Simplify 2p × 3p2 ×4p3

Twenty-nine percent of americans say they are confident that passenger trips to the moon will occur in their lifetime. You rando

CaHeK987 [17]

Answer:

the probability that less than 33% of the people sampled will answer yes to the question =0.5676

Step-by-step explanation:

8 0

2 years ago

which equation illustrates how the expression 1.15^t can be rewritten to approximate the equivalent monthly interest rate if the

denis23 [38]

sorry if this answer wrong

8 0

2 years ago

An article in Knee Surgery, Sports Traumatology, Arthroscopy, "Arthroscopic meniscal repair with an absorbable screw: results an

kotegsom [21]

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

In a network of 40 computers, 5 hold a copy of a particular file. Suppose that 7 computers at random fail. Let F denote the numb

anzhelika [568]

Answer:

a. E(F)=0.875

b. 99.9976%

c. P(X=2)=0.1683

Step-by-step explanation:

a. We notice that this is a binomial distribution with the probability of success;

$p=\frac{5}{40}=0.125$

#We are given the sample size, n=7. The Expected value is calculated as:

$E(X)=np\\\\E(F)=np , n=7, p=0.125\\\\E(F)=7\times 0.125\\\\=0.875$

Hence the expectation, E(F)=0.875

b. To calculate the probability of the range of F, we need to calculate all possible outcomes of F in the given sample;

$P(X\leq 5)=1-P(X=6)-P(X=7)\\\\=1-{7\choose 6}(0.125)^6(1-0.125)^1-{7\choose 7}(0.125)^7(1-0.125)^0\\\\=1-0.000023365-0.000000476\\\\=0.999976158\\\\=99.9976\%$

Hence, the range of F is 99.9976%

c. The probability that F=2 is calculated using the binomial distribution function as:

$P(X=2)={7\choose 2}(0.125)^2(1-0.125)^5\\\\=0.1683$

Hence, the probability of F=2 is 0.1683

3 0

2 years ago

According to some internet research, 85.2% of adult Americans have some form of medical insurance, and 75.9% of adult Americans

Anastasy [175]

Answer:

0.717 or 71.7%

Step-by-step explanation:

P(M) = 0.852

P(D) = 0.759

P(M or D) = 0.894

The probability that a randomly selected American has both medical and dental insurance is given by the probability of having medical insurance, added to the probability of having dental insurance, minus the probability of having either insurance:

$P(M\ and\ D) = P(M)+P(D)-P(M\ or\ D)\\P(M\ and\ D) =0.852+0.759-0.894\\P(M\ and\ D) =0.717=71.7\%$

The probability is 0.717 or 71.7%.

8 0

2 years ago

Simplify 2p × 3p2 ×4p3​

Simplify 2p × 3p2 ×4p3