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Ugo [173]
3 years ago
15

What is the sum? Thank you!

Mathematics
2 answers:
Daniel [21]3 years ago
5 0

Answer:

1835014

Step-by-step explanation:

Nady [450]3 years ago
4 0

Answer:

2097150

Step-by-step explanation:

sum=(2(1-1048576))/(1-2)

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PLEASE HELP ME WITH THIS! ASAP please and thank you
galben [10]

Just plug in

4a / (2b - c)

= 4(5) / (2*4 -  6)

= 20 / 2

= 10

Answer

10

5 0
3 years ago
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Select all the equations that have {4} as its solution set. 5 + 2x = 3x + 37−x=12x79−3x=12x+19x2+3=10x−21
Salsk061 [2.6K]

Answer:

Option 3 (79−3x=12x+19)

Step-by-step explanation:

When replacing x with 4 you will get the answer 67 = 67, which is true! The other three options are not true when replacing x with 4.

6 0
2 years ago
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(4.1.4) Let X and Y be Bernoulli random variables. Let Z = X + Y. a. Show that if X and Y cannot both be equal to 1, then Z is a
Fynjy0 [20]

Step-by-step explanation:

Given that,

a)

X ~ Bernoulli (p_x) and Y ~ Bernoulli (y_x)

X + Y = Z

The possible value for Z are Z = 0 when X = 0 and Y = 0

and Z = 1 when X = 0 and Y = 1 or when X = 1 and Y = 0

If X and Y can not be both equal to 1 , then the probability mass function of the random variable Z takes on the value of 0 for any value of Z other than 0 and 1,

Therefore Z is a Bernoulli random variable

b)

If X and Y can not be both equal to  1

then,

p_z = P(X=1 or Y=1)\\

p_z = P(X=1)+P(Y=1)-P(=1 and Y =1)

p_z = P(x=1)+P(Y=1)\\\\p_z=p_x+p_y

c)

If both X = 1 and Y = 1 then Z = 2

The possible values of the random variable Z are 0, 1 and 2.

since a  Bernoulli variable should be take on only values 0 and 1 the random variable Z does not have Bernoulli distribution

7 0
3 years ago
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valentina_108 [34]

Answer:

The answer is 5w+12.

Step-by-step explanation:

Combine like terms:

(3w+7)+(2w+5)

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3 years ago
How much change will you get back if you bought 3 $0.99 chocolate bars and paid with $5 bill
Molodets [167]
You will get $2.03 back in change 
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3 years ago
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