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3 years ago

Find the sum that amounts to Rs 9,144 in 3 years at 9% per year simple interest.

Mathematics

1 answer:

Hunter-Best [27]3 years ago

8 0

Given:

Amount = Rs. 9,144

Time = 3 years.

Rate of simple interest = 9%

To find:

The principal value.

Solution:

The formula for simple interest is:

$I=\dfrac{P\times r\times t}{100}$

Where, P is principal, r is the simple rate of interest, and t is the number of years.

Putting $r=9,t=3$ in the above formula, we get

$I=\dfrac{P\times 9\times 3}{100}$

$I=\dfrac{27P}{100}$

$I=0.27P$

We know that,

$\text{Amount}=\text{Principal + Interest}$

$9144=P+0.27P$

$9144=1.27P$

Divide both sides by 1.27, we get

$\dfrac{9144}{1.27}=P$

$7200=P$

Therefore, the principal value is Rs. 7200.

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Answer:

a) Null hypothesis: $\mu \leq 15$

Alternative hypothesis: $\mu > 15$

b) $t=\frac{17-15}{\frac{4}{\sqrt{35}}}=2.958$

c) $p_v =P(t_{34}>2.958)=0.0028$

d) If we compare the p value and the significance level given $\alpha=0.01$ we see that $p_v$ so we can conclude that we have enough evidence to reject the null hypothesis, so we can conclude that the true mean is higher than 15 at 1% of signficance.

Step-by-step explanation:

Data given and notation

The sample mean is given by:

$\bar X = \frac{\sum_{i=1}^n X_i}{n}$

And the sample deviation is:

$s = \sqrt{\frac{\sum_{i=1}^n (X_i -\bar X)^2}{n-1}}$

$\bar X=17$ represent the sample mean

$s=4$ represent the sample standard deviation

$n=35$ sample size

$\mu_o =15$ represent the value that we want to test

$\alpha=0.01$ represent the significance level for the hypothesis test.

z would represent the statistic (variable of interest)

$p_v$ represent the p value for the test (variable of interest)

Part a State the null and alternative hypotheses.

We need to conduct a hypothesis in order to check if the mean is less or equal than 15, the system of hypothesis would be:

Null hypothesis: $\mu \leq 15$

Alternative hypothesis: $\mu > 15$

Since we don't know the population deviation, is better apply a t test to compare the actual mean to the reference value, and the statistic is given by:

$t=\frac{\bar X-\mu_o}{\frac{s}{\sqrt{n}}}$ (1)

t-test: "Is used to compare group means. Is one of the most common tests and is used to determine if the mean is (higher, less or not equal) to an specified value".

Part b Calculate the statistic

We can replace in formula (1) the info given like this:

$t=\frac{17-15}{\frac{4}{\sqrt{35}}}=2.958$

Part c P-value

The degrees of freedom are given by:

$df = n-1= 35-1=34$

Since is a right tailed test the p value would be:

$p_v =P(t_{34}>2.958)=0.0028$

Part d Conclusion

If we compare the p value and the significance level given $\alpha=0.01$ we see that $p_v$ so we can conclude that we have enough evidence to reject the null hypothesis, so we can conclude that the true mean is higher than 15 at 1% of signficance.

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A group of teachers and students will take a bus trip to a museum, and each teacher will lead a group of no more than 4 students

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Answer:

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Step-by-step explanation:

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You can work this out by rearranging the equation and isolating the x.

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3 years ago

Find the sum that amounts to Rs 9,144 in 3 years at 9% per year simple interest.​

Find the sum that amounts to Rs 9,144 in 3 years at 9% per year simple interest.