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zysi [14]
3 years ago
15

Ms. Young acquired a loan of $40,000 at her local bank. The loan has a simple interest rate of 6% per year. What is the amount o

f interest that Ms. Young will be charged on the loan at the end of 5 years.
Mathematics
1 answer:
soldier1979 [14.2K]3 years ago
8 0

Answer:

12.000

Step-by-step explanation:

40000x6=  240000/100=  2400x5

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A line with points (-4.0) and (-3.1)<br> has a slope of?
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Slope is the change in y over the change in x

Slope = (1-0) /( -3 - -4)

Slope = 1/1

Slope = 1

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Can y’all help me on question 23?!
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Answer:

Option D. 5x + 12y + 4z is the corrct one.

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Keshawn is asked to compare and contrast the domain and range for the two functions.
scoray [572]
For the function 5x, the range is the value of f(x) which is five times the x and the domain is the value of x which can be the ratio of f(x) and 5. For the second function which is g(x) = 5, the range is 5 all through out the graph while the domain is infinity.
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3 years ago
What is the simplet form of 7<br> —<br> 12?
inysia [295]

Answer:

-5

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6 0
3 years ago
Suppose that the sample standard deviation was s = 5.1. Compute a 98% confidence interval for μ, the mean time spent volunteerin
NISA [10]

Answer:

The 95% confidence interval would be given by (5.139;5.861)  

Step-by-step explanation:

1) Previous concepts

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".

The margin of error is the range of values below and above the sample statistic in a confidence interval.

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

\bar X represent the sample mean for the sample  

\mu population mean (variable of interest)

s represent the sample standard deviation

n represent the sample size  

2) Confidence interval

Assuming that \bar X =5.5 and the ranfom sample n=1086.

The confidence interval for the mean is given by the following formula:

\bar X \pm t_{\alpha/2}\frac{s}{\sqrt{n}}   (1)

In order to calculate the critical value t_{\alpha/2} we need to find first the degrees of freedom, given by:

df=n-1=1086-1=1085

Since the Confidence is 0.98 or 98%, the value of \alpha=0.02 and \alpha/2 =0.01, and we can use excel, a calculator or a table to find the critical value. The excel command would be: "=-T.INV(0.01,1085)".And we see that t_{\alpha/2}=2.33

Now we have everything in order to replace into formula (1):

5.5-2.33\frac{5.1}{\sqrt{1086}}=5.139    

5.5+2.333\frac{5.1}{\sqrt{1086}}=5.861

So on this case the 95% confidence interval would be given by (5.139;5.861)    We are 98% confident that the mean time spent volunteering for the population of parents of school-aged children is between these two values.

5 0
3 years ago
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