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Afina-wow [57]
3 years ago
13

If you were creating a lamp by hand what type of material would you choose for the wiring in what material would you choose to c

ut the wiring Why?
Mathematics
2 answers:
grin007 [14]3 years ago
8 0

The material for the lamp of the wiring would be cutting with pliers.

dangina [55]3 years ago
6 0
I would choose copper for the wiring because copper conducts electricity well. I would cut the wire with wire cutters because they are made for cutting wires.
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What is the meaning of the unknown factor and quotient
konstantin123 [22]
Unknown factor is multiplication and then quotient is division 
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3 years ago
What is 2.33 written as a mixed number in simplest form?
aleksandr82 [10.1K]
2.33.......ok, 2 is ur whole number....33 is ur fraction...the last digit is in the hundredths place....so put it over 100

ur mixed number is 2 33/100...and this does not reduce
8 0
4 years ago
Read 2 more answers
PLEASE HELP.
ohaa [14]
I hope you're reading this because I didn't answer your question.
6 0
4 years ago
How many 4\5 cup servings are in 2\5 of a cup of icecream in simplist fraction form
DaniilM [7]

Divide total ice cream by serving size:

2/5 / 4/5

When you divide fractions, change division to multiplication and flip the second fraction over:

2/5 x 5/4

Now multiply ht top numbers and the bottom numbers:

(2 x 5) / (5 x 4) = 10/20 = 1/2

There is 1/2 serving.

3 0
3 years ago
Two dice are rolled. Let the random variable X denote the number that falls uppermost on the first die and let Y denote the numb
Vlada [557]

Step-by-step explanation:

a) Lets start by giving the answer:

The probability distribution of X is:

P(X=x)=\frac{1}{6}

For every x = 1, 2, 3, 4, 5, 6

Same for Y:

P(Y=y)=\frac{1}{6}

For every y = 1, 2, 3, 4, 5, 6

This is because each variable has 6 possibilities which are equiprobable. In other words, every number (1 to 6) has the same probability of falling uppermost on each dice.

Now if the question refers to a joint probability distribution, then we have a joint random viariable (X,Y). This probability is given by:

P[(X,Y)=(x,y)] = \frac{1}{36}

For every x = 1, 2, 3, 4, 5, 6 and y = 1, 2, 3, 4, 5, 6

The reason for this is that the combination of the result of both dices produces 36 equiprobable possibilities.

b) For the probability distribution of the sum of the variables we may define:

Z=X+Y

The for the random variable Z the next results are possible:

Z       P(Z)

2       1/36

3       2/36

4       3/36

5       4/36

6       5/36

7       6/36

8       5/36

9       4/36

10      3/36

11       2/36

12      1/36

This is our probability distribution for the sum of X+Y. Now to understand the results lets see some examples:

For z = 2 (x=1 and y=1) we obtain the probability P(Z=2)=\frac{1}{36}, this is because we have 1 possibility to obtain (1,1) among 36 possibilities.

For z = 4 that is for (x=1, y=3) and (x=2, y=2) and (x=3, y=1) we obtain P(Z=4)=\frac{3}{36} this is because we have 3 possibilities to obtain any of the combinations mentioned among 36 possibilities.

For z = 10 that is for (x=4, y=6) and (x=5, y=5) and (x=6, y=6) we obtain P(Z=4)=\frac{3}{36} because we have 3 possibilities to obtain any of the combinations mentioned among 36 possibilities.

8 0
4 years ago
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