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2 years ago

9

If $7500 is invested at an interest rate of 5% each year, what is the value of the investment in 4 years? Write an exponential f

unction and solve.

1 answer:

Vanyuwa [196]2 years ago

3 0

The answer is 9,116.3
7500*(1+0.05)^4

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The population can be modeled by P(t) = 82.5 − 67.5cos⎡ ⎣(π/6)t ⎤ ⎦, where t is time in months (t = 0 represents January 1) and

Fed [463]

Answer:

The intervals in which the population is less than 20,000 include

(0 ≤ t < 0.74) and (11.26 < t ≤ 12)

Step-by-step explanation:

P(t) = 82.5 - 67.5 cos [(π/6)t]

where

P = population in thousands.

t = time in months.

During a year, in what intervals is the population less than 20,000?

That is, during (0 ≤ t ≤ 12), when is (P < 20)

82.5 - 67.5 cos [(π/6)t] < 20

- 67.5 cos [(π/6)t] < 20 - 82.5

-67.5 cos [(π/6)t] < -62.5

Dividing both sides by (-67.5) changes the inequality sign

cos [(π/6)t] > (62.5/67.5)

Cos [(π/6)t] > 0.9259

Note: cos 22.2° = 0.9259 = cos (0.1233π) or cos 337.8° = cos (1.8767π) = 0.9259

If cos (0.1233π) = 0.9259

Cos [(π/6)t] > cos (0.1233π)

Since (cos θ) is a decreasing function, as θ increases in the first quadrant

(π/6)t < 0.1233π

(t/6) < 0.1233

t < 6×0.1233

t < 0.74 months

If cos (1.8767π) = 0.9259

Cos [(π/6)t] > cos (1.8767π)

cos θ is an increasing function, as θ increases in the 4th quadrant,

[(π/6)t] > 1.8767π (as long as (π/6)t < 2π, that is t ≤ 12)

(t/6) > 1.8767

t > 6 × 1.8767

t > 11.26

Second interval is 11.26 < t ≤ 12.

Hope this Helps!!!

3 0

3 years ago

What is the reason for each step in the solution of the equation?

iVinArrow [24]

4x-1=-2(x+1)
Distribute
4x-1=-2x-2
Move variable to left and change signs
4x+2x=-2+1
Combine like terms
6x=-1
Divide both sides by 6
X=-1/6

4 0

2 years ago

A runner jogged 5 1/4 miles in 45 minutes. At this rate how far will she run in 2 hours?

Anon25 [30]

<span>(21/2) / (3/4)

21/2 * 4/3
=14 miles </span>

5 0

3 years ago

Determine the margin of error for a 90% confidence interval to estimate the population mean when s = 40 for the sample sizes bel

Ann [662]

Answer:

a) The margin of error for a 90% confidence interval when n = 14 is 18.93.

b) The margin of error for a 90% confidence interval when n=28 is 12.88.

c) The margin of error for a 90% confidence interval when n = 45 is 10.02.

Step-by-step explanation:

The t-distribution is used to solve this question:

a) n = 14

The first step to solve this problem is finding how many degrees of freedom, we have. This is the sample size subtracted by 1. So

df = 14 - 1 = 13

90% confidence interval

Now, we have to find a value of T, which is found looking at the t table, with 13 degrees of freedom(y-axis) and a confidence level of $1 - \frac{1 - 0.9}{2} = 0.95$ . So we have T = 1.7709

The margin of error is:

$M = T\frac{s}{\sqrt{n}} = 1.7709\frac{40}{\sqrt{14}} = 18.93$

In which s is the standard deviation of the sample and n is the size of the sample.

The margin of error for a 90% confidence interval when n = 14 is 18.93.

b) n = 28

27 df, T = 1.7033

$M = T\frac{s}{\sqrt{n}} = 1.7033\frac{40}{\sqrt{28}} = 12.88$

The margin of error for a 90% confidence interval when n=28 is 12.88.

c) The margin of error for a 90% confidence interval when n = 45 is

44 df, T = 1.6802

$M = T\frac{s}{\sqrt{n}} = 1.6802\frac{40}{\sqrt{45}} = 10.02$

The margin of error for a 90% confidence interval when n = 45 is 10.02.

3 0

2 years ago

A computer can be classified as either cutting dash edge or ancient. Suppose that 86% of computers are classified as ancient.

Leno4ka [110]

Answer:

(a) The probability that both computers are ancient is 0.7396

(b) The probability that all seven computers are ancient is 0.3479

(c) The probability that at least one of seven randomly selected computers is cutting dash edge is 0.6520.

Because the probability is about 65% is it not unusual that at least one of seven randomly selected computers is cutting dash edge, it's more likely than not.

Step-by-step explanation:

We know that 86% of computers are classified as ancient. This means, if one computer is chosen at random, there is an 86% chance that it will be classified as ancient.

$P(ancient)=0.86$

(a) To find the probability that two computers are chosen at random and both are ancient you must,

The probability that the first computer is ancient is $P(ancient)=0.86$ and the probability that the second computer is ancient is $P(ancient)=0.86$

These events are independent; the selection of one computer does not affect the selection of another computer.

When calculating the probability that multiple independent events will all occur, the probabilities are multiplied, this is known as the rule of product.

Let A be the event "the first computer is ancient" and B the event "the second computer is ancient".

$P(A\:and \:B)=P(A)\cdot P(B)=0.86\cdot 0.86=0.86^2= 0.7396$

(b) To find the probability that seven computers are chosen at random and all are ancient you must,

Following the same logic in part (a) we have

Let A be the event "the first computer is ancient",

B the event "the second computer is ancient",

C the event "the third computer is ancient",

D the event "the fourth computer is ancient",

E the event "the fifth computer is ancient",

F the event "the sixth computer is ancient", and

G the event "the seventh computer is ancient"

$P(A\:and \:B\:and \:C\:and \:D\:and \:E\:and \:F\:and \:G)=\\P(A)\cdot P(B)\cdot P(C)\cdot P(D)\cdot P(E)\cdot P(F)\cdot P(G) =(0.86)^7=0.3479$

(c) To find the probability that at least one of seven randomly selected computers is cutting dash edge you must

Use the concept of complement. The complement of an event is the subset of outcomes in the sample space that are not in the event.

Let C the event "the computer is cutting dash edge".

Let A the event "the seven computers are ancient".

$P(C)=1-P(A)=1-0.3479=0.6520$

Because the probability is about 65% is it not unusual that at least one of seven randomly selected computers is cutting dash edge, it's more likely than not.

5 0

3 years ago