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2 years ago

USE THE ATTACHED IMAGE BELOW TO HELP DO MY HOMEWORK PLEASE !!!!

Mathematics

1 answer:

BartSMP [9]2 years ago

8 0

Part A

Sue

A = P*(1+r/n)^(n*t) is the compound interest formula

A = 2300*(1+0.024/1)^(1*3)

A = 2469.6061952

A = 2469.61

A - P = 2469.61-2300 = 169.61

<h3>Sue gets 169.61 pounds in interest</h3>

-----------

Bill

A = P*(1+r/n)^(n*t)

A = 1800*(1+0.034/1)^(1*3)

A = 1989.9131472

A = 1989.91

A-P = 1989.91-1800 = 189.91

<h3>Bill earns 189.91 pounds in interest</h3>

Bill has earned more in interest.

=====================================================

Part B

By year 2, Bill has 1924.48 pounds in his account based on the work shown below

A = P*(1+r/n)^(n*t)

A = 1800*(1+0.034/1)^(1*2)

A = 1924.4808

A = 1924.48

This amount is the new deposit, so to speak, when we change the interest rate. Now r = 0.034 changes to r = 0.04. We only go for one year so t = 1

A = P*(1+r/n)^(n*t)

A = 1924.48*(1+0.04/1)^(1*1)

A = 2001.4592

A = 2001.46

Bill has 2001.46 pounds in his account after 3 years if the interest rate for the 1234 account changes to 4% in the third year.

Now subtract off the original amount Bill deposited to get

2001.46-1800 = 201.46

For this scenario, Bill earns 201.46 pounds in interest.

Therefore, Bill has earned the most interest for both cases of the interest rate staying at 3.4% or changing to 4% for that third year.

<h3>Answer: Bill</h3>

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1 year ago

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A 5-card hand is dealt from a perfectly shuffled deck. Define the events: A: the hand is a four of a kind (all four cards of one

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In a hand of 5 cards, you want 4 of them to be of the same rank, and the fifth can be any of the remaining 48 cards. So if the rank of the 4-of-a-kind is fixed, there are $\binom44\binom{48}1=48$ possible hands. To account for any choice of rank, we choose 1 of the 13 possible ranks and multiply this count by $\binom{13}1=13$ . So there are 624 possible hands containing a 4-of-a-kind. Hence A occurs with probability

$\dfrac{\binom{13}1\binom44\binom{48}1}{\binom{52}5}=\dfrac{624}{2,598,960}\approx0.00024$

There are 4 aces in the deck. If exactly 1 occurs in the hand, the remaining 4 cards can be any of the remaining 48 non-ace cards, contributing $\binom41\binom{48}4=778,320$ possible hands. Exactly 2 aces are drawn in $\binom42\binom{48}3=103,776$ hands. And so on. This gives a total of

$\displaystyle\sum_{a=1}^4\binom4a\binom{48}{5-a}=886,656$

possible hands containing at least 1 ace, and hence B occurs with probability

$\dfrac{\sum\limits_{a=1}^4\binom4a\binom{48}{5-a}}{\binom{52}5}=\dfrac{18,472}{54,145}\approx0.3412$

The product of these probability is approximately 0.000082.

A and B are independent if the probability of both events occurring simultaneously is the same as the above probability, i.e. $P(A\cap B)=P(A)P(B)$ . This happens if

the hand has 4 aces and 1 non-ace, or
the hand has a non-ace 4-of-a-kind and 1 ace

The above "sub-events" are mutually exclusive and share no overlap. There are 48 possible non-aces to choose from, so the first sub-event consists of 48 possible hands. There are 12 non-ace 4-of-a-kinds and 4 choices of ace for the fifth card, so the second sub-event has a total of 12*4 = 48 possible hands. So $A\cap B$ consists of 96 possible hands, which occurs with probability