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yarga [219]
3 years ago
13

The drama club ordered 15 pizzas for the cast party. If the 25 members ate 2 & 1/2 pizzas out of the 15, how much pizza did

each cast member eat?
Mathematics
1 answer:
maria [59]3 years ago
4 0

like 10????? i dont understand that much but i tryed

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3(x-12)=15 what number is x fist to answer gets brainlest
Bingel [31]

Answer:

x = 17

Step-by-step explanation:

3(x-12)=15

->

x-12+12=5+12

7 0
2 years ago
Use elimination to solve 3x-2y=24 x+2y=48
MakcuM [25]

Answer:

X= 18; Y=15

Step-by-step explanation:

As a student myself working this out I cannot explain how I got the answer but I know it is correct.

5 0
3 years ago
Please answer the following questions: <br><br> 1. 1/2*8=<br> 2. 1/4*4=<br> 3. 1/3*6=
yan [13]
Hey there!

When we multiply fractions and whole numbers, we simply take that whole number and put it over them, and just multiply across. That gives us:

\frac{1}{2}x*8 = \frac{8}{2} = 4

Notice how it's four, because when we have a fraction, we divide. 8 divided by 2 equals four. Next:

\frac{1}{4} *4 = \frac{4}{4} = 1

4/4 is equal to 1 because 4 divided by 4 is one. 4 goes into 4 once. Next, we have:

\frac{1}{3}*6 = \frac{6}{3} = 2

That's because 6/3 = 2.

Hope this helps!

7 0
3 years ago
Read 2 more answers
9+19 - 12 + 12 + 12323 - 39
Alexxandr [17]

Answer : 12312

Step-by-step explanation : According to BODMAS, Addition (+) comes first

Hence --  9 + 19 + 12 + 12323 = 12363

               12363 - 12 - 39  = 12363 - 51

                                          = 12312

       

5 0
3 years ago
There are three power plants [X, Y, Z] that at any given time each one either generates electricity or idles. Event A is that pl
insens350 [35]

We're told that

P(A\cap B)=0.15

P(A\cup B)^C=0.06\implies P(A\cup B)=0.94

P(B\mid A)=P(B^C\mid A)=0.5

where the last fact is due to the law of total probability:

P(A)=P(A\cap B)+P(A\cap B^C)

\implies P(A)=P(B\mid A)P(A)+P(B^C\mid A)P(A)

\implies 1=P(B\mid A)+P(B^C\mid A)

so that B\mid A and B^C\mid A are complementary.

By definition of conditional probability, we have

P(B\mid A)=P(B^C\mid A)

\implies\dfrac{P(A\cap B)}{P(A)}=\dfrac{P(A\cap B^C)}{P(A)}

\implies P(A\cap B)=P(A\cap B^C)

We make use of the addition rule and complementary probabilities to rewrite this as

P(A\cap B)=P(A\cap B^C)

\implies P(A)+P(B)-P(A\cup B)=P(A)+P(B^C)-P(A\cup B^C)

\implies P(B)-[1-P(A\cup B)^C]=[1-P(B)]-P(A\cup B^C)

\implies2P(B)=2-[P(A\cup B)^C+P(A\cup B^C)]

\implies2P(B)=[1-P(A\cup B)^C]+[1-P(A\cup B^C)]

\implies2P(B)=P(A\cup B)+P(A\cup B^C)^C

\implies2P(B)=P(A\cup B)+P(A^C\cap B)\quad(*)

By the law of total probability,

P(B)=P(A\cap B)+P(A^C\cap B)

\implies P(A^C\cap B)=P(B)-P(A\cap B)

and substituting this into (*) gives

2P(B)=P(A\cup B)+[P(B)-P(A\cap B)]

\implies P(B)=P(A\cup B)-P(A\cap B)

\implies P(B)=0.94-0.15=\boxed{0.79}

8 0
3 years ago
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