AB\neq BC$ and hence, the given room is not a square.

8 0

3 years ago