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3 years ago

6

Shelley is fresh out of college and looking for housing in Boston. She makes $37,000 a year and may have to move within two year

s to her company’s office in Dallas. What should Shelley value when looking for housing?
1. flexibility and short term cost
2. tax incentives and home appreciation
3. maintenance and homeowner’s fees
4. stability and long term costs

thank you for you help and time!!!

2 answers:

34kurt3 years ago

6 0

answer is 1

as Shelley has to live in Boston for only 2 years, she needs to consider that the owner is flexible enough to make 2 year rent agreement and the short term rent of the house.

Other choices are irrelevant. they may be applied while purchasing home with long term aim to stay in the place.

Aleonysh [2.5K]3 years ago

5 0

Id go with option 4 or 1

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Answer:

X represents The value of players on the field

Step-by-step explanation:

the numbers 0-11 are showing how many people would be on the field should an official game be played

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3 years ago

One roll of carpet covers 900 square feet. how many rolls of carpet would be needed for a room that is 40ft. x 60ft.?

harina [27]

3 rolls of carpet

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4 years ago

Read 2 more answers

Seventy-two percent of the light aircraft that disappear while in flight in a certain country are subsequently discovered. Of th

Anton [14]

Answer:

a) 0.105 = 10.5% probability that it will not be discovered if it has an emergency locator.

b) 0.522 = 52.2% probability that it will be discovered if it does not have an emergency locator.

c) 0.064 = 6.4% probability that 7 of them are discovered.

Step-by-step explanation:

For itens a and b, we use conditional probability.

For item c, we use the binomial distribution along with the conditional probability.

Conditional Probability

We use the conditional probability formula to solve this question. It is

$P(B|A) = \frac{P(A \cap B)}{P(A)}$

In which

P(B|A) is the probability of event B happening, given that A happened.

$P(A \cap B)$ is the probability of both A and B happening.

P(A) is the probability of A happening.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

In which $C_{n,x}$ is the number of different combinations of x objects from a set of n elements, given by the following formula.

$C_{n,x} = \frac{n!}{x!(n-x)!}$

And p is the probability of X happening.

a) If it has an emergency locator, what is the probability that it will not be discovered?

Event A: Has an emergency locator

Event B: Not located.

Probability of having an emergency locator:

66% of 72%(Are discovered).

20% of 100 - 72 = 28%(not discovered). So

$P(A) = 0.66*0.72 + 0.2*0.28 = 0.5312$

Probability of having an emergency locator and not being discovered:

20% of 28%. So

$P(A cap B) = 0.2*0.28 = 0.056$

Probability:

$P(B|A) = \frac{P(A \cap B)}{P(A)} = \frac{0.056}{0.5312} = 0.105$

0.105 = 10.5% probability that it will not be discovered if it has an emergency locator.

b) If it does not have an emergency locator, what is the probability that it will be discovered?

Probability of not having an emergency locator:

0.5312 of having. So

$P(A) = 1 - 0.5312 = 0.4688$

Probability of not having an emergency locator and being discovered:

34% of 72%. So

$P(A \cap B) = 0.34*0.72 = 0.2448$

Probability:

$P(B|A) = \frac{P(A \cap B)}{P(A)} = \frac{0.2448}{0.4688} = 0.522$

0.522 = 52.2% probability that it will be discovered if it does not have an emergency locator.

c) If we consider 10 light aircraft that disappeared in flight with an emergency recorder, what is the probability that 7 of them are discovered?

p is the probability of being discovered with the emergency recorder:

0.5312 probability of having the emergency recorder.

Probability of having the emergency recorder and being located:

66% of 72%. So

$P(A \cap B) = 0.66*0.72 = 0.4752$

Probability of being discovered, given that it has the emergency recorder:

$p = P(B|A) = \frac{P(A \cap B)}{P(A)} = \frac{0.4752}{0.5312} = 0.8946$

This question asks for P(X = 7) when $n = 10$ . So

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 7) = C_{10,7}.(0.8946)^{7}.(0.1054)^{3} = 0.064$

0.064 = 6.4% probability that 7 of them are discovered.

8 0

3 years ago

Which of the following geometric series converges?

Artist 52 [7]

All three series converge, so the answer is D.

The common ratios for each sequence are (I) -1/9, (II) -1/10, and (III) -1/3.

Consider a geometric sequence with the first term <em>a</em> and common ratio |<em>r</em>| < 1. Then the <em>n</em>-th partial sum (the sum of the first <em>n</em> terms) of the sequence is

$S_n=a+ar+ar^2+\cdots+ar^{n-2}+ar^{n-1}$

Multiply both sides by <em>r</em> :

$rS_n=ar+ar^2+ar^3+\cdots+ar^{n-1}+ar^n$

Subtract the latter sum from the first, which eliminates all but the first and last terms:

$S_n-rS_n=a-ar^n$

Solve for $S_n$ :

$(1-r)S_n=a(1-r^n)\implies S_n=\dfrac a{1-r}-\dfrac{ar^n}{1-r}$

Then as gets arbitrarily large, the term $r^n$ will converge to 0, leaving us with

$S=\displaystyle\lim_{n\to\infty}S_n=\frac a{1-r}$

So the given series converge to

(I) -243/(1 + 1/9) = -2187/10

(II) -1.1/(1 + 1/10) = -1

(III) 27/(1 + 1/3) = 18

8 0

3 years ago

You are playing Monopoly with a friend. She owns and has developed Park Place and Boardwalk with hotels. If you roll an 8, you w

Mariulka [41]

Okay so this friend has a 8, 9, 10 as her options that she pays at.

8: 2,6 9: 6,3 10: 5,5
6,2 3,6 5,5
4,4
4,4

These are the possibilities of rolling these numbers. = 4 possibilities

4/11 (11 because that is the number of possibilities you get for two dice)

that leaves 7/11 possibilities of rolling and not paying!

36% lands on a spot that she pays at and about 64% possibilities of not paying.

Hope this helped!

:)

6 0

3 years ago