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just olya [345]
2 years ago
8

Martin has 12 times as many baseball cards as Josie. Josie has 29 more cards than Kal. How many baseball cards does Martin have?

Mathematics
1 answer:
s344n2d4d5 [400]2 years ago
3 0
I could work out this question but not exactly sure because I think that you need to find out how many kal has because kal is the problem here. once u figure that out. times the number by 12 and yeah. that would be your answer but idk how many kal has.
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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
2 years ago
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