An investment of $2000 in a bank account doubles every five years. The function that models the growth of this investment is f(x)=2000•2^x, where x is the number of doubling periods, or 10 years?

Answers

Answer 1
Answer: x=1 after the first 5 years;
x=2 after the another 5 years (or better said after the first 10 years in the bank);
x=3 after another 5 years ( or better said after the first 15 years in the bank)
and so on...

x is the number of 5 years intervals which have passed up to a moment in time :)
Answer 2
Answer:

Answer:

4150

Step-by-step explanation:


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What is the area of the trapezoid?

Answers

I hope this helps you

1.) What is the equation given a slope of ½ and a y-intercept of 4?2.) What is the slope of an equation y=2x-5

3.) What is the y-intercept of an equation y=7x+8?​

Answers

1) y=1/2x+4
2) slope is 2
3) 8

Find out the expected value and the standard deviation of the number of aces obtained in 60 rolls of a fair 6-face die. Ditto for 600 rolls. Use them to explain why the observed count of aces obtained is more likely to be within 2 from the expected value with 60 rolls than with 600 rolls.

Answers

Answer:

  • E(X) = 60*1/6 = 10
  • sd(X) = √8.666 = 2.886
  • E(Y) = 600*1/6 = 100
  • sd(Y) = √86.666 = 9.1287

Step-by-step explanation:

Lets call X the amount of aces obtained in 60 rolls, and Y the amount of aces obtained in 600 rolls.

Note that both X and Y are obtained from counting the amount of successful tries from repetitions of independent experiments that have 1/6 of probability of success. Thus, both X and Y are random variables with binomial distribution, with n = 60 and 600 respectively and probability 1/6.

Remember that if Z is a random variable, Z ≈ Bi(n,p), then

  • E(Z) = np, where E(Z) denotes the expected value of Z
  • V(Z) = np(1-p), where V(Z) denotes the variance of Z. Hence, the standard deviation is the square root of V(Z), √(np(1-p)).

As a result

  • E(X) = 60*1/6 = 10
  • V(X) = 10*(1-1/6) = 50/6 ≅ 8.666
  • sd(X) = √8.666 = 2.886
  • E(Y) = 600*1/6 = 100
  • V(Y) = 100*(1-1/6) = 500/6 ≅ 86.666
  • sd(Y) = √86.666 = 9.1287

The observed amount of aces is more likely to be closer from the expected value with 60 rolls because, since we have less rolls, it is more difficult to obtain spread results.

You can also notice that X and Y can be obtained by summing independent variables with distribution BI(1,p) (also called Bernoulli(p) ). When you sum independent variables with the same distribution you have this property:

  • E(r1+r2+...+rn) = n*E(r1)
  • V(r1+r2+...+rn) = n*V(r1)
  • sd(r1+r2+...+rn) = √n*sd(r1)

X can be obtained by summing 60 independent variables r1, ...., r60 with mean 1/6 and variance 1/6*(5/6) = 5/36. So we obtain that V(X) = 60*5/36, and sd(X) = √60 * √(5/36). While for the same argument sd(Y) = √600*√(5/36). The higher the number of rolls, the more spread the results are.

I hope this helped you!

The expected number of aces from 60 rolls of a fair die is 10 with a standard deviation of approximately 3.72. For 600 rolls, the expected number is 100 with a standard deviation of about 11.79. The observed count of aces is more likely to be closer to the expected value with fewer rolls due to the smaller standard deviation relative to the number of trials.

The expected value for the number of aces in a fair die roll is computed by multiplying the probability of rolling an ace ((1)/(6)) by the number of rolls. For 60 rolls, the expected number is 60 * ((1)/(6)) = 10 aces. For 600 rolls, the expected number is 600 * ((1)/(6)) = 100 aces

The standard deviation for the number of aces is calculated using the formula for the standard deviation of a binomial distribution, \sqrt(n* p* (1-p)), where n is the number of trials, p the probability of success (((1)/(6)) for an ace). For 60 rolls, it is \sqrt(60* ((1)/(6))* ((5)/(6))) \approx 3.72. For 600 rolls, it's \sqrt(60* ((1)/(6))* ((5)/(6))) \approx 11.79.

When you roll the die 60 times, the chances of the observed count of aces being within 2 from the expected value (10) is higher because the standard deviation is smaller relative to the number of trials than when you roll the die 600 times.

As the number of trials increases, the expected standard deviation grows larger, and the observed count is more likely to be within a wider range from the expected value (100).

Learn more about standard deviation here:

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Maggie spent $4.05 on cheese and fruit at the farmers market. she bought 1/8 pound of apples, 1/4 pound of pears, and 1.25 pounds of bananas. If fruit cost $0.80 per pound, how much did Maggie spend on cheese?

Answers

Answer: $2.75

Step-by-step explanation: Maggie spent $4.05 on cheese and fruit at the farmer's market. She bought 1/8 pound of apples, 1/4 pound of pears, and 1.25 pounds of bananas. We need to calculate how much she spent on cheese.

To solve this, we need to first calculate the total cost of the fruit. Since the fruit is weighed in fractions of pounds, we need to convert them to a similar form that allows us to add them together.

Converting the fractions to decimals:

1/8 = 0.125

1/4 = 0.25

1.25 = 1.25

The cost of the fruit can be calculated as follows:

(Total pounds of fruit bought) x (Cost per pound) = (0.125 + 0.25 + 1.25) x 0.8 = 1.625 x 0.8 = $1.30

To find the amount spent on cheese, we subtract the amount spent on fruit from the total amount spent.

Total amount spent - Amount spent on fruit = $4.05 - $1.30 = $2.75

Therefore, Maggie spent $2.75 on cheese.

Algebra is a precursor for calculus because it will help students to be successful in the latter course.Which word is closest in meaning to the underlined word? precursor is the underlined word

a.complex training

b.preceding requirement

c.brief introduction

d.predominant trait

Answers

I believe it is B, a preceding requirement. 

F(x) = 3x - 3
g(x) = 4x + 5

find (f+g) (3)

Answers

Answer:

23

Step-by-step explanation:

(f + g)(3) = f(3) + g(3)

f(3) = 3(3) - 3 = 9 - 3 = 6

g(3) = 4(3) + 5 = 12 + 5 = 17

Then

(f + g)(3) = 6 + 17 = 23