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Sidana [21]
3 years ago
5

Billy takes out a $2400 discounted loan using a simple interest rate of 8% for a period of 18 months. What is the effective inte

rest rate?
Mathematics
2 answers:
Nitella [24]3 years ago
7 0

Answer:

<h2>$3456</h2>

Step-by-step explanation:

($2400 x 8)÷100=$192

$192 x 18=$3456

kaheart [24]3 years ago
5 0

Answer:

9.1

Step-by-step explanation:

To calculate the total interest payable, we use the formula

I=P0rt,

and substituting our values yields

I=$2,400×0.08×1812=$288.

Therefore the total amount he receives at loan drawdown is $2,400−$288=$2,112. To calculate the effective interest rate, we use to formula

A=P0(1+ret).

In this instance we have A=$2,400,P0=$2,112 (redefined from the value above) and t=1.5. We substitute into the formula to get

$2,400=$2,112(1+1.5re).

Solve for re.

24002112=1+1.5re

24002112−1=1.5re

24002112−11.5=re

This gives re=0.0909⋯=9.0909…%,  which is 9.1%

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A data set includes 103 body temperatures of healthy adult humans having a mean of 98.9degreesf and a standard deviation of 0.67
trasher [3.6K]

Answer:

Step-by-step explanation:

Confidence interval is written in the form,

(Sample mean - margin of error, sample mean + margin of error)

The sample mean, x is the point estimate for the population mean.

Margin of error = z × s/√n

Where

s = sample standard deviation = 0.67

n = number of samples = 103

From the information given, the population standard deviation is unknown hence, we would use the t distribution to find the z score

In order to use the t distribution, we would determine the degree of freedom, df for the sample.

df = n - 1 = 103 - 1 = 102

Since confidence level = 99% = 0.99, α = 1 - CL = 1 – 0.99 = 0.01

α/2 = 0.01/2 = 0.005

the area to the right of z0.005 is 0.005 and the area to the left of z0.005 is 1 - 0.005 = 0.995

Looking at the t distribution table,

z = 2.6249

Margin of error = 2.6249 × 0.67/√103

= 0.173

Confidence interval = 98.6 ± 0.173

This suggests that the mean body temperature could very possibly be

98.6degrees°F.

3 0
4 years ago
Can someone please explain to me how to solve these types of equations? I'm supposed to solve for x in 4^(-5x-6)=4^(9x+4)
Ksenya-84 [330]

Answer:-5/7 i think that is wrong

Step-by-step explanation:

8 0
3 years ago
Average precipitation for the first 7 months of the year, the average precipitation in toledo, ohio, is 19.32 inches. if the ave
Colt1911 [192]
Part A:

The probability that a normally distributed data with a mean, μ and standard deviation, σ is greater than a given value, a is given by:

P(x\ \textgreater \ a)=1-P(x\ \textless \ a)=1-P\left(z\ \textless \  \frac{a-\mu}{\sigma}\right)

Given that the average precipitation in Toledo, Ohio for the past 7 months is 19.32 inches with a standard deviation of 2.44 inches, the probability that <span>a randomly selected year will have precipitation greater than 18 inches for the first 7 months is given by:

P(x\ \textgreater \ 18)=1-P(x\ \textless \ 18) \\  \\ =1-P\left(z\ \textless \ \frac{18-19.32}{2.44}\right) \\  \\ =1-P(z\ \textless \ -0.5410) \\  \\ =1-0.29426=\bold{0.7057}



Part B:

</span>The probability that an n randomly selected samples of a normally distributed data with a mean, μ and standard deviation, σ is greater than a given value, a is given by:

P(x\ \textgreater \ a)=1-P(x\ \textless \ a)=1-P\left(z\ \textless \ \frac{a-\mu}{\frac{\sigma}{\sqrt{n}}}\right)

Given that the average precipitation in Toledo, Ohio for the past 7 months is 19.32 inches with a standard deviation of 2.44 inches, the probability that <span>5 randomly selected years will have precipitation greater than 18 inches for the first 7 months is given by:

</span>P(x\ \textgreater \ 18)=1-P(x\ \textless \ 18) \\ \\ =1-P\left(z\ \textless \ \frac{18-19.32}{\frac{2.44}{\sqrt{5}}}\right) \\ \\ =1-P(z\ \textless \ -1.210) \\ \\ =1-0.1132=\bold{0.8868}
7 0
3 years ago
Exponential form of 16807
mylen [45]
<span>It is: 7 to the power of 5 = 16807</span>
8 0
4 years ago
Read 2 more answers
Can someone help me with this question please.
suter [353]

Answer:

Between 3 and 4 hours

Step-by-step explanation:

Based on the graph when it went from 3 to 4 hours no snow had fallen at all.

6 0
3 years ago
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