20×+4=4(5x+1) how many solution are in this equation

Mathematics

1 answer:

Tomtit [17]3 years ago

7 0

Simplify both sides of the equation

(20)(4)= 4(5x+1)

Simplify

80 = 20x+4

Flip the equation

20x+4 = 80

Subtract 4 from both sides

20x+4-4 = 80-4

20x = 76

Divide both sides by 20

20x/20 = 76/20

x= 3.8

There is one solution in this equation

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Consider a game in which a participant pays $2 to roll a die. The participant receives $3 if they roll a 1 (i.E. They go up by a

Sauron [17]

Answer:

The expected monetary value of a single roll is $1.17.

Step-by-step explanation:

The sample space of rolling a die is:

S = {1, 2, 3, 4, 5 and 6}

The probability of rolling any of the six numbers is same, i.e.

P (1) = P (2) = P (3) = P (4) = P (5) = P (6) = $\frac{1}{6}$

The expected pay for rolling the numbers are as follows:

E (X = 1) = $3

E (X = 2) = $0

E (X = 3) = $0

E (X = 4) = $0

E (X = 5) = $0

E (X = 6) = $4

The expected value of an experiment is:

$E(X)=\sum x\cdot P(X=x)$

Compute the expected monetary value of a single roll as follows:

$E(X)=\sum x\cdot P(X=x)\\=[E(X=1)\times \frac{1}{6}]+[E(X=2)\times \frac{1}{6}]+[E(X=3)\times \frac{1}{6}]\\+[E(X=4)\times \frac{1}{6}]+[E(X=5)\times \frac{1}{6}]+[E(X=6)\times \frac{1}{6}]\\=[3\times \frac{1}{6}]+[0\times \frac{1}{6}]+[0\times \frac{1}{6}]\\+[0\times \frac{1}{6}]+[0\times \frac{1}{6}]+[4\times \frac{1}{6}]\\=1.17$