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3 years ago

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A man sold a calculator at a loss of 12℅. If selling price was Rs.7040, what was the cost price?

1 answer:

Rashid [163]3 years ago

4 0

Answer:

Rs.6195.20

Step-by-step explanation:

So first you need to find 12 % of Rs.7040. Thus:

$\frac{12}{100}$ × 7040 = 844.80

Then you subtract 844.80 from 7040 from which you get Rs.6195.20

Rs.6195.20 was the cost price.

HOPE THIS HELPED

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I'm not sure what this means, please help me I'll give you 10 points ;)

Karo-lina-s [1.5K]

<u>Answer:</u>

So all the possible solutions are:

$\bold{523,~400,~379~and,~378.6}$

<u />

<u>Solution Steps:</u>

<em>First you need to solve the real inequality to understand how to find the rest of the possible equations. </em>

<u>Add 78 to both sides:</u>

<u /> $78+78=$ Cancels Out

$300+78=378$

<em>So now we know the real answer, but it ask for all possible answers. </em>

(Means anything larger than 300 when you plug it into x - 78 > 300.)

<u>Numbers that are greater than 378:</u>

1.) $377-78=299 > 300$ (False)

2.) $523-78=445>300$ (True)

3.) $0-78=-78>300$ (False)

4.) $-62-78=-140>300$ (False)

5.) $400-78=322>300$ (True)

6.) $379-78=301>300$ (True)

7.) $222-78=144>300$ (False)

8.) $378.6-78=300.6>300$ (True)

______________________________

$\bold{Hope~this~helps!}\\\bold{If~you~need~help~with~anything~else,~feel~free~to~ask!}\\\\\bold{~~~~~-TotallyNotTrillex}$

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3 years ago

A simple random sample of 100 8th graders at a large suburban middle school indicated that 81% of them are involved with some ty

Sophie [7]

Answer:

The interval is $0.7187 < p < 2.421$

Step-by-step explanation:

From the question we are told that

The sample size is $n = 100$

The population proportion is $p = 0.81$

The confidence level is C = 98%

The level of significance is mathematically evaluated as

$\alpha = 100 -98$

$\alpha = 2%$ %

$\alpha = 0.02$

Here this level of significance represented the left and the right tail

The degree of freedom is evaluated as

$df = n-1$

substituting value

$df = 100 - 1$

$df = 99$

Since we require the critical value of one tail in order to evaluate the 98% confidence interval that estimates the proportion of them that are involved in an after school activity. we will divide the level of significance by 2

The critical value of $\frac{\alpha}{2}$ and the evaluated degree of freedom is

$t_{df , \alpha } = t_{99 , \frac{0.02}{2} } = 2.33$

this is obtained from the critical value table

The standard error is mathematically evaluated as

$SE = \sqrt{\frac{p(1-p )}{n} }$

substituting value

$SE = \sqrt{\frac{0.81(1-0.81 )}{100} }$

$SE = 0.0392$

The 98% confidence interval is evaluated as

$p - t_{df , \frac{\alpha }{2} } * SE < p < p + t_{df , \frac{\alpha }{2} }$

substituting value

$0.81 - 2.33 * 0.0392 < p < 0.81 +2.33 * 0.0392$

$0.7187 < p < 2.421$

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3 years ago