Ask question

Ask question

All categories

3 years ago

13

John Brought 75 shares of HCC stock and 150 Shares

1 answer:

mestny [16]3 years ago

5 0

You have to take $285.00 times 75 (total number of shares) plus 150 (2nd total number of shares) multiplied by $2.00 . Your total should come out to $21,375.00

You might be interested in

Solve -7/2=-28 please solve and show work

Assoli18 [71]

Step-by-step explanation:

-7/2 = -28

-7 = -28 × 2

-7 = -56

5 0

3 years ago

The population of a certain town was 10,000 in 1990. The rate of change of the population, measured in people per year, is model

Katena32 [7]

The question is incomplete. The complete question is :

The population of a certain town was 10,000 in 1990. The rate of change of a population, measured in hundreds of people per year, is modeled by P prime of t equals two-hundred times e to the 0.02t power, where t is measured in years since 1990. Discuss the meaning of the integral from zero to twenty of P prime of t, d t. Calculate the change in population between 1995 and 2000. Do we have enough information to calculate the population in 2020? If so, what is the population in 2020?

Solution :

According to the question,

The rate of change of population is given as :

$\frac{dP(t)}{dt}=200e^{0.02t}$ in 1990.

Now integrating,

$\int_0^{20}\frac{dP(t)}{dt}dt=\int_0^{20}200e^{0.02t} \ dt$

$=\frac{200}{0.02}\left[e^{0.02(20)}-1\right]$

$=10,000[e^{0.4}-1]$

$=10,000[0.49]$

=4900

$\frac{dP(t)}{dt}=200e^{0.02t}$

$\int1.dP(t)=200e^{0.02t}dt$

$P=\frac{200}{0.02}e^{0.02t}$

$P=10,000e^{0.02t}$

$P=P_0e^{kt}$

This is initial population.

k is change in population.

So in 1995,

$P=P_0e^{kt}$

$=10,000e^{0.02(5)}$

$=11051$

In 2000,

$P=10,000e^{0.02(10)}$

$=12,214$

Therefore, the change in the population between 1995 and 2000 = 1,163.

8 0

3 years ago

Bannas cost $1.00 for 3 pounds.How much for 1 pound

Leno4ka [110]

Answer:

__

0.333

Step-by-step explanation:

you divide 1.00 by 3. The 3s never end.

8 0

3 years ago

Read 2 more answers

John is buying food for his cat, Fluffy. Fluffy LOVES a fancy wet food which costs $1.50 per day. Fluffy will also tolerate dry

Vedmedyk [2.9K]

Answer:

The answer is below

Step-by-step explanation:

Let x represent the number of days that fluffy eats wet food in a week and y represents the number of days that fluffy eats dry food in a week.

Hence:

x + y = 7 (1)

Also, John wants to spend at most $9.00 on cat food each week. Hence:

1.5x + 0.75y ≤ 9 (2)

The list of possible points after solving graphically are:

(0,7), (6,0), (0,12) and (5, 2). If x,y > 0, then the point that satisfies the inequality is:

(5, 2) i.e. 5 wet food and 2 dry food

5 0

3 years ago

Two number cubes are rolled. What is the probability that the sum of the numbers rolled is either 3 or 9?

MariettaO [177]

$|\Omega|=6^2=36\\ |A|=\underbrace{2}_{\text{the sum is 3}}+\underbrace{4}_{\text{the sum is 9}}=6\\\\ P(A)=\dfrac{6}{36}=\dfrac{1}{6}\approx17\%$

7 0

4 years ago