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3 years ago

if a business man makes a profit of of 20% on the selling price then this profit percentage on the cost price will be

Mathematics

1 answer:

netineya [11]3 years ago

3 0

Let selling price = SP and cost price = CP

He makes 20 % on selling price means = >

(sp-cp) / sp = 20 % = 20/100 = 0.2

(sp - cp) / sp = 0.2

1 - cp/sp = 0.2

1 - 0.2 = cp/sp

0.8 = cp/sp

Reversing 1 / 0.8 = sp/cp = 1.25

We need to find profit on cost price means =

=(sp-cp)/ cp

= sp/cp - 1 = 1.25 - 1 = 0.25

Means 25 % profit on cost price : Answer

Hope that will help :)

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Consider a chemical company that wishes to determine whether a new catalyst, catalyst XA-100, changes the mean hourly yield of i

kolezko [41]

Answer:

Null hypothesis: $\mu = 750$

Alternative hypothesis: $\mu \neq 750$

$t=\frac{811-750}{\frac{19.647}{\sqrt{5}}}=6.943$

$p_v =2*P(t_{4}>6.943)=0.00226$

If we compare the p value and a significance level assumed $\alpha=0.05$ we see that $p_v$ so we can conclude that we reject the null hypothesis, and the actual true mean is significantly different from 750 pounds per hour.

Step-by-step explanation:

Data given and notation

Data: 801, 814, 784, 836,820

We can calculate the sample mean and sample deviation with the following formulas:

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

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

$\bar X=811$ represent the sample mean

$s=19.647$ represent the standard deviation for the sample

$n=5$ sample size

$\mu_o =750$ represent the value that we want to test

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

t would represent the statistic (variable of interest)

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

State the null and alternative hypotheses to be tested

We need to conduct a hypothesis in order to determine if the mean is different from 750 pounds per hour, the system of hypothesis would be:

Null hypothesis: $\mu = 750$

Alternative hypothesis: $\mu \neq 750$

Compute the test statistic

We don't know the population deviation, so for this case 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".

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

$t=\frac{811-750}{\frac{19.647}{\sqrt{5}}}=6.943$

Now we need to find the degrees of freedom for the t distirbution given by:

$df=n-1=5-1=4$