Please help answer would be appreciated

Find the amount of interest payable if $400 is borrowed at 14% interest and is to be repaid in 6 months

Shtirlitz [24]

Answer:

The amount of interest payable after 6 months is $28.

Step-by-step explanation:

Here, the Principal amount borrowed = $400

Rate of Interest = 14%

Time = 6 months = 6 / 12 years = 0.5 year

Now, SIMPLE INTEREST = $\frac{P \times R \times T}{100}$

So, here SI = $SI = \frac{400 \times 14 \times 0.5}{100} = 28$

or, SI = $28

Hence, the amount of interest payable after 6 months is $28.

4 0

3 years ago

What is the mean,median,mode(s) an range if the numbers are 15,30,17,15,24,31,15

Phoenix [80]

The mean is 21. The median is 15. The mode is 15. The range is 16.

7 0

3 years ago

Read 2 more answers

At a school trivia competition, contestants can answer two kinds of questions: easy questions and hard

lorasvet [3.4K]

Answer:

B.

Step-by-step explanation:

so 8*5 is 40 which leaves 39 points to be accounted for.

39/3 is 13 (easy questions)

8+13=21 which is the amount of questions Chantrea answered correctly so yee

5 0

3 years ago

Multiply and place the decimal point in the product of each problem -0.7 x 0.2

Elina [12.6K]

Your answer would be -0.14

8 0

2 years ago

Read 2 more answers

A union of restaurant and foodservice workers would like to estimate the mean hourly wage, μ, of foodservice workers in the U.S.

pav-90 [236]

Answer:

$n=(\frac{1.960(2.25)}{0.35})^2 =158.76 \approx 159$

So the answer for this case would be n=159 rounded up to the nearest integer

Step-by-step explanation:

Assuming this complete question: A union of restaurant and foodservice workers would like to estimate the mean hourly wage, , of foodservice workers in the U.S. The union will choose a random sample of wages and then estimate using the mean of the sample. What is the minimum sample size needed in order for the union to be 95% confident that its estimate is within $0.35 of ? Suppose that the standard deviation of wages of foodservice workers in the U.S. is about $2.15 .

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)

$\sigma=2.25$ represent the population standard deviation

n represent the sample size

Solution to the problem

Since the Confidence is 0.95 or 95%, the value of $\alpha=0.05$ The margin of error is given by this formula:

$ME=z_{\alpha/2}\frac{s}{\sqrt{n}}$ (a)

And on this case we have that ME =0.35 and we are interested in order to find the value of n, if we solve n from equation (a) we got:

$n=(\frac{z_{\alpha/2} \sigma}{ME})^2$ (b)

The critical value for 95% of confidence interval now can be founded using the normal distribution. And in excel we can use this formla to find it:"=-NORM.INV(0.025;0;1)", and we got $z_{\alpha/2}=1.960$ , replacing into formula (b) we got:

$n=(\frac{1.960(2.25)}{0.35})^2 =158.76 \approx 159$

So the answer for this case would be n=159 rounded up to the nearest integer

3 0

2 years ago