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Elina [12.6K]
3 years ago
11

If i roll 2 dice what is the probability that neither of them are prime nor even?

Mathematics
1 answer:
Eduardwww [97]3 years ago
4 0

Answer:

There are 3 prime numbers shown on the die: 2, 3 and 5. The probability of showing a prime number on a single die is 1/2, hence the probability of not showing a prime number is also 1/2. Total possible outcomes when two dice are rolled = 6*6 = 36. Total possible outcomes when two dice are rolled = 6*6 = 36.

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Which correctly describes how the graph of the inequality 5x + 2y = 13 is shaded?
exis [7]

Answer:

The answer is above the solid line.

Step-by-step explanation:

6 0
2 years ago
Solve this equation with variables on both sides- 3x=5x+18 show all steps
Dima020 [189]
3x=5x+18
3x-3x=5x-3x+18
0=2x+18
0-18=2x+18-18
-18=2x
-18÷2=2x÷2
-9=x
Hope this helps!
Vote me brainliest!
8 0
3 years ago
Read 2 more answers
For what value of x is (2x^2+8x)/(x+4)(x^2-9) undefined?
irakobra [83]

Answer:

The value at x =3 is undefined for the given expression

Step-by-step explanation:

Given expression is \frac{2x^2+8x}{(x+4)(x^2-9)}

To find for what value of x is undefined in the given expression :

\frac{2x^2+8x}{(x+4)(x^2-9)}=\frac{2x(x+4)}{(x+4)(x^2-9)}

=\frac{2x}{x^2-9}

\frac{2x^2+8x}{(x+4)(x^2-9)}=\frac{2x}{x^2-9}

If we put x=3 in the above expression we get

\frac{2x}{x^2-9)}=\frac{2(3)}{3^2-9}

=\frac{6}{9-9}

=\frac{6}{0}

Therefore \frac{2x^2+8x}{(x+4)(x^2-9)}=\frac{6}{0}

The value at x =3 is undefined for the given expression

7 0
3 years ago
Which is faster a car driving at a rate of 40 kph or a car at driving at a rate of 15 metres a second
LuckyWell [14K]
40 kph is equal to abt 11 mps so that means 40kph is faster
7 0
3 years ago
At a farm, there are two huge pits for storing hay. 90 tons of hay is stored in the first pit, 75 tons in the second pit. Then,
Travka [436]

Answer:

63 tons

Step-by-step explanation:

The problem statement asks for the tons of hay removed from the first pit. It is convenient to let a variable (x) represent that amount. This is said to be 3 times the amount removed from the second pit, so that amount must be x/3.

The amount remaining in the first pit is 90-x.

The amount remaining in the second pit is 75 -x/3.

Since the first pit remaining amount is half the second pit remaining amount, we can write the equation ...

... 90 -x = (1/2)(75 -x/3)

... 180 -2x = 75 -x/3 . . . . multiply by 2

... 105 - 2x = -x/3 . . . . . . subtract 75

... 315 -6x = -x . . . . . . . . multiply by 3

... 315 = 5x . . . . . . . . . . . add 6x

... 63 = x . . . . . . . . . . . . . divide by 5

63 tons of hay were taken from the first pit.

_____

<em>Check</em>

After removing 63 tons from the first pit, there are 27 tons remaining. After removing 63/3 = 21 tons from the second pit, there are 54 tons remaining. 27 is half of 54, so the answer checks OK.

6 0
3 years ago
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