The starting balance is $42.51 please help me with math

Margaret deposits $1,000 in a savings account that pays 1.4% interest compounded

ozzi

Answer:

The balance after 1 year is;

$1,014.05

Step-by-step explanation:

To do this, we use the compound interest formula

That will be ;

A =P (1 + r/n)^nt

A is the amount generated which we want to calculate

r is the rate = 1.4% = 0.014

P is the amount deposited = $1,000

n is the number of times it is compounded annually which is 2 (semi-annually means 2 times in a year)

this the number of years which is 1

we have this as:

A = 1,000( 1 + 0.014/2)^(2*1)

A = 1,000(1 + 0.007)^2

A = 1,000(1.007)^2

A = $1,014.05

8 0

3 years ago

Prachi, the office manager at a private practice, identified several quantitative relationships surrounding a patient's visit an

Vitek1552 [10]

Answer:

A

Step-by-step explanation:

W(P(t)) represents a patients's waiting time W() as function of the time of day P(t)

5 0

3 years ago

What is 2x+y=3 and x+y=3<br> 1.{-1.5,0}<br> 2.{0,-1.5]<br> 3.{0,3}<br> 4.{3,0}

Yakvenalex [24]

The answer would be 3. {0,3}.

Remember: {x,y}

2(0)+3=3
0+3=3

0+3=3

4 0

3 years ago

Suppose that an airline uses a seat width of 16.5 in. Assume men have hip breadths that are normally distributed with a mean of

Alexxx [7]

Answer:

a) 0.018

b) 0

Step-by-step explanation:

We are given the following information in the question:

Mean, μ = 14.4 in

Standard Deviation, σ = 1 in

We are given that the distribution of breadths is a bell shaped distribution that is a normal distribution.

Formula:

$z_{score} = \displaystyle\frac{x-\mu}{\sigma}$

a) P(breadth will be greater than 16.5 in)

P(x > 16.5)

$P( x > 16.5) = P( z > \displaystyle\frac{16.5 - 14.4}{1}) = P(z > 2.1)$

$= 1 - P(z \leq 2.1)$

Calculation the value from standard normal z table, we have,

$P(x > 16.5) = 1 - 0.982 = 0.018 = 1.8\%$

0.018 is the probability that if an individual man is randomly selected, his hip breadth will be greater than 16.5 in.

b) P( with 123 randomly selected men, these men have a mean hip breadth greater than 16.5 in)

Formula:

$z_{score} = \displaystyle\frac{x-\mu}{\frac{\sigma}{\sqrt{n}}}$

P(x > 16.5)

$P( x > 16.5) = P( z > \displaystyle\frac{16.5-14.4}{\frac{1}{\sqrt{123}}}) = P(z > 23.29)$

$= 1 - P(z \leq 23.29)$

Calculation the value from standard normal z table, we have,

$P(x > 16.5) = 1 - 1 = 0$

There is 0 probability that 123 randomly selected men have a mean hip breadth greater than 16.5 in

4 0

3 years ago

Fibonacci posed the following problem: Suppose that rabbits live forever and that every month each pair produces a new pair whic

docker41 [41]

Answer:

Let f_n be the number of rabbit pairs at the beginning of each month. We start with one pair, that is f_1 = 1. After one month the rabbits still do not produce a new pair, which means f_2 = 1. After two months a new born pair appears, that is f_3 = 2, and so on. Let now n $\geq$ 3 be any natural number. We have that f_n is equal to the previous amount of pairs f_n-1 plus the amount of new born pairs. The last amount is f_n-2, since any two month younger pair produced its first baby pair. Finally we have

f_1 = f_2 = 1,f_n = f_n-1 + f_n-2 for any natural n $\geq$ 3.

5 0

3 years ago