Jim invested $4,000 in a bond at a yearly rate of 4.5%. He earned $540 in interest.How long was the money invested?

A grandsons age in years is equal to his grandfather's age in years. The difference of the grandsons age in years and the grandf

garik1379 [7]

The age of the grandson is 7 years and the grandfather is 84 years if the grandson's age in months is equal to the age of grandpa in years and the difference in age is 77.

<h3>What is a linear equation?</h3>

It is defined as the relation between two variables, if we plot the graph of the linear equation we will get a straight line.

If in the linear equation, one variable is present, then the equation is known as the linear equation in one variable.

Let's suppose the age of the grandsons is x years and

age of the grandfather's age is y years

Here some data are misprinted, so we are assuming the grandson's age in months is equal to the age of grandpa in years and the difference in age is 77.

Then according to the problem:

12x = y (as in 1 year, there are 12 months)

y -x = 77 (as difference in age is 77)

After solving both equations, we get:

x = 7 years and y = 84 years

Thus, the age of the grandson is 7 years and the grandfather is 84 years if the grandson's age in months is equal to the age of grandpa in years and the difference in age is 77.

Learn more about the linear equation here:

brainly.com/question/11897796

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4 0

2 years ago

A coin, having probability p of landing heads, is continually flipped until at least one head and one tail have been flipped. (a

Natali [406]

Answer:

(a)

The probability that you stop at the fifth flip would be

$p^4 (1-p) + (1-p)^4 p$

(b)

The expected numbers of flips needed would be

$\sum\limits_{n=1}^{\infty} n p(1-p)^{n-1} = 1/p$

Therefore, suppose that $p = 0.5$ , then the expected number of flips needed would be 1/0.5 = 2.

Step-by-step explanation:

(a)

Case 1

Imagine that you throw your coin and you get only heads, then you would stop when you get the first tail. So the probability that you stop at the fifth flip would be

$p^4 (1-p)$

Case 2

Imagine that you throw your coin and you get only tails, then you would stop when you get the first head. So the probability that you stop at the fifth flip would be

$(1-p)^4p$

Therefore the probability that you stop at the fifth flip would be

$p^4 (1-p) + (1-p)^4 p$

(b)

The expected numbers of flips needed would be

$\sum\limits_{n=1}^{\infty} n p(1-p)^{n-1} = 1/p$

Therefore, suppose that $p = 0.5$ , then the expected number of flips needed would be 1/0.5 = 2.

7 0

2 years ago

Lance used two gallons of paint to pain 785 square feet. How many gallons will he need to purchase to cover 2,500 square feet?

Digiron [165]

Answer:

7 gallon

Step-by-step explanation:

Divide the space that needs paints by how much space has been painted

2500/785 = 3.18

Then double that because it takes 2 gallons to paint every 785 square feet

3.18 x 2 = 6.36

You would need 6.36 gallons of paint, which would round up to 7 because you cannot buy 0.36 of a gallon

8 0

2 years ago

Find the sum of the first 18 terms of the arithmetic sequence whose nth term is an = 3n - 1.

vodka [1.7K]

Answer:

495

Step-by-step explanation:

In order to find the sum of the first 18 terms you have to find the 18th term and the first term using the equation given.

a1=3(1)-1 a1=2

a18=3(18)-1 a18= 53

Then plug in 53 for an, 18 for n, and 2 in for a1 in the sum equation: Sn=n/2(a1+an)

Sn=18/2(2+53) Solve for sn= 495

7 0

3 years ago

Adrian's CD player can hold six disks at a time and shuffles all of the albums and their songs. If he has thirteen CD's, how man

Crank

Adrian can put a total of 1716 combinations in the player.

The formula to be used is as follows:

total combinations = n!/[r! *(n-r)!]
n = number of disks available = 13
r = number of disks that be held = 6

= 13! = (13• 12•11•10•9•8•7•6•5•4•3•2•1)/(6! <span>• 7!)
</span>=1716

Thank you for posting your question. Feel free to ask me more.

6 0

3 years ago