This is an exponential decay problem.
Using the equation Y = a *(1-rate)^time
where Y is the future value given as 12,000 and a is the starting value given as 13,000.
The rate is also given as 5%.
The equation becomes:
12,000 = 13,000(1-0.05)^x
12,000 = 13,000(0.95)^x
Divide each side by 13000:
12000/13000 = 0.95^x/13000
12/13 = 0.95^x
Use the natural log function:
x = ln(12/13) / ln(0.95)
x = 1.56 years. ( this will equal 12,000
Round to 2 years it will be less than 12000.
9 or -9 ...if I’m wrong sorry
Answer: 5 minutes.
Step-by-step explanation:
120 words = 1 min
600 words = 5 mins
(divide the 120 by 600 and you’ll end up getting 5 minutes)
1. Y= 5/4x -1.4
2.Y=-2/4x -7
Formula is Y=MX+B
To solve these equations you have to make the Y separate so add or subtract the x value depending on the charge remember what you do to one side you do to the other after you divide everything by the number in the Y value and for the x you put the y value on-top and make it a fraction
Answer:
0.3 years
Step-by-step explanation:
With problems like these, I always like to start by breaking down the information into smaller pieces.
μ = 13.6
σ = 3.0
Survey of 100 self-employed people
(random variable) X = # of years of education
So now we have some notation, where μ represents population mean and σ represents population standard deviation. Hopefully, you already know that the sample mean of x-bar is the same as the population mean, so x-bar = 13.6. Now, the question asks us what the standard deviation is. Since the sample here is random, we can use the Central Limit Theorem, which allows us to guess that a distribution will be approximately normal for large sample sizes (that is, n ≥ 30). In this case, our sample size is 100, so that is satisfied. We're also told our sample is random, so we're good there, too. Now all we have to do is plug some stuff in.
The Central Limit Theorem says that for large values of n, x-bar follows an approximately normal distribution with sample mean = μ and sample standard deviation = σ/√n. So, with that info, all we need to do to find the standard deviation of x-bar is to plug our σ and n into the above formula.
σ(x-bar) = σ/√n
σ(x-bar) = 3.0/√100
σ(x-bar) = 0.3
So your answer here is .3 years.