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3 years ago

5

Lucy's house cost £200 000 at the start of 2014, the value of the house increased every year by 5%, work out the value of her ho

use at start of 2017

1 answer:

liq [111]3 years ago

8 0

Answer:

212000

Step-by-step explanation:

120000 - 5% = 190000

hence 5% is 10000

120000+ 30000= 212000

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A jet ski depreciates at 11% of its original value each year. If the jet ski was $8,000 at its time of purchase, what is the val

dsp73

Compounded depreciation formula:

A = P(1 - r)ⁿ , where P = original price, r= rate of depreciation, n = number of years and A = actual value (after depreciation):

A= $8000(1 - 11%)⁵ = 8000(0.89)⁵ = 4,467.24 ≈$4,467

4 0

3 years ago

Read 2 more answers

Question 22 Petrolyn motor oil is a combination of natural oil and synthetic oil. It contains 8 liters of natural oil for every

Setler [38]

Answer:

897 liters

Step-by-step explanation:

The motor oil is composed of 8 liters of natural oil and 5 liters of synthetic oil. This means there is a ratio of 8/5. We have 552 liters of natural oil to y liters of synthetic. This is the ratio 552/y. Create a proportion that compares the two ratios.

$\frac{8}{5}=\frac{552}{y}$

Solve by cross multiplying.

8y = 5(552)

8y = 2,760

y= 345

345 liters of synthetic oil.

Together 552+345= 897 liters of motor oil.

4 0

3 years ago

Which shows the expression below in simplified form?

statuscvo [17]

Answer:

B. 9x10^4

Step-by-step explanation:

$\frac{4.5 \times {10}^{8} }{5 \times {10}^{3} } \\ = 0.9 \times {10}^{8 - 3} \\ = 0.9 \times {10}^{5} \\ = 9 \times {10}^{4}$

5 0

3 years ago

In the first year of ownership, a new car can lose 20% of its value. If a car lost $4200 of value in the first year, how much di

Sever21 [200]

$5250
$5250-20%=$4200

5 0

3 years ago

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Assume that a simple random sample has been selected from a normally distributed population. State the hypotheses, find the test

BaLLatris [955]

Solution

Hypotheses:

- The population mean is 132. In order to test the claim that the mean is 132, we should check for if the mean is not 132.

- Thus, the Hypotheses are:

$\begin{gathered} H_0:\mu=132 \\ H_1:\mu\ne132 \end{gathered}$

Test statistic:

- The test statistic has to be a t-statistic because the sample size (n) is less than 30.

- The formula for finding the t-statistic is:

$\begin{gathered} t=\frac{\bar{X}-\mu}{\frac{s}{\sqrt{n}}} \\ \\ where, \\ \bar{X}=\text{ Sample mean} \\ \mu=\text{ Population mean} \\ s=\text{ Standard deviation} \\ n=\text{ Sample size} \end{gathered}$

- Applying the formula, we have:

$\begin{gathered} t=\frac{137-132}{\frac{14.2}{\sqrt{20}}} \\ \\ t=\frac{5}{3.1752} \\ \\ t\approx1.5747 \end{gathered}$

Critical value:

- The critical value t-critical, is gotten by reading off the t-distribution table.

- For this, we need the degrees of freedom (df) which is gotten by the formula:

$\begin{gathered} df=n-1 \\ df=20-1=19 \end{gathered}$

- And then we also use the significance level of 0.1 and the fact that it is a two-tailed test to trace out the t-critical. (Note: significance level of 0.1 implies 10% significance level)

- This is done below:

- The critical value is 1.729

P-value:

- To find the p-value, we simply check the table for where the t-statistic falls.

- The t-statistic given is 1.5747. We simply check which values this falls between in the t-distribution table. It falls between 1.328 and 1.729. We can simply choose a value between 0.1 and 0.05 and multiply the result by 2 since it is a two-tailed test.

- However, we can also use a t-distribution calculator, we have:

- Thus, the p-value is 0.13183

Final Conclusion:

- The p-value is 0.13183, and comparing this to the significance level of 0.1, we can see that 0.13183 is outside the rejection region.

- Thus, the result is not significant and we fail to reject the null hypothesis

7 0

1 year ago