What is the surface area of the rectangular prism shown by the net?

Adia collects 435 cents in nickels. How many nickels does she collect?

julia-pushkina [17]

Answer:

Adia collects 87 nickels.

Step-by-step explanation:

Each nickel is worth 5 cents.

435 / 5 = 87

4 0

3 years ago

Given the Arithmetic series A1+A2+A3+A4

andreev551 [17]

The answer would be 185 bc if you add all of them up that’s what u get

3 0

3 years ago

A pilot is allowed to fly a maximum of 60,000 minutes per year How many hours is a pilot allowed to fly in a year? (Round to the

enot [183]

The answer is 1000 hours

8 0

3 years ago

Some parts of California are particularly earthquake-prone. Suppose that in one metropolitan area, 33% of all homeowners are ins

Lina20 [59]

Answer:

(a) The probability mass function of X is:

$P(X=x)={4\choose x}\ (0.33)^{x}\ (1-0.33)^{4-x};\ x=0,1,2,3...$

(b) The most likely value for X is 1.32.

(c) The probability that at least two of the four selected have earthquake insurance is 0.4015.

Step-by-step explanation:

The random variable X is defined as the number among the four homeowners who have earthquake insurance.

The probability that a homeowner has earthquake insurance is, p = 0.33.

The random sample of homeowners selected is, n = 4.

The event of a homeowner having an earthquake insurance is independent of the other three homeowners.

(a)

All the statements above clearly indicate that the random variable X follows a Binomial distribution with parameters n = 4 and p = 0.33.

The probability mass function of X is:

$P(X=x)={4\choose x}\ (0.33)^{x}\ (1-0.33)^{4-x};\ x=0,1,2,3...$

(b)

The most likely value of a random variable is the expected value.

The expected value of a Binomial random variable is:

$E(X)=np$

Compute the expected value of X as follows:

$E(X)=np$

$=4\times 0.33\\=1.32$

Thus, the most likely value for X is 1.32.

(c)

Compute the probability that at least two of the four selected have earthquake insurance as follows:

P (X ≥ 2) = 1 - P (X < 2)

= 1 - P (X = 0) - P (X = 1)

$=1-{4\choose 0}\ (0.33)^{0}\ (1-0.33)^{4-0}-{4\choose 1}\ (0.33)^{1}\ (1-0.33)^{4-1}\\\\=1-0.20151121-0.39700716\\\\=0.40148163\\\\\approx 0.4015$

Thus, the probability that at least two of the four selected have earthquake insurance is 0.4015.

3 0

3 years ago

At a 20th high school reunion, all the classmates were asked the number of children they had. The probability of having a partic

finlep [7]

Answer:

a) We need to check two conditions:

1) $\sum_{i=1}^n P_i = 1$

$0.05+0.14+0.34+0.24+0.11+0.07+0.02+0.02+0.01= 1$

2) $P_i \geq 0 , \forall i=1,2,...,n$

So we satisfy the two conditions so then we have a probability distribution

b) $P(C \geq 1)$

And we can use the complement rule and we got:

$P(C \geq 1)= 1-P(C$

c) $P(C=0) = 0.05$

d) For this case we see that the result from part b use the probability calculated from part c using the complement rule.

Step-by-step explanation:

For this case we have the following probability distribution given:

C 0 1 2 3 4 5 6 7 8

P 0.05 0.14 0.34 0.24 0.11 0.07 0.02 0.02 0.01

And we assume the following questions:

a) Verify that this is a probability distribution

We need to check two conditions:

1) $\sum_{i=1}^n P_i = 1$

$0.05+0.14+0.34+0.24+0.11+0.07+0.02+0.02+0.01= 1$

2) $P_i \geq 0 , \forall i=1,2,...,n$

So we satisfy the two conditions so then we have a probability distribution

b) What is the probability one randonmly chosen classmate has at least one child

For this case we want this probability:

$P(C \geq 1)$

And we can use the complement rule and we got:

$P(C \geq 1)= 1-P(C$

c) What is the probability one randonmly chosen classmate has no children

For this case we want this probability:

$P(C=0) = 0.05$

d) Look at the answers for parts b and c and explain their relationship

For this case we see that the result from part b use the probability calculated from part c using the complement rule.

5 0

3 years ago