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3 years ago

Q5. A car is bought for £5000 and sold for £7500. What is the percentage profit?

Mathematics

1 answer:

Nitella [24]3 years ago

8 0

Answer:

50% profit

Step-by-step explanation:

Cost price = £ 5000

Selling price = £ 7500

Profit = Selling price - cost price

= 7500 - 5000

= £ 2500

Profit percentage = $\frac{2500}{5000}*100$

= 50 %

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You roll an eight-sided die. What is the chance that it will land on either a 2 or a 5? Write your answer as a fraction.

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2/8 and the simplified version is 1/4

Step-by-step explanation:

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I Think Its 14 Feet.

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What is the definition of a line

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A sample with a mean of m = 40 and a variance of s2 = 20 has an estimated standard error of 1 point. how many scores are in the

tamaranim1 [39]

The correct option is 20.

The score found for the sample is 20.

<h3>What is Standard error?</h3>

A standard error is a statistic that is applied to test the distribution of data. This metric is comparable to standard deviation. We can calculate the standard error if we know the sample size & standard deviation. It assesses the mean's precision.

Now, according to the question;

Sample mean; $\bar{x}=40$

Sample variance; $s^{2}=20$

Thus, $s=\sqrt{20}=4.47$

Standard error SE = 1

The amount of scores with in sample must be determined here.

The standard error formula is as follows:

$S E=\frac{s}{\sqrt{n}}$

We might rearrange the formula as follows:

$\begin{aligned}\sqrt{n} &=\frac{s}{S E} \\n &=\left(\frac{s}{S E}\right)^{2}\end{aligned}$

Substituting the values;

$\begin{aligned}&n=\left(\frac{4.47}{1}\right)^{2} \\&n=19.98\end{aligned}$

The sample's number of scores n = 19.98 = 20 (round up)

Therefore, the scores are in the sample is 20.

To know more about the Standard error, here

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2 years ago

Simplify f+g / f-g when f(x)= x-4 / x+9 and g(x)= x-9 / x+4

steposvetlana [31]

$f(x)=\dfrac{x-4}{x+9};\ g(x)=\dfrac{x-9}{x+4}\\\\f(x)+g(x)=\dfrac{x-4}{x+9}+\dfrac{x-9}{x+4}=\dfrac{(x-4)(x+4)+(x-9)(x+9)}{(x+9)(x+4)}\\\\\text{use}\ a^2-b^2=(a-b)(a+b)\\\\=\dfrac{x^2-4^2+x^2-9^2}{(x+9)(x+4)}=\dfrac{2x^2-16-81}{(x+9)(x+4)}=\dfrac{2x^2-97}{(x+9)(x+4)}\\\\f(x)-g(x)=\dfrac{x-4}{x+9}-\dfrac{x-9}{x+4}=\dfrac{(x-4)(x+4)-(x-9)(x+9)}{(x+9)(x+4)}\\\\\text{use}\ a^2-b^2=(a-b)(a+b)\\\\=\dfrac{x^2-4^2-(x^2-9^2)}{(x+9)(x+4)}=\dfrac{x^2-16-x^2+81}{(x+9)(x+4)}=\dfrac{65}{(x+9)(x+4)}$

$\dfrac{f+g}{f-g}=(f+g):(f-g)=\dfrac{2x^2-97}{(x+9)(x+4)}:\dfrac{65}{(x+9)(x+4)}\\\\=\dfrac{2x^2-97}{(x+9)(x+4)}\cdot\dfrac{(x+9)(x+4)}{65}\\\\Answer:\ \boxed{\dfrac{f+g}{f-g}=\dfrac{2x^2-97}{65}}$

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