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Dahasolnce [82]
3 years ago
8

Find the value of d.

Mathematics
1 answer:
Elena-2011 [213]3 years ago
5 0

I’m pretty sure the answer is 16, I’m very sorry if this is incorrect.

You might be interested in
Si en 11 cajas de chocolates hay 220 chocolates en total. Calcula cuántos chocolates habrá en total en 5 cajas de chocolates.
pochemuha

Answer:

Habrá 100 chocolates en total en 5 cajas de chocolates.

Step-by-step explanation:

Esta pregunta es resolvida usando proporciones, con lo uso de regla de tres .

11 cajas, 220 chocolates:

Entonces:

\frac{220}{11} = 20

20 chocolates por caja.

Calcula cuántos chocolates habrá en total en 5 cajas de chocolates.

20 chocolates por caja, entonces:

5*20 = 100

Habrá 100 chocolates en total en 5 cajas de chocolates.

6 0
3 years ago
Ху
Galina-37 [17]

Answer:

a

Step-by-step explanation:

-2 over 3 is it cuz -4 over 6 simplifies to that and all the others do too

8 0
3 years ago
Solve for x. 3x/2 = 15
defon

3x : 2 = 15

3x = 15 * 2

3x = 30

x = 30 : 3

x = 10

--------------------------------

3(10)  :2 = 15

6 0
3 years ago
Identify which statement about the mean of a discrete random variable is not true or state that they are all true.Choose the cor
pashok25 [27]

Answer:

A, C are true .  B is not true.

Step-by-step explanation:

Mean of a discrete random variable can be interpreted as the average outcome if the experiment is repeated many times. Expected value or average of the distribution is analogous to mean of the distribution.

The mean can be found using summation from nothing to nothing x times Upper P (x) , i.e ∑x•P(x).

Example : If two outcomes 100 & 50 occur with probabilities 0.5 each. Expected value (Average) (Mean) : ∑x•P(x) = (0.5)(100) + (0.5)(50) = 50 + 25 = 75

The mean may not be a possible value of the random variable.

Example : Mean of possible no.s on a die = ( 1 + 2 + 3 + 4 + 5 + 6 ) / 6 = 21/6 = 3.5, which is not a possible value of the random variable 'no. on a die'

4 0
3 years ago
How to solve this problem
Agata [3.3K]
So first, you have to get two of the same variables to cancel out. Let's do this for x. In order for the x's to cancel out, we could multiply the bottom problem by 2.
(2) 3x-6y=24
After multiplying all the numbers by 2, you get the equation 6x-12y=48
The set of equations is now
-6x+2y=12
6x-12y=48
Now you can add them. The x variables cancel out, so you are left with the y variable.
2y+-12y=-10y and 12+48=60
Then you would divide 60 by -10 to get y=-6.
You would plug the answer for y into one of the original equations, lets do the top one. -6x+2y=12 becomes -6x+2(-6)=12
You'd multiply the 2 and -6 to get -12 so the equation is
-6x-12=12
The negative 12 turn positive and you add to both sides to get the -6x alone.
-6x-12=12
+12=12
-6x=24
Then divide 24 by -6
X=4

(-4,-6) is your final answer.
7 0
3 years ago
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