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Ludmilka [50]
2 years ago
9

What’s the 38th term of the arithmetic sequence 31,40,49,58?

Mathematics
1 answer:
Trava [24]2 years ago
3 0

Answer:

a_3_8=364

Step-by-step explanation:

we know that

In an <u><em>Arithmetic Sequence</em></u> the difference between one term and the next is a constant, and this constant is called the common difference

we have

31,40,49,58,...

Let

a_1=31\\a_2=40\\a_3=49\\a_4=58

we have that

a_2-a_1=40-31=9

a_3-a_2=49-40=9

a_4-a_3=58-49=9

so

The common difference is equal to 9

We can write an Arithmetic Sequence as a rule:

a_n=a_1+d(n-1)

where

a_n is the nth term                                                              

a_1 is the first term

d is the common difference                        

n is the number of terms

Find the 38th term of the arithmetic sequence

we have                  

a_1=31\\d=9\\n=38              

substitute the values

a_3_8=31+9(38-1)

a_3_8=31+9(37)

a_3_8=364

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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

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

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

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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:

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= 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

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The variance of the number of successful surgeries is 0.3111.

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