Ask question

Ask question

All categories

4 years ago

12

Corey deposits $300 in a bank that pays 4% annual interest. How much interest does Corey earn in one year? What would the intere

st be if the rate were 6%?

1 answer:

hammer [34]4 years ago

5 0

$300/100= $3 1%=$3 $3x4=$12 $300+$12=4%$312
$3x6=$18 $300+$18=$318

You might be interested in

Olk-onm <br> Find the value of u.

damaskus [11]

Answer:

olk 69 -_-

Step-by-step explanation:

5 0

3 years ago

If it is 5 degrees outside, and the temperature will drop 12

kolezko [41]

Answer:

-7 degrees

Step-by-step explanation:

Dropping 12 degrees means that it will get colder by 12 degrees. The colder it gets, the lower the temperature. 5 degrees - the current temperature, is going to be dropped by 12 degrees. To calculate this: 5-12 = -7 (degrees)

7 0

3 years ago

suppose the line segment whose endpoints are H(5,0) and I(-6,-3) is reflected over the y axis? what are the coordinates of H’I’?

padilas [110]

Answer:

see explanation

Step-by-step explanation:

Under a reflection in the y- axis

a point (x, y ) → (- x, y ), thus

H(5, 0 ) → H'(- 5, 0 )

I(- 6, - 3 ) → I'(6, - 3 )

7 0

3 years ago

Using proportional reasoning, about how many of 180 seventh grade students in school would like to snorkel

amm1812

Answer:

We need some aditional data. For example, the average number of students who like to snorkel in certain population.

Step-by-step explanation:

3 0

3 years ago

Situation:A researcher in North America discoversa fossile that contains 65% of its originalamount of C-14..-ktN=NoeNo inital am

zheka24 [161]

SOLUTION

We have been given the equation of the decay as

$\begin{gathered} N=N_0e^{-kt} \\ where\text{ } \\ N_0=initial\text{ amount of C-14 at time t} \\ N=amount\text{ of C-14 at time t = 65\% of N}_0=0.65N_0 \\ k=0.0001 \\ t=time\text{ in years = ?} \end{gathered}$

So we are looking for the time

Plugging the values into the equation, we have

$\begin{gathered} N=N_0e^{-kt} \\ 0.65N_0=N_0e^{-0.0001t} \\ e^{-0.0001t}=\frac{0.65N_0}{N_0} \\ e^{-0.0001t}=0.65 \end{gathered}$

Taking Ln of both sides, we have

$\begin{gathered} ln(e^{-0.000t})=ln(0.65) \\ -0.0001t=ln(0.65) \\ t=\frac{ln(0.65)}{-0.0001} \\ t=4307.82916 \end{gathered}$

Hence the answer is 4308 to the nearest year

8 0

1 year ago