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Deffense [45]
3 years ago
15

Find the annual rate of interest. Principal = 4600 rupees, Period = 5 years, Total amount = 6440 rupees, Annual rate of interest

= __%
Mathematics
1 answer:
Len [333]3 years ago
6 0

The annual rate of interest per year is 8%

<u>Solution:</u>

Given:- Principal (p) = 4600 rupees, Time –Period (t) = 5 years, Total amount(A) = 6440 rupees

First we will calculate the Interest and then using formula of simple interest we will calculate the rate of interest

Interest = Amount – Principal

Interest = 6440  – 4600 = 1840

Now using the formula of simple Interest and on putting values we get,

\text {Simple Interest }=\frac{P \times R \times T}{100}

Where "P" is the principal and "R" is the rate of interest per annum and "T" is the time period

1840=\frac{4600 \times R \times 5}{100}

\begin{aligned} \mathrm{R} &=\frac{1840 \times 100}{4600 \times 5} \\\\ \mathrm{R} &=\frac{1840}{230} \\\\ \mathrm{R} &=8 \% \end{aligned}

Hence, the required rate of interest per year is 8%

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Answer:

The hypothesis statements are: H0: p = 0.88 versus HA : p < 0.88

The p-value of the test is 0.2483 > 0.1, which means that the data does not provide sufficient evidence to conclude that the proportion of times that luggage is returned within 24 hours is less than 0.88.

Step-by-step explanation:

Test if the proportion of times that luggage is returned within 24 hours is less than 0. 88

At the null hypothesis, we test if the proportion is of 0.88, that is:

H_0: p = 0.88

At the alternate hypothesis, we test if this proportion is less than 0.88, that is:

H_a: p < 0.88

The hypothesis statements are: H0: p = 0.88 versus HA : p < 0.88

The test statistic is:

z = \frac{X - \mu}{\frac{\sigma}{\sqrt{n}}}

In which X is the sample mean, \mu is the value tested at the null hypothesis, \sigma is the standard deviation and n is the size of the sample.

0.88 is tested at the null hypothesis:

This means that \mu = 0.88, \sigma = \sqrt{0.88*0.12}

A consumer group who surveyed a large number of air travelers found that 138 out of 160 people who lost luggage on that airline were reunited with the missing items by the next day.

This means that n = 160, X = \frac{138}{160} = 0.8625

Test statistic:

z = \frac{X - \mu}{\frac{\sigma}{\sqrt{n}}}

z = \frac{0.8625 - 0.88}{\frac{\sqrt{0.88*0.12}}{\sqrt{160}}}

z = -0.68

P-value of the test:

The p-value of the test is the probability of finding a sample proportion below 0.8625, which is the p-value of z = -0.68.

Looking at the z-table, the p-value of z = -0.68 is of 0.2483.

The p-value of the test is 0.2483 > 0.1, which means that the data does not provide sufficient evidence to conclude that the proportion of times that luggage is returned within 24 hours is less than 0.88.

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