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KIM [24]
4 years ago
13

1) Suppose you put money into two different bank accounts. In account #1 you deposit $500 and you will be earning 6% interest co

mpounded quarterly. In account #2 you deposit $600 and you will be earning 5% interest compounded annually. Which statement below best describes the relationship between the amount in Account #1 and Account #2 after 10 years have passed? Assume that during these years you do not withdraw any money.
a) Account #1 will have approximately $397 less than Account #2
b) Account #1 will have approximately $82 more than Account #2
c) Account #1 will have approximately $70 less than Account #2
d) Account #1 and Account #2 will have approximately the same amount of money in them.

2) In a single-elimination tournament, half of the remaining teams are eliminated in each round. If the tournament starts with 128 teams, how many teams will be left after 5 rounds?
a) 16 teams
b) 12 teams
c) 8 teams
d) 4 teams
Thank you!
Mathematics
2 answers:
Elis [28]4 years ago
6 0
The answer for number 1 is D they will have the same amount of money.

Hoochie [10]4 years ago
3 0

To question 1:

The answer will be <em>C</em>.

  • I got this from the comment section on the other answer posted here.

To question 2:

The answer is <em>D</em>.

  • I figured this out by putting it into a table on the website Desmos.

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The total resistance R of two resistors connected in parallel circuit is given by 1/R = 1/R_1 + 1/R_2. Approximate the change in
Naily [24]

Answer:

a) Approximate the change in R is 0.5 ohm.

b) The actual change in R is 0.04 ohm.

Step-by-step explanation:

Given : The total resistance R of two resistors connected in parallel circuit is given by \frac{1}{R}=\frac{1}{R_1}+\frac{1}{R_2}

To find :

a) Approximate the change in R ?

b) Compute the actual change.

Solution :

a) Approximate the change in R

R_1=12\ ohm and R_2=10\ ohm

R_1 is decreased from 12 ohms to 11 ohms.

i.e. \triangle R_1=21-11=1\ ohm

R_2 is increased from 10 ohms to 11 ohms.

i.e. \triangle R_2=11-10=1\ ohm

The change in R is given by,

\frac{1}{\triangle R}=\frac{1}{\triangle R_1}+\frac{1}{\triangle R_2}

\frac{1}{\triangle R}=\frac{\triangle R_2+\triangle R_1}{(\triangle R_1)(\triangle R_2)}

\triangle R=\frac{(\triangle R_1)(\triangle R_2)}{\triangle R_2+\triangle R_1}

\triangle R=\frac{(1)(1)}{1+1}

\triangle R=\frac{1}{2}

\triangle R=0.5\ ohm

b) The actual change in Resistance

\frac{1}{R}=\frac{1}{R_1}+\frac{1}{R_2}

\frac{1}{R}=\frac{1}{12}+\frac{1}{10}

R=\frac{10\times 12}{10+12}

R=\frac{120}{22}

R=5.46\ ohm

When resistances are charged, R_1=R_2=11

\frac{1}{R'}=\frac{1}{R_1}+\frac{1}{R_2}

\frac{1}{R'}=\frac{1}{11}+\frac{1}{11}

R'=\frac{11}{2}

R'=5.5\ ohm

Change in resistance is given by,

C=R'-R

C=5.5-5.46

C=0.04\ ohm

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