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scoundrel [369]
2 years ago
14

What is the equation of -2(x-3)+4x=-(-x+1)

Mathematics
2 answers:
BabaBlast [244]2 years ago
8 0

Answer:

Exact form: -5/3

Decimal form: -1.66 (repeating)

Mixed number form: -1\frac{2}{3}

Katyanochek1 [597]2 years ago
3 0

Answer:

x = -7

Step-by-step explanation:

<u>Step 1:  Distribute </u>

-2(x - 3) + 4x = -(-x + 1)

<em>-2x + 6 + 4x = x - 1</em>

<em />

<u>Step 2:  Combine like terms</u>

-2x + 6 + 4x = x - 1

<em>2x + 6 = x - 1</em>

<em />

<u>Step 3:  Subtract x from both sides</u>

2x + 6 - x = x - 1 - x

<em>x + 6 = -1</em>

<em />

<u>Step 4:  Subtract 6 from both sides</u>

x + 6 - 6 = -1 - 6

<em>x = -7</em>

<em />

Answer:  x = -7

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Answer:

The mean number of successful surgeries is 1.57.

The variance of the number of successful surgeries is 0.3111.

Step-by-step explanation:

STEP 1

If the tear on the left knee has a rim width of less than 3mm, the probability that the surgery on the left knee will be successful (ls) is 0.90.

That isP(Is)=0.90

.

The probability that the surgery on the left knee will fail (lf) is 0.10. That isP(V)=0.10

.

If the tear on the right knee has a rim width of 3-6 mm, the probability that the surgery on the right knee will be successful (rs) is 0.67. That isP(rs)=0.67

.

The probability that the surgery on the right knee will fail (rf) is 0.33.

That isP(vf)= 0.33

.

Let the random variable X denote the number of successful surgeries. The range of X is{0,1,2)

.

Now find the probabilities associated with the possible values of X. The number of successful surgeries is equal to 0 if the surgeries on both knees fail. Since the surgeries are independent, we have:

P(X = 0)=P(lf and rf)

= P()P(rf)

=(0.10)(0.33)

= 0.033

The number of successful surgeries is equal to 1 if the surgery on one knee is successful and the surgery on the other knee fails, that is

P(X =1)= P((ls and rf) or (if and rs))

= P(Is)P(rf)+P(V)P(rs)

= (0.90)(0.33)+(0.10)(0.67)

= 0.364

The number of successful surgeries is equal to 2 if the surgeries on both knees are successful, that is

P(X = 2)=P(ls and rs)

= P(Is) P(rs)

=(0.90)(0.67)

= 0.603

So, the probability mass function of X is the following

X        0              1               2

f(x)     0.033      0.364      0.603

The mean number of successful surgeries is,

E(X)=∑xP(x)

=(0×0.033) +(1×0.364)+(2×0.603)

=1.57

the mean number of successful surgeries is 1.57

The expected value is obtained by taking the summation of the product of each possible value of the random variable X with its corresponding probabilities. Thus, the mean number of successful surgeries is 1.57.

STEP 2

The variance of the number of successful surgeries is,

Var(X)=∑x²P(x)-(E(x²)

=(0×0.033) + (1 × 0.364) +(2 × 0.603) - (1.57)²

= 2.776 -(1.57)²

=0.3111

The variance of the number of successful surgeries is 0.3111.

6 0
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