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2 years ago

Riya purchased a machine for Rs36000 and spent Rs2400 on its repair. She then sold it for Rs42000. Find the gain or loss percent

Mathematics

2 answers:

Natasha_Volkova [10]2 years ago

7 0

Step-by-step explanation:

Cost Price= ₹36,000

Repair Price=₹2,400

Actual Cost Price= CP + Repair Price

= ₹36,000 + ₹2,400

= ₹38400

Selling Price = ₹42000

Profit = SP - CP

= ₹42000 - ₹38400

= ₹3600

Profit% = Profit/CP × 100

= ₹3600/₹38400 × 100

= 9.375%

Scorpion4ik [409]2 years ago

5 0

Answer:

gain%=9.37%(approximately)

Step-by-step explanation:

C.P.= 36000+2400

C.P.=Rs38400

S.P.=Rs42000

gain=S.P.-C.P.

gain=42000-38400

gain=Rs3600

gain%=(gain/C.P.)*100

gain%=(3600/38400)*100

gain%=9.37%(approximately)

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Step-by-step explanation:

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For this case, the first thing we are going to do is write the generic equation of motion for the vertical axis.

We have then:

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$h2 = \frac {1} {2} gt ^ 2 - \frac {11} {12} * t + 50$

By the time both bodies have the same height we have:

$h1 = h2\\$

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Rewriting we have:

$\frac {8} {12} * t + 21 = - \frac {11} {12} * t + 50$

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Step-by-step explanation:

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Answer:

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For this case since the confidence interval for the difference of means contains the 0 we can conclude that we don't have significant differences at 10% of significance between the two means analyzed.

Step-by-step explanation:

Previous concepts

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".

The margin of error is the range of values below and above the sample statistic in a confidence interval.

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

$\bar X_1 =2.0233$ represent the sample mean 1

$\bar X_2 =3.048$ represent the sample mean 2

n1=15 represent the sample 1 size

n2=15 represent the sample 2 size

$s_1 =4.893387$ population sample deviation for sample 1

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H0: $\mu_1 = \mu_2$

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$(\bar X_1 -\bar X_2) \pm t_{\alpha/2}\sqrt{\frac{s^2_1}{n_1}+\frac{s^2_2}{n_2}}$ (1)

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The degrees of freedom are given by:

$df = n_1 +n_2 -2 = 15+15-2=28$

Since the Confidence is 0.90 or 90%, the value of $\alpha=0.1$ and $\alpha/2 =0.05$ , and we can use excel, a calculator or a table to find the critical value. The excel command would be: "=-T.INV(0.05,28)".And we see that $t_{\alpha/2}=\pm 1.701$

Now we have everything in order to replace into formula (1):

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