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loris [4]
3 years ago
5

Where would you put the decimal point in 201 x 15 to make it equal 30.15 201 x 15 = 30.15

Mathematics
2 answers:
Norma-Jean [14]3 years ago
4 0

Answer:

2.01 x 15 = 30.15

Step-by-step explanation:

2.01 x 15 = 30.15

or

201 * .15 = 30.15

zlopas [31]3 years ago
4 0

Answer:

201 changes to 2.01 or 15 changes to 0.15

2.01 · 15 = 30.15

201 · 0.15 = 30.15

Step-by-step explanation:

Since 201 · 15 = 3015, and we want 30.15

We move the decimal two place to the left.

Since we did this on one side of the equation, we must do it on the other as well.

201 becomes 2.01. You could also change 15

15 becomes 0.15

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1) Two coins are to be flipped. The first coin will land on heads with probability .6, the second with probability .7. Assume th
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Answer:

(a) P(X=1)=0.46

(b) E[X]=1.3

Step-by-step explanation:

(a)

Let A be the event that first coin will land on heads and B be the event that second coin will land on heads.

According to the given information

P(A)=0.6

P(B)=0.7

P(A')=1-P(A)=1-0.6=0.4

P(B')=1-P(B)=1-0.7=0.3

P(X=1) is the probability of getting exactly one head.

P(X=1) = P(1st heads and 2nd tails ∪ 1st tails and 2nd heads)

          = P(1st heads and 2nd tails) + P(1st tails and 2nd heads)

Since the two events are disjoint, therefore we get

P(X=1)=P(A)P(B')+P(A')P(B)

P(X=1)=(0.6)(0.3)+(0.4)(0.7)

P(X=1)=0.18+0.28

P(X=1)=0.46

Therefore the value of P(X=1) is 0.46.

(b)

Thevalue of E[X] is

E[X]=\sum_{x}xP(X=x)

E[X]=0P(X=0)+1P(X=1)+2P(X=2)

E[X]=P(X=1)+2P(X=2)                      ..... (1)

First we calculate  the value of P(X=2).

P{X = 2} = P(1st heads and 2nd heads)

             = P(1st heads)P(2nd heads)

P(X=2)=P(A)P(B)

P(X=2)=(0.6)(0.7)

P(X=2)=0.42

Substitute P(X=1)=0.46 and P(X=2)=0.42 in equation (1).

E[X]=0.46+2(0.42)

E[X]=1.3

Therefore the value of E[X] is 1.3.

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